id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-6600 | Every Ultrafilter Converges implies Every Filter has Adherent Point | Let $T = \struct {S, \tau}$ be a topological space.
Let every ultrafilter on $S$ be convergent.
Then every filter on $S$ has a adherent point. | Let $\FF$ be a filter on $S$.
By the Ultrafilter Lemma, there exists an ultrafilter $\FF'$ such that $\FF \subseteq \FF'$.
We have {{hypothesis}} that $\FF'$ converges to some $x \in S$.
This, by Adherent Point of Filter iff Superfilter Converges, implies that $x$ is a adherent point of $\FF$.
{{qed}}
{{BPI|Ultrafilte... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let every [[Definition:Ultrafilter on Set|ultrafilter]] on $S$ be [[Definition:Convergent Filter|convergent]].
Then every [[Definition:Filter on Set|filter]] on $S$ has a [[Definition:Adherent Point of Filter|adherent point]]. | Let $\FF$ be a [[Definition:Filter on Set|filter]] on $S$.
By the [[Ultrafilter Lemma]], there exists an [[Definition:Ultrafilter on Set|ultrafilter]] $\FF'$ such that $\FF \subseteq \FF'$.
We have {{hypothesis}} that $\FF'$ [[Definition:Convergent Filter|converges]] to some $x \in S$.
This, by [[Adherent Point of ... | Every Ultrafilter Converges implies Every Filter has Adherent Point | https://proofwiki.org/wiki/Every_Ultrafilter_Converges_implies_Every_Filter_has_Adherent_Point | https://proofwiki.org/wiki/Every_Ultrafilter_Converges_implies_Every_Filter_has_Adherent_Point | [
"Filters on Sets",
"Ultrafilters on Sets",
"Adherent Points",
"Convergent Filters"
] | [
"Definition:Topological Space",
"Definition:Ultrafilter on Set",
"Definition:Convergent Filter",
"Definition:Filter on Set",
"Definition:Adherent Point/Filter"
] | [
"Definition:Filter on Set",
"Ultrafilter Lemma",
"Definition:Ultrafilter on Set",
"Definition:Convergent Filter",
"Adherent Point of Filter iff Superfilter Converges",
"Definition:Adherent Point/Filter"
] |
proofwiki-6601 | Left Distributive and Commutative implies Distributive | Let $\struct {S, \circ, *}$ be an algebraic structure.
Let the operation $\circ$ be left distributive over the operation $*$.
Let $\circ$ be commutative.
Then $\circ$ is distributive over $*$. | Let $a, b, c \in S$.
Then
{{begin-eqn}}
{{eqn | l = \paren {a * b} \circ c
| r = c \circ \paren {a * b}
| c = $\circ$ is commutative
}}
{{eqn | r = \paren {c \circ a} * \paren {c \circ b}
| c = $\circ$ is left distributive over $*$
}}
{{eqn | r = \paren {a \circ c} * \paren {b \circ c}
| c = $\c... | Let $\struct {S, \circ, *}$ be an [[Definition:Algebraic Structure with Two Operations|algebraic structure]].
Let the [[Definition:Binary Operation|operation]] $\circ$ be [[Definition:Left Distributive Operation|left distributive]] over the [[Definition:Binary Operation|operation]] $*$.
Let $\circ$ be [[Definition:Co... | Let $a, b, c \in S$.
Then
{{begin-eqn}}
{{eqn | l = \paren {a * b} \circ c
| r = c \circ \paren {a * b}
| c = $\circ$ is [[Definition:Commutative Operation|commutative]]
}}
{{eqn | r = \paren {c \circ a} * \paren {c \circ b}
| c = $\circ$ is [[Definition:Left Distributive Operation|left distributive]... | Left Distributive and Commutative implies Distributive | https://proofwiki.org/wiki/Left_Distributive_and_Commutative_implies_Distributive | https://proofwiki.org/wiki/Left_Distributive_and_Commutative_implies_Distributive | [
"Commutativity",
"Distributive Operations"
] | [
"Definition:Algebraic Structure/Two Operations",
"Definition:Operation/Binary Operation",
"Definition:Distributive Operation/Left",
"Definition:Operation/Binary Operation",
"Definition:Commutative/Operation",
"Definition:Distributive Operation"
] | [
"Definition:Commutative/Operation",
"Definition:Distributive Operation/Left",
"Definition:Commutative/Operation",
"Definition:Distributive Operation/Right",
"Definition:Distributive Operation/Left",
"Definition:Distributive Operation/Right",
"Definition:Distributive Operation",
"Category:Commutativity... |
proofwiki-6602 | Filter on Product Space Converges iff Projections Converge | Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of non-empty topological spaces where $I$ is an arbitrary index set.
Let $\ds X := \prod_{i \mathop \in I} X_i$ be the corresponding product space.
Let $\pr_i: X \to X_i$ denote the projection from $X$ onto $X_i$.
Let $\FF \subset \powerset X$ be a filter on $X... | === Sufficient Condition ===
Let $\FF$ converge.
Then there is a point $x \in X$ such that $\FF$ converges to $x$.
By Filter on Product Space Converges to Point iff Projections Converge to Projections of Point:
:$\forall i \in I: \map {\pr_i} \FF$ converges to $x_i$
Thus
:$\forall i \in I: \map {\pr_i} \FF$ converges. | Let $\family {X_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Non-Empty Set|non-empty]] [[Definition:Topological Space|topological spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
Let $\ds X := \prod_{i \mathop \in I} X_i$ be the corresponding [[Defin... | === Sufficient Condition ===
Let $\FF$ converge.
Then there is a point $x \in X$ such that $\FF$ converges to $x$.
By [[Filter on Product Space Converges to Point iff Projections Converge to Projections of Point]]:
:$\forall i \in I: \map {\pr_i} \FF$ converges to $x_i$
Thus
:$\forall i \in I: \map {\pr_i} \FF$ con... | Filter on Product Space Converges iff Projections Converge | https://proofwiki.org/wiki/Filter_on_Product_Space_Converges_iff_Projections_Converge | https://proofwiki.org/wiki/Filter_on_Product_Space_Converges_iff_Projections_Converge | [
"Topology",
"Filter Theory",
"Projections"
] | [
"Definition:Indexing Set/Family",
"Definition:Non-Empty Set",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Product Space (Topology)",
"Definition:Projection (Mapping Theory)",
"Definition:Filter on Set",
"Definition:Convergent Filter",
"Definition:Image Filter"
] | [
"Filter on Product Space Converges to Point iff Projections Converge to Projections of Point",
"Filter on Product Space Converges to Point iff Projections Converge to Projections of Point"
] |
proofwiki-6603 | Tychonoff's Theorem for Hausdorff Spaces | Let $I$ be an indexing set.
Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of non-empty Hausdorff spaces.
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the corresponding product space.
Then $X$ is compact {{iff}} each $X_i$ is compact. | First assume that $X$ is compact.
From Projection from Product Topology is Continuous, the projections:
:$\pr_i : X \to X_i$
are continuous.
From Continuous Image of Compact Space is Compact, it follows that the $X_i$ are compact.
Assume now that each $X_i$ is compact.
From Topological Space is Compact iff Every Ultr... | Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {X_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Non-Empty Set|non-empty]] [[Definition:Hausdorff Space|Hausdorff spaces]].
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the corresponding [[Definition:Produc... | First assume that $X$ is [[Definition:Compact Topological Space|compact]].
From [[Projection from Product Topology is Continuous]], the [[Definition:Projection (Mapping Theory)|projections]]:
:$\pr_i : X \to X_i$
are [[Definition:Continuous Mapping (Topology)|continuous]].
From [[Continuous Image of Compact Space i... | Tychonoff's Theorem for Hausdorff Spaces | https://proofwiki.org/wiki/Tychonoff's_Theorem_for_Hausdorff_Spaces | https://proofwiki.org/wiki/Tychonoff's_Theorem_for_Hausdorff_Spaces | [
"Tychonoff's Theorem",
"Hausdorff Spaces"
] | [
"Definition:Indexing Set",
"Definition:Indexing Set/Family",
"Definition:Non-Empty Set",
"Definition:T2 Space",
"Definition:Product Space (Topology)",
"Definition:Compact Topological Space",
"Definition:Compact Topological Space"
] | [
"Definition:Compact Topological Space",
"Projection from Product Topology is Continuous",
"Definition:Projection (Mapping Theory)",
"Definition:Continuous Mapping (Topology)",
"Continuous Image of Compact Space is Compact",
"Definition:Compact Topological Space",
"Definition:Compact Topological Space",
... |
proofwiki-6604 | Filter on Product of Hausdorff Spaces Converges iff Projections Converge | Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of non-empty Hausdorff spaces where $I$ is an arbitrary index set.
Let $\ds X := \prod_{i \mathop \in I} X_i$ be the corresponding product space.
Let $\pr_i: X \to X_i$ denote the projection from $X$ onto $X_i$.
Let $\FF \subset \powerset X$ be a filter on $X$.... | === Sufficient Condition ===
Let $\FF$ converge.
Then there is a point $x \in X$ such that $\FF$ converges to $x$.
By Filter on Product Space Converges to Point iff Projections Converge to Projections of Point:
:$\forall i \in I: \map {\pr_i} \FF$ converges to $x_i$.
Thus, for each $i \in I$, $\map {\pr_i} \FF$ converg... | Let $\family {X_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Non-Empty Set|non-empty]] [[Definition:Hausdorff Space|Hausdorff spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
Let $\ds X := \prod_{i \mathop \in I} X_i$ be the corresponding [[Definitio... | === Sufficient Condition ===
Let $\FF$ converge.
Then there is a point $x \in X$ such that $\FF$ converges to $x$.
By [[Filter on Product Space Converges to Point iff Projections Converge to Projections of Point]]:
:$\forall i \in I: \map {\pr_i} \FF$ converges to $x_i$.
Thus, for each $i \in I$, $\map {\pr_i} \FF$... | Filter on Product of Hausdorff Spaces Converges iff Projections Converge | https://proofwiki.org/wiki/Filter_on_Product_of_Hausdorff_Spaces_Converges_iff_Projections_Converge | https://proofwiki.org/wiki/Filter_on_Product_of_Hausdorff_Spaces_Converges_iff_Projections_Converge | [
"Topology",
"Filter Theory",
"Projections"
] | [
"Definition:Indexing Set/Family",
"Definition:Non-Empty Set",
"Definition:T2 Space",
"Definition:Indexing Set",
"Definition:Product Space (Topology)",
"Definition:Projection (Mapping Theory)",
"Definition:Filter on Set",
"Definition:Convergent Filter",
"Definition:Image Filter"
] | [
"Filter on Product Space Converges to Point iff Projections Converge to Projections of Point",
"Filter on Product Space Converges to Point iff Projections Converge to Projections of Point"
] |
proofwiki-6605 | Positive Real has Real Square Root | Let $x \in \R_{>0}$ be a (strictly) positive real number.
Then:
:$\exists y \in \R: x = y^2$ | Let $f: \R \to \R$ be defined as:
:$\forall x \in \R: \map f x = x^2$
We have that $f$ is the pointwise product of the identity mapping with itself.
By Product Rule for Continuous Real Functions and Identity Mapping is Continuous, $f$ is continuous.
By Power Function is Unbounded Above:
:$\exists q \in \R: \map f q > x... | Let $x \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
Then:
:$\exists y \in \R: x = y^2$ | Let $f: \R \to \R$ be defined as:
:$\forall x \in \R: \map f x = x^2$
We have that $f$ is the [[Definition:Pointwise Multiplication of Real-Valued Functions|pointwise product]] of the [[Definition:Identity Mapping|identity mapping]] with itself.
By [[Product Rule for Continuous Real Functions]] and [[Identity Mapping... | Positive Real has Real Square Root | https://proofwiki.org/wiki/Positive_Real_has_Real_Square_Root | https://proofwiki.org/wiki/Positive_Real_has_Real_Square_Root | [
"Real Numbers"
] | [
"Definition:Strictly Positive/Real Number"
] | [
"Definition:Pointwise Multiplication of Real-Valued Functions",
"Definition:Identity Mapping",
"Combination Theorem for Continuous Functions/Real/Product Rule",
"Identity Mapping is Continuous",
"Definition:Continuous Real Function",
"Limit at Infinity of x^n",
"Intermediate Value Theorem",
"Category:... |
proofwiki-6606 | Join is Idempotent | Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Then $\vee$ is idempotent. | Let $a \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a \vee a
| r = \sup \set {a, a}
| c = {{Defof|Join (Order Theory)|Join}}
}}
{{eqn | r = \sup \set a
| c = {{Defof|Set}}
}}
{{eqn | r = a
| c = Supremum of Singleton
}}
{{end-eqn}}
Hence the result.
{{qed}} | Let $\struct {S, \vee, \preceq}$ be a [[Definition:Join Semilattice|join semilattice]].
Then $\vee$ is [[Definition:Idempotent Operation|idempotent]]. | Let $a \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a \vee a
| r = \sup \set {a, a}
| c = {{Defof|Join (Order Theory)|Join}}
}}
{{eqn | r = \sup \set a
| c = {{Defof|Set}}
}}
{{eqn | r = a
| c = [[Supremum of Singleton]]
}}
{{end-eqn}}
Hence the result.
{{qed}} | Join is Idempotent | https://proofwiki.org/wiki/Join_is_Idempotent | https://proofwiki.org/wiki/Join_is_Idempotent | [
"Join Operation",
"Examples of Idempotence"
] | [
"Definition:Join Semilattice",
"Definition:Idempotence/Operation"
] | [
"Supremum of Singleton"
] |
proofwiki-6607 | Product of Positive Strictly Increasing Mappings is Strictly Increasing | Let $A$ be an ordered set.
Let $B$ be an ordered field.
Let $f, g: A \to B$ be strictly increasing mappings with positive values.
Let $h: A \to B$ be defined by $\map h x = \map f x \map g x$.
Then $h$ is strictly increasing. | Let $x, y \in A$ such that $x < y$.
If $\map h x = 0$.
By the definition of strictly increasing:
:$\map f y > \map f x \ge 0$
and:
:$\map g y > \map g x \ge 0$
So:
:$\map h y > 0 = \map h x$
If $\map h x \ne 0$, then $\map h x > 0$
so:
:$\map f x > 0$
and:
:$\map g x > 0$
Also:
:$\map f x < \map f y$
and:
:$\map g x < ... | Let $A$ be an ordered set.
Let $B$ be an ordered field.
Let $f, g: A \to B$ be strictly increasing mappings with positive values.
Let $h: A \to B$ be defined by $\map h x = \map f x \map g x$.
Then $h$ is strictly increasing. | Let $x, y \in A$ such that $x < y$.
If $\map h x = 0$.
By the definition of strictly increasing:
:$\map f y > \map f x \ge 0$
and:
:$\map g y > \map g x \ge 0$
So:
:$\map h y > 0 = \map h x$
If $\map h x \ne 0$, then $\map h x > 0$
so:
:$\map f x > 0$
and:
:$\map g x > 0$
Also:
:$\map f x < \map f y$
and:
:$\map ... | Product of Positive Strictly Increasing Mappings is Strictly Increasing | https://proofwiki.org/wiki/Product_of_Positive_Strictly_Increasing_Mappings_is_Strictly_Increasing | https://proofwiki.org/wiki/Product_of_Positive_Strictly_Increasing_Mappings_is_Strictly_Increasing | [] | [] | [] |
proofwiki-6608 | Supremum of Singleton | Let $\struct {S, \preceq}$ be an ordered set.
Then for all $a \in S$:
:$\sup \set a = a$
where $\sup$ denotes supremum. | Since $a \preceq a$, $a$ is an upper bound of $\set a$.
Let $b$ be another upper bound of $\set a$.
Then necessarily $a \preceq b$.
It follows that indeed:
:$\sup \set a = a$
as desired.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Then for all $a \in S$:
:$\sup \set a = a$
where $\sup$ denotes [[Definition:Supremum of Set|supremum]]. | Since $a \preceq a$, $a$ is an [[Definition:Upper Bound of Set|upper bound]] of $\set a$.
Let $b$ be another [[Definition:Upper Bound of Set|upper bound]] of $\set a$.
Then necessarily $a \preceq b$.
It follows that indeed:
:$\sup \set a = a$
as desired.
{{qed}} | Supremum of Singleton | https://proofwiki.org/wiki/Supremum_of_Singleton | https://proofwiki.org/wiki/Supremum_of_Singleton | [
"Suprema",
"Singletons"
] | [
"Definition:Ordered Set",
"Definition:Supremum of Set"
] | [
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set"
] |
proofwiki-6609 | Infimum of Singleton | Let $\struct {S, \preceq}$ be an ordered set.
Then for all $a \in S$:
:$\inf \set a = a$
where $\inf$ denotes infimum. | Since $a \preceq a$, $a$ is a lower bound for $\set a$.
Let $b$ be another lower bound for $\set a$.
Then necessarily $b \preceq a$.
It follows that indeed:
:$\inf \set a = a$
as desired.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Then for all $a \in S$:
:$\inf \set a = a$
where $\inf$ denotes [[Definition:Infimum of Set|infimum]]. | Since $a \preceq a$, $a$ is a [[Definition:Lower Bound of Set|lower bound]] for $\set a$.
Let $b$ be another lower bound for $\set a$.
Then necessarily $b \preceq a$.
It follows that indeed:
:$\inf \set a = a$
as desired.
{{qed}} | Infimum of Singleton | https://proofwiki.org/wiki/Infimum_of_Singleton | https://proofwiki.org/wiki/Infimum_of_Singleton | [
"Infima",
"Singletons"
] | [
"Definition:Ordered Set",
"Definition:Infimum of Set"
] | [
"Definition:Lower Bound of Set"
] |
proofwiki-6610 | Meet is Idempotent | Let $\struct {S, \wedge, \preceq}$ be a meet semilattice.
Then $\wedge$ is idempotent. | Let $a \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a \wedge a
| r = \inf \set {a, a}
| c = {{Defof|Meet (Order Theory)}}
}}
{{eqn | r = \inf \set a
| c = {{Defof|Set}}
}}
{{eqn | r = a
| c = Infimum of Singleton
}}
{{end-eqn}}
Hence the result.
{{qed}} | Let $\struct {S, \wedge, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
Then $\wedge$ is [[Definition:Idempotent Operation|idempotent]]. | Let $a \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a \wedge a
| r = \inf \set {a, a}
| c = {{Defof|Meet (Order Theory)}}
}}
{{eqn | r = \inf \set a
| c = {{Defof|Set}}
}}
{{eqn | r = a
| c = [[Infimum of Singleton]]
}}
{{end-eqn}}
Hence the result.
{{qed}} | Meet is Idempotent | https://proofwiki.org/wiki/Meet_is_Idempotent | https://proofwiki.org/wiki/Meet_is_Idempotent | [
"Meet Operation",
"Examples of Idempotence"
] | [
"Definition:Meet Semilattice",
"Definition:Idempotence/Operation"
] | [
"Infimum of Singleton"
] |
proofwiki-6611 | GCD from Prime Decomposition | Let $a, b \in \Z$.
From Expression for Integers as Powers of Same Primes, let:
{{begin-eqn}}
{{eqn | l = a
| r = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}
}}
{{eqn | l = b
| r = p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r}
}}
{{eqn | q = \forall i \in \set {1, 2, \dotsc, r}
| l = p_i
| o = \divides
| r... | Note that if one of the primes $p_i$ does not appear in the prime decompositions of either one of $a$ or $b$, then its corresponding index $k_i$ or $l_i$ will be zero.
Let $d \divides a$.
Then:
:$d$ is of the form $p_1^{h_1} p_2^{h_2} \ldots p_r^{h_r}, \forall i: 1 \le i \le r, 0 \le h_i \le k_i$
:$d \divides a \iff \f... | Let $a, b \in \Z$.
From [[Expression for Integers as Powers of Same Primes]], let:
{{begin-eqn}}
{{eqn | l = a
| r = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}
}}
{{eqn | l = b
| r = p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r}
}}
{{eqn | q = \forall i \in \set {1, 2, \dotsc, r}
| l = p_i
| o = \divides
... | Note that if one of the [[Definition:Prime Number|primes]] $p_i$ does not appear in the [[Definition:Prime Decomposition|prime decompositions]] of either one of $a$ or $b$, then its corresponding index $k_i$ or $l_i$ will be [[Definition:Zero (Number)|zero]].
Let $d \divides a$.
Then:
:$d$ is of the form $p_1^{h_1}... | GCD from Prime Decomposition | https://proofwiki.org/wiki/GCD_from_Prime_Decomposition | https://proofwiki.org/wiki/GCD_from_Prime_Decomposition | [
"Greatest Common Divisor",
"Prime Decompositions",
"GCD from Prime Decomposition"
] | [
"Expression for Integers as Powers of Same Primes",
"Definition:Prime Number",
"Definition:Prime Decomposition",
"Definition:Divisor (Algebra)/Integer",
"Definition:Greatest Common Divisor/Integers"
] | [
"Definition:Prime Number",
"Definition:Prime Decomposition",
"Definition:Zero (Number)",
"Definition:Prime Number"
] |
proofwiki-6612 | LCM from Prime Decomposition | Let $a, b \in \Z$.
From Expression for Integers as Powers of Same Primes, let:
{{begin-eqn}}
{{eqn | l = a
| r = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}
}}
{{eqn | l = b
| r = p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r}
}}
{{eqn | q = \forall i \in \set {1, 2, \dotsc, r}
| l = p_i
| o = \divides
| r... | {{begin-eqn}}
{{eqn | l = \lcm \set {a, b}
| r = \frac {a b} {\gcd \set {a, b} }
| c = Product of GCD and LCM
}}
{{eqn | r = \frac {\paren {p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r} } \paren {p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r} } } {p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \cdots p_r^{\min \set {... | Let $a, b \in \Z$.
From [[Expression for Integers as Powers of Same Primes]], let:
{{begin-eqn}}
{{eqn | l = a
| r = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}
}}
{{eqn | l = b
| r = p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r}
}}
{{eqn | q = \forall i \in \set {1, 2, \dotsc, r}
| l = p_i
| o = \divides
... | {{begin-eqn}}
{{eqn | l = \lcm \set {a, b}
| r = \frac {a b} {\gcd \set {a, b} }
| c = [[Product of GCD and LCM]]
}}
{{eqn | r = \frac {\paren {p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r} } \paren {p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r} } } {p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \cdots p_r^{\min \s... | LCM from Prime Decomposition/Proof 1 | https://proofwiki.org/wiki/LCM_from_Prime_Decomposition | https://proofwiki.org/wiki/LCM_from_Prime_Decomposition/Proof_1 | [
"Lowest Common Multiple",
"Prime Numbers",
"LCM from Prime Decomposition"
] | [
"Expression for Integers as Powers of Same Primes",
"Definition:Prime Number",
"Definition:Prime Decomposition",
"Definition:Divisor (Algebra)/Integer",
"Definition:Lowest Common Multiple/Integers"
] | [
"Product of GCD and LCM",
"GCD from Prime Decomposition",
"Sum Less Minimum is Maximum"
] |
proofwiki-6613 | LCM from Prime Decomposition | Let $a, b \in \Z$.
From Expression for Integers as Powers of Same Primes, let:
{{begin-eqn}}
{{eqn | l = a
| r = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}
}}
{{eqn | l = b
| r = p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r}
}}
{{eqn | q = \forall i \in \set {1, 2, \dotsc, r}
| l = p_i
| o = \divides
| r... | Note that if one of the primes $p_i$ does not appear in the prime decompositions of either one of $a$ or $b$, then its corresponding index $k_i$ or $l_i$ will be zero.
Let $a \divides m$.
Then:
:$m$ is of the form $p_1^{h_1} p_2^{h_2} \ldots p_r^{h_r}, \forall i: 1 \le i \le r, 0 \le k_i \le h_i$
:$a \divides l \iff \f... | Let $a, b \in \Z$.
From [[Expression for Integers as Powers of Same Primes]], let:
{{begin-eqn}}
{{eqn | l = a
| r = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}
}}
{{eqn | l = b
| r = p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r}
}}
{{eqn | q = \forall i \in \set {1, 2, \dotsc, r}
| l = p_i
| o = \divides
... | Note that if one of the [[Definition:Prime Number|primes]] $p_i$ does not appear in the [[Definition:Prime Decomposition|prime decompositions]] of either one of $a$ or $b$, then its corresponding index $k_i$ or $l_i$ will be [[Definition:Zero (Number)|zero]].
Let $a \divides m$.
Then:
:$m$ is of the form $p_1^{h_1}... | LCM from Prime Decomposition/Proof 2 | https://proofwiki.org/wiki/LCM_from_Prime_Decomposition | https://proofwiki.org/wiki/LCM_from_Prime_Decomposition/Proof_2 | [
"Lowest Common Multiple",
"Prime Numbers",
"LCM from Prime Decomposition"
] | [
"Expression for Integers as Powers of Same Primes",
"Definition:Prime Number",
"Definition:Prime Decomposition",
"Definition:Divisor (Algebra)/Integer",
"Definition:Lowest Common Multiple/Integers"
] | [
"Definition:Prime Number",
"Definition:Prime Decomposition",
"Definition:Zero (Number)",
"Definition:Prime Number"
] |
proofwiki-6614 | Product of Positive Element and Element Greater than One | Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity $1_R$ and zero $0_R$.
Let $x, y \in R$.
Suppose that $x > 0_R$ and $y > 1_R$.
Then $x \circ y > x$ and $y \circ x > x$. | {{begin-eqn}}
{{eqn | l = y
| o = >
| r = 1_R
}}
{{eqn | l = x \circ y
| o = >
| r = x \circ 1_R
| c = $x > 0_R$, Properties of Ordered Ring: $(6)$
}}
{{eqn | l = x \circ y
| o = >
| r = x
| c = {{Defof|Unity of Ring}}
}}
{{end-eqn}}
A similar argument shows that $y \cir... | Let $\struct {R, +, \circ, \le}$ be an [[Definition:Ordered Ring|ordered ring]] [[Definition:Ring with Unity|with unity]] $1_R$ and [[Definition:Ring Zero|zero]] $0_R$.
Let $x, y \in R$.
Suppose that $x > 0_R$ and $y > 1_R$.
Then $x \circ y > x$ and $y \circ x > x$. | {{begin-eqn}}
{{eqn | l = y
| o = >
| r = 1_R
}}
{{eqn | l = x \circ y
| o = >
| r = x \circ 1_R
| c = $x > 0_R$, [[Properties of Ordered Ring|Properties of Ordered Ring: $(6)$]]
}}
{{eqn | l = x \circ y
| o = >
| r = x
| c = {{Defof|Unity of Ring}}
}}
{{end-eqn}}
A sim... | Product of Positive Element and Element Greater than One | https://proofwiki.org/wiki/Product_of_Positive_Element_and_Element_Greater_than_One | https://proofwiki.org/wiki/Product_of_Positive_Element_and_Element_Greater_than_One | [
"Ordered Rings"
] | [
"Definition:Ordered Ring",
"Definition:Ring with Unity",
"Definition:Ring Zero"
] | [
"Properties of Ordered Ring",
"Category:Ordered Rings"
] |
proofwiki-6615 | Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element | Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity.
Let $x \in R$ with $x > 1$ and $x > 0$.
Let $n \in \N_{>0}$.
Then:
:$\circ^n x \ge x$ | The proof proceeds by induction:
If $n = 1$, then $\circ^n x = x$.
So:
:$\circ^n x \ge x$
Suppose that $\circ^n x \ge x$.
Then since $x > 1$:
:$\circ^n x > 1$
By Product of Positive Element and Element Greater than One:
:$x \circ \paren {\circ^n x} > x$
Hence:
:$\circ^{n + 1} x \ge x$
{{qed}}
Category:Ordered Rings
gns... | Let $\struct {R, +, \circ, \le}$ be an [[Definition:Ordered Ring|ordered ring]] [[Definition:Ring with Unity|with unity]].
Let $x \in R$ with $x > 1$ and $x > 0$.
Let $n \in \N_{>0}$.
Then:
:$\circ^n x \ge x$ | The proof proceeds by [[Principle of Mathematical Induction|induction]]:
If $n = 1$, then $\circ^n x = x$.
So:
:$\circ^n x \ge x$
Suppose that $\circ^n x \ge x$.
Then since $x > 1$:
:$\circ^n x > 1$
By [[Product of Positive Element and Element Greater than One]]:
:$x \circ \paren {\circ^n x} > x$
Hence:
:$\circ^{... | Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element | https://proofwiki.org/wiki/Strictly_Positive_Power_of_Strictly_Positive_Element_Greater_than_One_Succeeds_Element | https://proofwiki.org/wiki/Strictly_Positive_Power_of_Strictly_Positive_Element_Greater_than_One_Succeeds_Element | [
"Ordered Rings"
] | [
"Definition:Ordered Ring",
"Definition:Ring with Unity"
] | [
"Principle of Mathematical Induction",
"Product of Positive Element and Element Greater than One",
"Category:Ordered Rings"
] |
proofwiki-6616 | Directed Set has Strict Successors iff Unbounded Above | Let $\struct {S, \le}$ be a directed set.
Then every element of $S$ has a strict successor in $S$ {{iff}} $S$ has no upper bound in $S$. | === Necessary Condition ===
Suppose that each element of $S$ has a strict successor in $S$.
If $x$ is any element of $S$, then $x$ has a strict successor.
Therefore, $x$ is not an upper bound of $S$.
{{qed|lemma}} | Let $\struct {S, \le}$ be a [[Definition:Directed Set|directed set]].
Then every element of $S$ has a [[Definition:Strict Successor|strict successor]] in $S$ {{iff}} $S$ has no [[Definition:Upper Bound of Set|upper bound]] in $S$. | === Necessary Condition ===
Suppose that each element of $S$ has a [[Definition:Strict Successor|strict successor]] in $S$.
If $x$ is any element of $S$, then $x$ has a [[Definition:Strict Successor|strict successor]].
Therefore, $x$ is not an [[Definition:Upper Bound of Set|upper bound]] of $S$.
{{qed|lemma}} | Directed Set has Strict Successors iff Unbounded Above | https://proofwiki.org/wiki/Directed_Set_has_Strict_Successors_iff_Unbounded_Above | https://proofwiki.org/wiki/Directed_Set_has_Strict_Successors_iff_Unbounded_Above | [
"Order Theory"
] | [
"Definition:Directed Preordering",
"Definition:Strictly Succeed",
"Definition:Upper Bound of Set"
] | [
"Definition:Strictly Succeed",
"Definition:Strictly Succeed",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set",
"Definition:Strictly Succeed"
] |
proofwiki-6617 | Complex Numbers are Uncountable | The set of complex numbers $\C$ is uncountably infinite. | For all $r \in \R$, we have $r = r + 0 i \in C$.
Thus the set of real numbers $\R$ can be considered a subset of $\C$.
As the Real Numbers are Uncountable, it follows from Sufficient Conditions for Uncountability, proposition $(1)$, that $\C$ is uncountably infinite.
{{qed}}
Category:Complex Numbers
Category:Set Theory... | The [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$ is [[Definition:Uncountable Set|uncountably infinite]]. | For all $r \in \R$, we have $r = r + 0 i \in C$.
Thus the [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] $\R$ can be considered a [[Definition:Subset|subset]] of $\C$.
As the [[Real Numbers are Uncountable]], it follows from [[Sufficient Conditions for Uncountability]], proposition $(1)$, that $\C$... | Complex Numbers are Uncountable | https://proofwiki.org/wiki/Complex_Numbers_are_Uncountable | https://proofwiki.org/wiki/Complex_Numbers_are_Uncountable | [
"Complex Numbers",
"Set Theory"
] | [
"Definition:Set",
"Definition:Complex Number",
"Definition:Uncountable/Set"
] | [
"Definition:Set",
"Definition:Real Number",
"Definition:Subset",
"Real Numbers are Uncountably Infinite",
"Sufficient Conditions for Uncountability",
"Definition:Uncountable/Set",
"Category:Complex Numbers",
"Category:Set Theory"
] |
proofwiki-6618 | Infimum of Infima | Let $\left({S, \preceq}\right)$ be an ordered set.
Let $\mathbb T$ be a collection of subsets of $S$.
Suppose all $T \in \mathbb T$ admit an infimum $\inf T$ in $S$.
Then:
:$\inf \bigcup \mathbb T = \inf \left\{{\inf T: T \in \mathbb T}\right\}$
as soon as one of these two quantities exists. | Suppose that $s = \inf \bigcup \mathbb T \in S$.
By Set is Subset of Union, $T \subseteq \bigcup \mathbb T$ for all $T \in \mathbb T$.
Hence by Infimum of Subset:
:$\forall T \in \mathbb T: s \preceq \inf T$
Suppose now that $a \in S$ satisfies:
:$\forall T \in \mathbb T: a \preceq \inf T$
Then by transitivity of $\pre... | Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
Let $\mathbb T$ be a collection of [[Definition:Subset|subsets]] of $S$.
Suppose all $T \in \mathbb T$ admit an [[Definition:Infimum of Set|infimum]] $\inf T$ in $S$.
Then:
:$\inf \bigcup \mathbb T = \inf \left\{{\inf T: T \in \mathbb T}... | Suppose that $s = \inf \bigcup \mathbb T \in S$.
By [[Set is Subset of Union]], $T \subseteq \bigcup \mathbb T$ for all $T \in \mathbb T$.
Hence by [[Infimum of Subset]]:
:$\forall T \in \mathbb T: s \preceq \inf T$
Suppose now that $a \in S$ satisfies:
:$\forall T \in \mathbb T: a \preceq \inf T$
Then by [[Defi... | Infimum of Infima | https://proofwiki.org/wiki/Infimum_of_Infima | https://proofwiki.org/wiki/Infimum_of_Infima | [
"Order Theory"
] | [
"Definition:Ordered Set",
"Definition:Subset",
"Definition:Infimum of Set"
] | [
"Set is Subset of Union",
"Infimum of Subset",
"Definition:Transitive Relation",
"Definition:Infimum of Set",
"Definition:Infimum of Set",
"Definition:Transitive Relation",
"Definition:Infimum of Set",
"Category:Order Theory"
] |
proofwiki-6619 | Join Semilattice is Ordered Structure | Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Then $\struct {S, \vee, \preceq}$ is an ordered structure.
That is, $\preceq$ is compatible with $\vee$. | For $\struct {S, \vee, \preceq}$ to be an ordered structure is equivalent to, for all $a, b, c \in S$:
:$a \preceq b \implies a \vee c \preceq b \vee c$
:$a \preceq b \implies c \vee a \preceq c \vee b$
Since Join is Commutative, it suffices to prove the first of these implications.
By definition of join:
:$a \vee c = ... | Let $\struct {S, \vee, \preceq}$ be a [[Definition:Join Semilattice|join semilattice]].
Then $\struct {S, \vee, \preceq}$ is an [[Definition:Ordered Structure|ordered structure]].
That is, $\preceq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\vee$. | For $\struct {S, \vee, \preceq}$ to be an [[Definition:Ordered Structure|ordered structure]] is equivalent to, for all $a, b, c \in S$:
:$a \preceq b \implies a \vee c \preceq b \vee c$
:$a \preceq b \implies c \vee a \preceq c \vee b$
Since [[Join is Commutative]], it suffices to prove the first of these implication... | Join Semilattice is Ordered Structure/Proof 1 | https://proofwiki.org/wiki/Join_Semilattice_is_Ordered_Structure | https://proofwiki.org/wiki/Join_Semilattice_is_Ordered_Structure/Proof_1 | [
"Join Semilattices",
"Ordered Structures",
"Join Semilattice is Ordered Structure"
] | [
"Definition:Join Semilattice",
"Definition:Ordered Structure",
"Definition:Relation Compatible with Operation"
] | [
"Definition:Ordered Structure",
"Join is Commutative",
"Definition:Join (Order Theory)",
"Definition:Supremum of Set",
"Join Succeeds Operands",
"Definition:Transitive Relation",
"Definition:Upper Bound of Set",
"Definition:Supremum of Set"
] |
proofwiki-6620 | Join Semilattice is Ordered Structure | Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Then $\struct {S, \vee, \preceq}$ is an ordered structure.
That is, $\preceq$ is compatible with $\vee$. | Let $a, b, c \in S$.
Let $a \preceq b$.
By the definition of join semilattice:
:$a \vee b = b$
Thus:
:$\paren {a \vee b} \vee c = b \vee c$
Since $\vee$ is associative, commutative, and idempotent:
:$\paren {a \vee c} \vee \paren {b \vee c} = b \vee c$
Therefore, $a \vee c \preceq b \vee c$.
From Join is Commutative, w... | Let $\struct {S, \vee, \preceq}$ be a [[Definition:Join Semilattice|join semilattice]].
Then $\struct {S, \vee, \preceq}$ is an [[Definition:Ordered Structure|ordered structure]].
That is, $\preceq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\vee$. | Let $a, b, c \in S$.
Let $a \preceq b$.
By the definition of [[Definition:Join Semilattice|join semilattice]]:
:$a \vee b = b$
Thus:
:$\paren {a \vee b} \vee c = b \vee c$
Since $\vee$ is [[Definition:Associative Operation|associative]], [[Definition:Commutative Operation|commutative]], and [[Definition:Idempotent ... | Join Semilattice is Ordered Structure/Proof 2 | https://proofwiki.org/wiki/Join_Semilattice_is_Ordered_Structure | https://proofwiki.org/wiki/Join_Semilattice_is_Ordered_Structure/Proof_2 | [
"Join Semilattices",
"Ordered Structures",
"Join Semilattice is Ordered Structure"
] | [
"Definition:Join Semilattice",
"Definition:Ordered Structure",
"Definition:Relation Compatible with Operation"
] | [
"Definition:Join Semilattice",
"Definition:Associative Operation",
"Definition:Commutative/Operation",
"Definition:Idempotence/Operation",
"Join is Commutative"
] |
proofwiki-6621 | Meet Semilattice is Ordered Structure | Let $\struct {S, \wedge, \preceq}$ be a meet semilattice.
Then $\struct {S, \wedge, \preceq}$ is an ordered structure. | For $\struct {S, \wedge, \preceq}$ to be an ordered structure is equivalent to, for all $a,b,c \in S$:
:$a \preceq b \implies a \wedge c \preceq b \wedge c$
:$a \preceq b \implies c \wedge a \preceq c \wedge b$
Since Meet is Commutative, it suffices to prove the first of these implications.
By definition of meet:
:$b \... | Let $\struct {S, \wedge, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
Then $\struct {S, \wedge, \preceq}$ is an [[Definition:Ordered Structure|ordered structure]]. | For $\struct {S, \wedge, \preceq}$ to be an [[Definition:Ordered Structure|ordered structure]] is equivalent to, for all $a,b,c \in S$:
:$a \preceq b \implies a \wedge c \preceq b \wedge c$
:$a \preceq b \implies c \wedge a \preceq c \wedge b$
Since [[Meet is Commutative]], it suffices to prove the first of these imp... | Meet Semilattice is Ordered Structure | https://proofwiki.org/wiki/Meet_Semilattice_is_Ordered_Structure | https://proofwiki.org/wiki/Meet_Semilattice_is_Ordered_Structure | [
"Meet Semilattices",
"Ordered Structures"
] | [
"Definition:Meet Semilattice",
"Definition:Ordered Structure"
] | [
"Definition:Ordered Structure",
"Meet is Commutative",
"Definition:Meet (Order Theory)",
"Definition:Infimum of Set",
"Meet Precedes Operands",
"Definition:Transitive Relation",
"Definition:Lower Bound of Set",
"Definition:Infimum of Set"
] |
proofwiki-6622 | Join Semilattice is Semilattice | Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Then $\struct {S, \vee}$ is a semilattice. | Recall the definition of join semilattice:
{{:Definition:Join Semilattice}}
By definition of join semilattice, $\vee$ is closed.
The other three defining properties for a semilattice follow respectively from:
:Join is Commutative
:Join is Associative
:Join is Idempotent
Hence $\struct {S, \vee}$ is a semilattice.
{{qed... | Let $\struct {S, \vee, \preceq}$ be a [[Definition:Join Semilattice|join semilattice]].
Then $\struct {S, \vee}$ is a [[Definition:Semilattice|semilattice]]. | Recall the definition of [[Definition:Join Semilattice|join semilattice]]:
{{:Definition:Join Semilattice}}
By definition of [[Definition:Join Semilattice|join semilattice]], $\vee$ is [[Definition:Closed Operation|closed]].
The other three defining properties for a [[Definition:Semilattice|semilattice]] follow respe... | Join Semilattice is Semilattice | https://proofwiki.org/wiki/Join_Semilattice_is_Semilattice | https://proofwiki.org/wiki/Join_Semilattice_is_Semilattice | [
"Join Semilattices",
"Semilattices"
] | [
"Definition:Join Semilattice",
"Definition:Semilattice"
] | [
"Definition:Join Semilattice",
"Definition:Join Semilattice",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Semilattice",
"Join is Commutative",
"Join is Associative",
"Join is Idempotent",
"Definition:Semilattice"
] |
proofwiki-6623 | Meet Semilattice is Semilattice | Let $\struct {S, \wedge, \preceq}$ be a meet semilattice.
Then $\struct {S, \wedge}$ is a semilattice. | By definition of meet semilattice, $\wedge$ is closed.
The other three defining properties for a semilattice follow respectively from:
:Meet is Commutative
:Meet is Associative
:Meet is Idempotent
Hence $\struct {S, \wedge}$ is a semilattice.
{{qed}} | Let $\struct {S, \wedge, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
Then $\struct {S, \wedge}$ is a [[Definition:Semilattice|semilattice]]. | By definition of [[Definition:Meet Semilattice|meet semilattice]], $\wedge$ is [[Definition:Closed Operation|closed]].
The other three defining properties for a [[Definition:Semilattice|semilattice]] follow respectively from:
:[[Meet is Commutative]]
:[[Meet is Associative]]
:[[Meet is Idempotent]]
Hence $\struct {S... | Meet Semilattice is Semilattice | https://proofwiki.org/wiki/Meet_Semilattice_is_Semilattice | https://proofwiki.org/wiki/Meet_Semilattice_is_Semilattice | [
"Meet Semilattices",
"Semilattices"
] | [
"Definition:Meet Semilattice",
"Definition:Semilattice"
] | [
"Definition:Meet Semilattice",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Semilattice",
"Meet is Commutative",
"Meet is Associative",
"Meet is Idempotent",
"Definition:Semilattice"
] |
proofwiki-6624 | Semilattice Induces Ordering | Let $\struct {S, \circ}$ be a semilattice.
Let $\RR$ be the relation on $S$ defined by, for all $a, b \in S$:
:$a \mathrel \RR b$ {{iff}} $a \circ b = b$
Then $\RR$ is an ordering. | Let us verify that $\RR$ satisfies the three conditions for an ordering. | Let $\struct {S, \circ}$ be a [[Definition:Semilattice|semilattice]].
Let $\RR$ be the [[Definition:Relation|relation]] on $S$ defined by, for all $a, b \in S$:
:$a \mathrel \RR b$ {{iff}} $a \circ b = b$
Then $\RR$ is an [[Definition:Ordering|ordering]]. | Let us verify that $\RR$ satisfies the three conditions for an [[Definition:Ordering|ordering]]. | Semilattice Induces Ordering | https://proofwiki.org/wiki/Semilattice_Induces_Ordering | https://proofwiki.org/wiki/Semilattice_Induces_Ordering | [
"Semilattices"
] | [
"Definition:Semilattice",
"Definition:Relation",
"Definition:Ordering"
] | [
"Definition:Ordering",
"Definition:Ordering"
] |
proofwiki-6625 | Elementary Row Operations Associate with Matrix Multiplication | Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\mathbf A = \sqbrk a_{m n}$ be an $m \times n$ matrix over $R$.
Let $\mathbf B = \sqbrk b_{n p}$ be an $n \times p$ matrix over $R$.
Let $\hat o_1, \ldots, \hat o_{\hat n}$ be a finite sequence of elementary row operations that can be performed on a matrix over $R... | Proof by induction over $\hat n \in \N$, the number of elementary row operations. | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Operation|commutative]] [[Definition:Ring (Abstract Algebra)|ring]].
Let $\mathbf A = \sqbrk a_{m n}$ be an [[Definition:Matrix|$m \times n$ matrix]] over $R$.
Let $\mathbf B = \sqbrk b_{n p}$ be an [[Definition:Matrix|$n \times p$ matrix]] over $R$.
Let $\h... | Proof by [[Principle of Mathematical Induction|induction]] over $\hat n \in \N$, the number of [[Definition:Elementary Row Operation|elementary row operations]]. | Elementary Row Operations Associate with Matrix Multiplication | https://proofwiki.org/wiki/Elementary_Row_Operations_Associate_with_Matrix_Multiplication | https://proofwiki.org/wiki/Elementary_Row_Operations_Associate_with_Matrix_Multiplication | [
"Elementary Row Operations",
"Conventional Matrix Multiplication",
"Commutativity"
] | [
"Definition:Commutative/Operation",
"Definition:Ring (Abstract Algebra)",
"Definition:Matrix",
"Definition:Matrix",
"Definition:Finite Sequence",
"Definition:Elementary Operation/Row",
"Definition:Matrix/Row",
"Definition:Matrix Product (Conventional)"
] | [
"Principle of Mathematical Induction",
"Definition:Elementary Operation/Row",
"Definition:Elementary Operation/Row",
"Definition:Elementary Operation/Row"
] |
proofwiki-6626 | Union of Relations Compatible with Operation is Compatible | Let $\struct {S, \circ}$ be a closed algebraic structure.
Let $\FF$ be a family of relations on $S$.
Let each element of $\FF$ be compatible with $\circ$.
Let $\QQ = \bigcup \FF$.
Then $Q$ is a relation compatible with $\circ$. | Let $x, y, z \in S$.
Let $x \mathrel \QQ y$.
Then for some $\RR \in \FF$:
:$x \mathrel \RR y$.
Since $\RR$ is a relation compatible with $\circ$:
:$\tuple {x \circ z} \mathrel \RR \tuple {y \circ z}$
Since $\RR \in \FF$:
:$\RR \subseteq \bigcup \FF = \QQ$
Thus
:$\tuple {x \circ z} \mathrel \QQ \tuple {y \circ z}$
We h... | Let $\struct {S, \circ}$ be a [[Definition:Closed Algebraic Structure|closed algebraic structure]].
Let $\FF$ be a family of [[Definition:Endorelation|relations]] on $S$.
Let each element of $\FF$ be [[Definition:Relation Compatible with Operation|compatible with $\circ$]].
Let $\QQ = \bigcup \FF$.
Then $Q$ is a [... | Let $x, y, z \in S$.
Let $x \mathrel \QQ y$.
Then for some $\RR \in \FF$:
:$x \mathrel \RR y$.
Since $\RR$ is a [[Definition:Relation Compatible with Operation|relation compatible with $\circ$]]:
:$\tuple {x \circ z} \mathrel \RR \tuple {y \circ z}$
Since $\RR \in \FF$:
:$\RR \subseteq \bigcup \FF = \QQ$
Thus
:$\... | Union of Relations Compatible with Operation is Compatible | https://proofwiki.org/wiki/Union_of_Relations_Compatible_with_Operation_is_Compatible | https://proofwiki.org/wiki/Union_of_Relations_Compatible_with_Operation_is_Compatible | [
"Compatible Relations"
] | [
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Endorelation",
"Definition:Relation Compatible with Operation",
"Definition:Relation Compatible with Operation"
] | [
"Definition:Relation Compatible with Operation",
"Definition:Relation Compatible with Operation",
"Category:Compatible Relations"
] |
proofwiki-6627 | Complement of Relation Compatible with Group is Compatible | Let $\struct {G, \circ}$ be a group.
Let $\RR$ be a relation on $G$.
Let $\RR$ be compatible with $\circ$.
Let $\QQ = \complement_{G \times G} \RR$, so that:
:$\forall a, b \in G: a \mathrel \QQ b \iff \neg \paren {a \mathrel \RR b}$
Then $\QQ$ is a relation compatible with $\circ$. | Let $x, y, z \in G$.
Suppose that $\neg \paren {\paren {x \circ z} \mathrel \QQ \paren {y \circ z} }$.
Then by the definition of $\QQ$:
:$\paren {x \circ z} \mathrel \RR \paren {y \circ z}$
Because $\RR$ is compatible with $\circ$:
:$\paren {x \circ z} \circ z^{-1} \mathrel \RR \paren {y \circ z} \circ z^{-1}$
By {{Gro... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\RR$ be a [[Definition:Endorelation|relation]] on $G$.
Let $\RR$ be [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Let $\QQ = \complement_{G \times G} \RR$, so that:
:$\forall a, b \in G: a \mathrel \QQ b \iff \neg \paren {a \... | Let $x, y, z \in G$.
Suppose that $\neg \paren {\paren {x \circ z} \mathrel \QQ \paren {y \circ z} }$.
Then by the definition of $\QQ$:
:$\paren {x \circ z} \mathrel \RR \paren {y \circ z}$
Because $\RR$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$:
:$\paren {x \circ z} \circ z^{-1} ... | Complement of Relation Compatible with Group is Compatible | https://proofwiki.org/wiki/Complement_of_Relation_Compatible_with_Group_is_Compatible | https://proofwiki.org/wiki/Complement_of_Relation_Compatible_with_Group_is_Compatible | [
"Compatible Relations",
"Group Theory"
] | [
"Definition:Group",
"Definition:Endorelation",
"Definition:Relation Compatible with Operation",
"Definition:Relation Compatible with Operation"
] | [
"Definition:Relation Compatible with Operation",
"Rule of Transposition",
"Definition:Relation Compatible with Operation",
"Category:Compatible Relations",
"Category:Group Theory"
] |
proofwiki-6628 | Relation Compatible with Group Operation is Strongly Compatible | $\RR$ is strongly compatible with $\circ$:
:$\forall x, y, z \in G:$
::$x \mathrel \RR y \iff x \circ z \mathrel \RR y \circ z$
::$x \mathrel \RR y \iff z \circ x \mathrel \RR z \circ y$ | Since $\RR$ is compatible with $\circ$:
:$\forall a, b, c \in G: a \mathrel \RR b \implies a \circ c \mathrel \RR b \circ c$
In particular, letting $a = x$, $b = y$, and $c = z$, we see that:
:$x \mathrel \RR y \implies x \circ z \mathrel \RR y \circ z$
On the other hand, letting $a = x \circ z$, $b = y \circ z$, and $... | $\RR$ is [[Definition:Relation Strongly Compatible with Operation|strongly compatible]] with $\circ$:
:$\forall x, y, z \in G:$
::$x \mathrel \RR y \iff x \circ z \mathrel \RR y \circ z$
::$x \mathrel \RR y \iff z \circ x \mathrel \RR z \circ y$ | Since $\RR$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$:
:$\forall a, b, c \in G: a \mathrel \RR b \implies a \circ c \mathrel \RR b \circ c$
In particular, letting $a = x$, $b = y$, and $c = z$, we see that:
:$x \mathrel \RR y \implies x \circ z \mathrel \RR y \circ z$
On the oth... | Relation Compatible with Group Operation is Strongly Compatible | https://proofwiki.org/wiki/Relation_Compatible_with_Group_Operation_is_Strongly_Compatible | https://proofwiki.org/wiki/Relation_Compatible_with_Group_Operation_is_Strongly_Compatible | [
"Relations Compatible with Group Operation"
] | [
"Definition:Relation Strongly Compatible with Operation"
] | [
"Definition:Relation Compatible with Operation",
"Definition:Associative Operation",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Relation Strongly Compatible with Operation",
"Category:Relations Compatible with Group Operation"
] |
proofwiki-6629 | Intersection of Relations Compatible with Operation is Compatible | Let $\struct {S, \circ}$ be a closed algebraic structure.
Let $\mathscr F$ be a indexed family of relations on $S$.
Suppose that each element of $\mathscr F$ is compatible with $\circ$.
Let $\QQ = \bigcap \mathscr F$ be the intersection of $\mathscr F$.
Then $\QQ$ is a relation compatible with $\circ$. | Let $x, y, z \in S$.
Suppose that $x \mathrel \QQ y$.
Then for each $\RR \in \mathscr F$:
:$x \mathrel \RR y$
Then since $\RR$ is a relation compatible with $\circ$:
:$\paren {x \circ z} \mathrel \RR \paren {y \circ z}$
Since this holds for each $\RR \in \mathscr F$:
:$\paren {x \circ z} \mathrel \QQ \paren {y \circ z}... | Let $\struct {S, \circ}$ be a [[Definition:Closed Algebraic Structure|closed algebraic structure]].
Let $\mathscr F$ be a [[Definition:Indexed Family|indexed family]] of [[Definition:Endorelation|relations]] on $S$.
Suppose that each element of $\mathscr F$ is [[Definition:Relation Compatible with Operation|compatibl... | Let $x, y, z \in S$.
Suppose that $x \mathrel \QQ y$.
Then for each $\RR \in \mathscr F$:
:$x \mathrel \RR y$
Then since $\RR$ is a [[Definition:Relation|relation]] [[Definition:Relation Compatible with Operation|compatible]] with $\circ$:
:$\paren {x \circ z} \mathrel \RR \paren {y \circ z}$
Since this holds for... | Intersection of Relations Compatible with Operation is Compatible | https://proofwiki.org/wiki/Intersection_of_Relations_Compatible_with_Operation_is_Compatible | https://proofwiki.org/wiki/Intersection_of_Relations_Compatible_with_Operation_is_Compatible | [
"Compatible Relations"
] | [
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Indexing Set/Family",
"Definition:Endorelation",
"Definition:Relation Compatible with Operation",
"Definition:Set Intersection/Family of Sets",
"Definition:Relation",
"Definition:Relation Compatible with Operation"
] | [
"Definition:Relation",
"Definition:Relation Compatible with Operation",
"Definition:Relation",
"Definition:Relation Compatible with Operation",
"Category:Compatible Relations"
] |
proofwiki-6630 | Inverse of Relation Compatible with Operation is Compatible | Let $\struct {S, \circ}$ be a closed algebraic structure.
Let $\RR$ be a relation on $S$ which is compatible with $\circ$.
Let $\QQ$ be the inverse relation of $\RR$.
Then $\QQ$ is compatible with $\circ$. | Let $x, y, z \in S$.
Suppose that $x \mathrel \QQ y$.
Then by the definition of $\QQ$:
:$y \mathrel \RR x$.
Since $\RR$ is compatible with $\circ$:
:$\paren {y \circ z} \mathrel \RR \paren {x \circ z}$
and
:$\paren {z \circ y} \mathrel \RR \paren {z \circ x}$
Thus by the definition of $\QQ$:
:$\paren {x \circ z} \mathr... | Let $\struct {S, \circ}$ be a [[Definition:Closed Algebraic Structure|closed algebraic structure]].
Let $\RR$ be a [[Definition:Endorelation|relation]] on $S$ which is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Let $\QQ$ be the [[Definition:Inverse Relation|inverse relation]] of $\RR$.... | Let $x, y, z \in S$.
Suppose that $x \mathrel \QQ y$.
Then by the definition of $\QQ$:
:$y \mathrel \RR x$.
Since $\RR$ is compatible with $\circ$:
:$\paren {y \circ z} \mathrel \RR \paren {x \circ z}$
and
:$\paren {z \circ y} \mathrel \RR \paren {z \circ x}$
Thus by the definition of $\QQ$:
:$\paren {x \circ z} \m... | Inverse of Relation Compatible with Operation is Compatible | https://proofwiki.org/wiki/Inverse_of_Relation_Compatible_with_Operation_is_Compatible | https://proofwiki.org/wiki/Inverse_of_Relation_Compatible_with_Operation_is_Compatible | [
"Compatible Relations"
] | [
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Endorelation",
"Definition:Relation Compatible with Operation",
"Definition:Inverse Relation"
] | [
"Category:Compatible Relations"
] |
proofwiki-6631 | Ordering Compatible with Group Operation is Strongly Compatible | {{begin-eqn}}
{{eqn | q = \forall x, y, z \in G
| l = x \preccurlyeq y
| o = \iff
| r = x \circ z \preccurlyeq y \circ z
}}
{{eqn | l = x \preccurlyeq y
| o = \iff
| r = z \circ x \preccurlyeq z \circ y
}}
{{eqn | l = x \prec y
| o = \iff
| r = x \circ z \prec y \circ z
}}
{{eq... | By definition of ordered group, $\preccurlyeq$ is a relation compatible with $\circ$.
Thus by Relation Compatible with Group Operation is Strongly Compatible:
{{begin-eqn}}
{{eqn | l = x \preccurlyeq y
| o = \iff
| r = x \circ z \preccurlyeq y \circ z
}}
{{eqn | l = x \preccurlyeq y
| o = \iff
|... | {{begin-eqn}}
{{eqn | q = \forall x, y, z \in G
| l = x \preccurlyeq y
| o = \iff
| r = x \circ z \preccurlyeq y \circ z
}}
{{eqn | l = x \preccurlyeq y
| o = \iff
| r = z \circ x \preccurlyeq z \circ y
}}
{{eqn | l = x \prec y
| o = \iff
| r = x \circ z \prec y \circ z
}}
{{eq... | By definition of [[Definition:Ordered Group|ordered group]], $\preccurlyeq$ is a [[Definition:Relation Compatible with Operation|relation compatible with $\circ$]].
Thus by [[Relation Compatible with Group Operation is Strongly Compatible]]:
{{begin-eqn}}
{{eqn | l = x \preccurlyeq y
| o = \iff
| r = x \c... | Ordering Compatible with Group Operation is Strongly Compatible | https://proofwiki.org/wiki/Ordering_Compatible_with_Group_Operation_is_Strongly_Compatible | https://proofwiki.org/wiki/Ordering_Compatible_with_Group_Operation_is_Strongly_Compatible | [
"Ordered Groups",
"Compatible Relations",
"Ordering Compatible with Group Operation is Strongly Compatible"
] | [] | [
"Definition:Ordered Group",
"Definition:Relation Compatible with Operation",
"Relation Compatible with Group Operation is Strongly Compatible",
"Reflexive Reduction of Relation Compatible with Group Operation is Compatible",
"Definition:Relation Compatible with Operation",
"Relation Compatible with Group ... |
proofwiki-6632 | Ordering Compatible with Group Operation is Strongly Compatible | {{begin-eqn}}
{{eqn | q = \forall x, y, z \in G
| l = x \preccurlyeq y
| o = \iff
| r = x \circ z \preccurlyeq y \circ z
}}
{{eqn | l = x \preccurlyeq y
| o = \iff
| r = z \circ x \preccurlyeq z \circ y
}}
{{eqn | l = x \prec y
| o = \iff
| r = x \circ z \prec y \circ z
}}
{{eq... | By the definition of an ordered group, $\preccurlyeq$ is a relation compatible with $\circ$.
Thus by {{Corollary|Relation Compatible with Group Operation is Strongly Compatible}}, we obtain the first four results.
By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $\prec$ is compatible wi... | {{begin-eqn}}
{{eqn | q = \forall x, y, z \in G
| l = x \preccurlyeq y
| o = \iff
| r = x \circ z \preccurlyeq y \circ z
}}
{{eqn | l = x \preccurlyeq y
| o = \iff
| r = z \circ x \preccurlyeq z \circ y
}}
{{eqn | l = x \prec y
| o = \iff
| r = x \circ z \prec y \circ z
}}
{{eq... | By the definition of an [[Definition:Ordered Group|ordered group]], $\preccurlyeq$ is a [[Definition:Relation Compatible with Operation|relation compatible]] with $\circ$.
Thus by {{Corollary|Relation Compatible with Group Operation is Strongly Compatible}}, we obtain the first four results.
By [[Reflexive Reduction... | Ordering Compatible with Group Operation is Strongly Compatible/Corollary/Proof 1 | https://proofwiki.org/wiki/Ordering_Compatible_with_Group_Operation_is_Strongly_Compatible | https://proofwiki.org/wiki/Ordering_Compatible_with_Group_Operation_is_Strongly_Compatible/Corollary/Proof_1 | [
"Ordered Groups",
"Compatible Relations",
"Ordering Compatible with Group Operation is Strongly Compatible"
] | [] | [
"Definition:Ordered Group",
"Definition:Relation Compatible with Operation",
"Reflexive Reduction of Relation Compatible with Group Operation is Compatible",
"Definition:Relation Compatible with Operation"
] |
proofwiki-6633 | Ordering Compatible with Group Operation is Strongly Compatible | {{begin-eqn}}
{{eqn | q = \forall x, y, z \in G
| l = x \preccurlyeq y
| o = \iff
| r = x \circ z \preccurlyeq y \circ z
}}
{{eqn | l = x \preccurlyeq y
| o = \iff
| r = z \circ x \preccurlyeq z \circ y
}}
{{eqn | l = x \prec y
| o = \iff
| r = x \circ z \prec y \circ z
}}
{{eq... | Each result follows from Ordering Compatible with Group Operation is Strongly Compatible.
For example, by Ordering Compatible with Group Operation is Strongly Compatible:
:$x \preccurlyeq y \iff x \circ x^{-1} \preccurlyeq y \circ x^{-1}$
Since $x \circ x^{-1} = e$:
:$x \preccurlyeq y \iff e \preccurlyeq y \circ x^{-1}... | {{begin-eqn}}
{{eqn | q = \forall x, y, z \in G
| l = x \preccurlyeq y
| o = \iff
| r = x \circ z \preccurlyeq y \circ z
}}
{{eqn | l = x \preccurlyeq y
| o = \iff
| r = z \circ x \preccurlyeq z \circ y
}}
{{eqn | l = x \prec y
| o = \iff
| r = x \circ z \prec y \circ z
}}
{{eq... | Each result follows from [[Ordering Compatible with Group Operation is Strongly Compatible]].
For example, by [[Ordering Compatible with Group Operation is Strongly Compatible]]:
:$x \preccurlyeq y \iff x \circ x^{-1} \preccurlyeq y \circ x^{-1}$
Since $x \circ x^{-1} = e$:
:$x \preccurlyeq y \iff e \preccurlyeq y ... | Ordering Compatible with Group Operation is Strongly Compatible/Corollary/Proof 2 | https://proofwiki.org/wiki/Ordering_Compatible_with_Group_Operation_is_Strongly_Compatible | https://proofwiki.org/wiki/Ordering_Compatible_with_Group_Operation_is_Strongly_Compatible/Corollary/Proof_2 | [
"Ordered Groups",
"Compatible Relations",
"Ordering Compatible with Group Operation is Strongly Compatible"
] | [] | [
"Ordering Compatible with Group Operation is Strongly Compatible",
"Ordering Compatible with Group Operation is Strongly Compatible"
] |
proofwiki-6634 | Relations Compatible with Operation Form Complete Distributive Lattice | Let $\struct {S, \circ}$ be an algebraic structure.
Let $C$ be the set of relations on $S$ compatible with $\circ$.
Then $\struct {C, \cap, \cup, \subseteq}$ is a complete distributive lattice. | {{ProofWanted}}
Category:Compatible Relations
Category:Complete Lattices
Category:Distributive Lattices
poocmfwwi4z2mb95eaedgnbgag5sxt3 | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $C$ be the [[Definition:Set|set]] of [[Definition:Endorelation|relations]] on $S$ compatible with $\circ$.
Then $\struct {C, \cap, \cup, \subseteq}$ is a [[Definition:Complete Distributive Lattice|complete d... | {{ProofWanted}}
[[Category:Compatible Relations]]
[[Category:Complete Lattices]]
[[Category:Distributive Lattices]]
poocmfwwi4z2mb95eaedgnbgag5sxt3 | Relations Compatible with Operation Form Complete Distributive Lattice | https://proofwiki.org/wiki/Relations_Compatible_with_Operation_Form_Complete_Distributive_Lattice | https://proofwiki.org/wiki/Relations_Compatible_with_Operation_Form_Complete_Distributive_Lattice | [
"Compatible Relations",
"Complete Lattices",
"Distributive Lattices"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Set",
"Definition:Endorelation",
"Definition:Complete Distributive Lattice"
] | [
"Category:Compatible Relations",
"Category:Complete Lattices",
"Category:Distributive Lattices"
] |
proofwiki-6635 | Relations Compatible with Group Form Complete Boolean Algebra | Let $\struct {S, \circ}$ be a group.
Let $C$ be the set of relations on $S$ which are compatible with $\circ$.
Then $\struct {C, \cap, \cup, \subseteq}$ is a complete Boolean lattice. | {{ProofWanted}}
Category:Compatible Relations
Category:Complete Lattices
Category:Boolean Lattices
tbnnoumtcegy19fm00qmai3k9d785za | Let $\struct {S, \circ}$ be a [[Definition:Group|group]].
Let $C$ be the set of [[Definition:Endorelation|relations]] on $S$ which are [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Then $\struct {C, \cap, \cup, \subseteq}$ is a [[Definition:Complete Boolean Lattice|complete Boolean latti... | {{ProofWanted}}
[[Category:Compatible Relations]]
[[Category:Complete Lattices]]
[[Category:Boolean Lattices]]
tbnnoumtcegy19fm00qmai3k9d785za | Relations Compatible with Group Form Complete Boolean Algebra | https://proofwiki.org/wiki/Relations_Compatible_with_Group_Form_Complete_Boolean_Algebra | https://proofwiki.org/wiki/Relations_Compatible_with_Group_Form_Complete_Boolean_Algebra | [
"Compatible Relations",
"Complete Lattices",
"Boolean Lattices"
] | [
"Definition:Group",
"Definition:Endorelation",
"Definition:Relation Compatible with Operation",
"Definition:Complete Boolean Lattice"
] | [
"Category:Compatible Relations",
"Category:Complete Lattices",
"Category:Boolean Lattices"
] |
proofwiki-6636 | Hypothetical Syllogism/Formulation 3 | :$\vdash \paren {\paren {p \implies q} \land \paren {q \implies r} } \implies \paren {p \implies r}$ | Let us use the following abbreviations:
{{begin-eqn}}
{{eqn | l = \phi
| o = \text{ for }
| r = p \implies q
| c =
}}
{{eqn | l = \psi
| o = \text{ for }
| r = q \implies r
| c =
}}
{{eqn | l = \chi
| o = \text{ for }
| r = p \implies r
| c =
}}
{{end-eqn}}
{{Beg... | :$\vdash \paren {\paren {p \implies q} \land \paren {q \implies r} } \implies \paren {p \implies r}$ | Let us use the following abbreviations:
{{begin-eqn}}
{{eqn | l = \phi
| o = \text{ for }
| r = p \implies q
| c =
}}
{{eqn | l = \psi
| o = \text{ for }
| r = q \implies r
| c =
}}
{{eqn | l = \chi
| o = \text{ for }
| r = p \implies r
| c =
}}
{{end-eqn}}
{{... | Hypothetical Syllogism/Formulation 3/Proof 1 | https://proofwiki.org/wiki/Hypothetical_Syllogism/Formulation_3 | https://proofwiki.org/wiki/Hypothetical_Syllogism/Formulation_3/Proof_1 | [
"Hypothetical Syllogism"
] | [] | [] |
proofwiki-6637 | Hypothetical Syllogism/Formulation 3 | :$\vdash \paren {\paren {p \implies q} \land \paren {q \implies r} } \implies \paren {p \implies r}$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations.
:<nowiki>$\begin{array}{|ccccccc|c|ccc|} \hline
((p & \implies & q) & \land & (q & \implies & r)) & \implies & (p & \implies & r) \\
\hline
\F & \T & \F & \T & \F & \T &... | :$\vdash \paren {\paren {p \implies q} \land \paren {q \implies r} } \implies \paren {p \implies r}$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|... | Hypothetical Syllogism/Formulation 3/Proof by Truth Table | https://proofwiki.org/wiki/Hypothetical_Syllogism/Formulation_3 | https://proofwiki.org/wiki/Hypothetical_Syllogism/Formulation_3/Proof_by_Truth_Table | [
"Hypothetical Syllogism"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-6638 | Hypothetical Syllogism/Formulation 4 | :$\vdash \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }$ | Let us use the following abbreviations
{{begin-eqn}}
{{eqn | l = \phi
| o = \text{ for }
| r = p \implies q
}}
{{eqn | l = \psi
| o = \text{ for }
| r = q \implies r
}}
{{eqn | l = \chi
| o = \text{ for }
| r = p \implies r
}}
{{end-eqn}}
{{BeginTableau|\paren {p \implies q} \implies... | :$\vdash \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }$ | Let us use the following abbreviations
{{begin-eqn}}
{{eqn | l = \phi
| o = \text{ for }
| r = p \implies q
}}
{{eqn | l = \psi
| o = \text{ for }
| r = q \implies r
}}
{{eqn | l = \chi
| o = \text{ for }
| r = p \implies r
}}
{{end-eqn}}
{{BeginTableau|\paren {p \implies q} \impl... | Hypothetical Syllogism/Formulation 4/Proof 1 | https://proofwiki.org/wiki/Hypothetical_Syllogism/Formulation_4 | https://proofwiki.org/wiki/Hypothetical_Syllogism/Formulation_4/Proof_1 | [
"Hypothetical Syllogism"
] | [] | [
"Hypothetical Syllogism/Formulation 3",
"Rule of Exportation"
] |
proofwiki-6639 | Hypothetical Syllogism/Formulation 4 | :$\vdash \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }$ | {{BeginTableau|\paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} } }}
{{Assumption|1|p \implies q}}
{{Assumption|2|p}}
{{Assumption|3|q \implies r}}
{{ModusPonens|4|1,2|q|1|2}}
{{ModusPonens|5|1,2,3|r|3|4}}
{{Implication|6|1,3|p \implies r|2|5}}
{{Implication|7|1|\paren {q \imp... | :$\vdash \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }$ | {{BeginTableau|\paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} } }}
{{Assumption|1|p \implies q}}
{{Assumption|2|p}}
{{Assumption|3|q \implies r}}
{{ModusPonens|4|1,2|q|1|2}}
{{ModusPonens|5|1,2,3|r|3|4}}
{{Implication|6|1,3|p \implies r|2|5}}
{{Implication|7|1|\paren {q \imp... | Hypothetical Syllogism/Formulation 4/Proof 2 | https://proofwiki.org/wiki/Hypothetical_Syllogism/Formulation_4 | https://proofwiki.org/wiki/Hypothetical_Syllogism/Formulation_4/Proof_2 | [
"Hypothetical Syllogism"
] | [] | [] |
proofwiki-6640 | Hypothetical Syllogism/Formulation 4 | :$\vdash \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }|instance 1 of a Hilbert proof system}}
{{Assumption|1|p}}
{{Assumption|2|p \implies q}}
{{ModusPonens|3|1, 2|q|1|2}}
{{Assumption|4|q \implies r}}
{{ModusPonens|5|1, 2, 4|r|3|4}}
{{TableauLine|n = 6
| p... | :$\vdash \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }|[[Definition:Hilbert Proof System/Instance 1|instance 1 of a Hilbert proof system]]}}
{{Assumption|1|p}}
{{Assumption|2|p \implies q}}
{{ModusPonens|3|1, 2|q|1|2}}
{{Assumption|4|q \implies r}}
{{ModusPon... | Hypothetical Syllogism/Formulation 4/Proof 3 | https://proofwiki.org/wiki/Hypothetical_Syllogism/Formulation_4 | https://proofwiki.org/wiki/Hypothetical_Syllogism/Formulation_4/Proof_3 | [
"Hypothetical Syllogism"
] | [] | [
"Definition:Hilbert Proof System/Instance 1",
"Definition:Assumption",
"Definition:Discharged Assumption",
"Definition:Assumption",
"Definition:Discharged Assumption",
"Definition:Assumption",
"Definition:Discharged Assumption"
] |
proofwiki-6641 | Hypothetical Syllogism/Formulation 5 | :$\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | Let us use the following abbreviations
{{begin-eqn}}
{{eqn | l = \phi
| o = \text{ for }
| r = p \implies q
| c =
}}
{{eqn | l = \psi
| o = \text{ for }
| r = q \implies r
| c =
}}
{{eqn | l = \chi
| o = \text{ for }
| r = p \implies r
| c =
}}
{{end-eqn}}
From H... | :$\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | Let us use the following abbreviations
{{begin-eqn}}
{{eqn | l = \phi
| o = \text{ for }
| r = p \implies q
| c =
}}
{{eqn | l = \psi
| o = \text{ for }
| r = q \implies r
| c =
}}
{{eqn | l = \chi
| o = \text{ for }
| r = p \implies r
| c =
}}
{{end-eqn}}
Fro... | Hypothetical Syllogism/Formulation 5/Proof 1 | https://proofwiki.org/wiki/Hypothetical_Syllogism/Formulation_5 | https://proofwiki.org/wiki/Hypothetical_Syllogism/Formulation_5/Proof_1 | [
"Hypothetical Syllogism"
] | [] | [
"Hypothetical Syllogism/Formulation 3",
"Hypothetical Syllogism/Formulation 3",
"Rule of Exportation"
] |
proofwiki-6642 | Hypothetical Syllogism/Formulation 5 | :$\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | {{BeginTableau |\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }|Instance 2 of the Hilbert-style systems}}
{{TableauLine
| n = 1
| f = \paren {q \implies r} \implies \paren {\paren {p \lor q} \implies \paren {p \lor r} }
| rlnk = Definition:Hilbert Proof System/Ins... | :$\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | {{BeginTableau |\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }|[[Definition:Hilbert Proof System/Instance 2|Instance 2 of the Hilbert-style systems]]}}
{{TableauLine
| n = 1
| f = \paren {q \implies r} \implies \paren {\paren {p \lor q} \implies \paren {p \lor r} ... | Hypothetical Syllogism/Formulation 5/Proof 2 | https://proofwiki.org/wiki/Hypothetical_Syllogism/Formulation_5 | https://proofwiki.org/wiki/Hypothetical_Syllogism/Formulation_5/Proof_2 | [
"Hypothetical Syllogism"
] | [] | [
"Definition:Hilbert Proof System/Instance 2"
] |
proofwiki-6643 | Hypothetical Syllogism/Formulation 5 | :$\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} } }}
{{Assumption|1|q \implies r}}
{{Assumption|2|p \implies q}}
{{Assumption|3|p}}
{{ModusPonens|4|2,3|q|2|3}}
{{ModusPonens|5|1,2,3|r|1|4}}
{{Implication|6|1,2|p \implies r|3|5}}
{{Implication|7|1|\paren ... | :$\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} } }}
{{Assumption|1|q \implies r}}
{{Assumption|2|p \implies q}}
{{Assumption|3|p}}
{{ModusPonens|4|2,3|q|2|3}}
{{ModusPonens|5|1,2,3|r|1|4}}
{{Implication|6|1,2|p \implies r|3|5}}
{{Implication|7|1|\paren ... | Hypothetical Syllogism/Formulation 5/Proof 3 | https://proofwiki.org/wiki/Hypothetical_Syllogism/Formulation_5 | https://proofwiki.org/wiki/Hypothetical_Syllogism/Formulation_5/Proof_3 | [
"Hypothetical Syllogism"
] | [] | [] |
proofwiki-6644 | Duality Principle (Order Theory)/Global Duality | The following are equivalent:
:$(1): \quad \Sigma$ is true for all ordered sets
:$(2): \quad \Sigma^*$ is true for all ordered sets | === $(1)$ implies $(2)$ ===
Let $\struct {S, \preceq}$ be an ordered set.
Let $\struct {S, \succeq}$ be its dual.
By assumption, $\Sigma$ is true for $\struct {S, \succeq}$.
By Local Duality, this implies $\Sigma^*$ is true for $\struct {S, \preceq}$.
Since $\struct {S, \preceq}$ was arbitrary, the result follows.
{{qe... | The following are [[Definition:Logical Equivalence|equivalent]]:
:$(1): \quad \Sigma$ is true for all [[Definition:Ordered Set|ordered sets]]
:$(2): \quad \Sigma^*$ is true for all [[Definition:Ordered Set|ordered sets]] | === $(1)$ implies $(2)$ ===
Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $\struct {S, \succeq}$ be its [[Definition:Dual Ordered Set|dual]].
By assumption, $\Sigma$ is true for $\struct {S, \succeq}$.
By [[Duality Principle (Order Theory)/Local Duality|Local Duality]], this implies $... | Duality Principle (Order Theory)/Global Duality | https://proofwiki.org/wiki/Duality_Principle_(Order_Theory)/Global_Duality | https://proofwiki.org/wiki/Duality_Principle_(Order_Theory)/Global_Duality | [
"Order Theory",
"Named Theorems"
] | [
"Definition:Logical Equivalence",
"Definition:Ordered Set",
"Definition:Ordered Set"
] | [
"Definition:Ordered Set",
"Definition:Dual Ordering/Dual Ordered Set",
"Duality Principle (Order Theory)/Local Duality"
] |
proofwiki-6645 | Duality Principle (Order Theory)/Local Duality | Let $\struct {S, \preceq}$ be an ordered set, and let $\struct {S, \succeq}$ be its dual.
Then the following are equivalent:
:$(1): \quad \Sigma$ is true for $\struct {S, \preceq}$
:$(2): \quad \Sigma^*$ is true for $\struct {S, \succeq}$ | === $(1)$ implies $(2)$ ===
By assumption, $\Sigma$ is true for $\struct {S, \preceq}$.
By Dual of Dual Ordering, the dual statement $\Sigma^*$ applied to $\struct {S, \succeq}$ is the same as $\Sigma$ applied to $\struct {S, \preceq}$.
Hence $\Sigma^*$ is true for $\struct {S, \succeq}$.
{{qed|lemma}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]], and let $\struct {S, \succeq}$ be its [[Definition:Dual Ordered Set|dual]].
Then the following are [[Definition:Logical Equivalence|equivalent]]:
:$(1): \quad \Sigma$ is true for $\struct {S, \preceq}$
:$(2): \quad \Sigma^*$ is true for $\struc... | === $(1)$ implies $(2)$ ===
By assumption, $\Sigma$ is true for $\struct {S, \preceq}$.
By [[Dual of Dual Ordering]], the [[Definition:Dual Statement (Order Theory)|dual statement]] $\Sigma^*$ applied to $\struct {S, \succeq}$ is the same as $\Sigma$ applied to $\struct {S, \preceq}$.
Hence $\Sigma^*$ is true for $... | Duality Principle (Order Theory)/Local Duality | https://proofwiki.org/wiki/Duality_Principle_(Order_Theory)/Local_Duality | https://proofwiki.org/wiki/Duality_Principle_(Order_Theory)/Local_Duality | [
"Order Theory",
"Named Theorems"
] | [
"Definition:Ordered Set",
"Definition:Dual Ordering/Dual Ordered Set",
"Definition:Logical Equivalence"
] | [
"Dual of Dual Ordering",
"Definition:Dual Statement (Order Theory)",
"Dual of Dual Ordering"
] |
proofwiki-6646 | Diagonal Complement Relation Compatible with Group Operation | Let $\left({G, \circ}\right)$ be a group.
Let $\Delta_G$ be the diagonal relation on $G$.
Then $\Delta_G^c = \complement_{G \times G} \Delta_G$ is a relation compatible with $\circ$.
{{explain|Find another way to write the above, it's ugly}}
In other words, $\ne$ is a relation compatible with $\circ$. | By Diagonal Relation is Universally Compatible, $\Delta_G$ is compatible with $\circ$.
By Complement of Relation Compatible with Group is Compatible , $\Delta_G^c$ is also compatible with $\circ$.
{{qed}}
Category:Compatible Relations
n07d4bm10j5n1oe24z8vocmgjasgc45 | Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]].
Let $\Delta_G$ be the [[Definition:Diagonal Relation|diagonal relation]] on $G$.
Then $\Delta_G^c = \complement_{G \times G} \Delta_G$ is a [[Definition:Relation|relation]] [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
{{expl... | By [[Diagonal Relation is Universally Compatible]], $\Delta_G$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
By [[Complement of Relation Compatible with Group is Compatible ]], $\Delta_G^c$ is also compatible with $\circ$.
{{qed}}
[[Category:Compatible Relations]]
n07d4bm10j5n1oe24z8vo... | Diagonal Complement Relation Compatible with Group Operation | https://proofwiki.org/wiki/Diagonal_Complement_Relation_Compatible_with_Group_Operation | https://proofwiki.org/wiki/Diagonal_Complement_Relation_Compatible_with_Group_Operation | [
"Compatible Relations"
] | [
"Definition:Group",
"Definition:Diagonal Relation",
"Definition:Relation",
"Definition:Relation Compatible with Operation"
] | [
"Diagonal Relation is Universally Compatible",
"Definition:Relation Compatible with Operation",
"Complement of Relation Compatible with Group is Compatible ",
"Category:Compatible Relations"
] |
proofwiki-6647 | Reflexive Closure of Relation Compatible with Operation is Compatible | Let $\struct {S, \circ}$ be a magma.
Let $\RR$ be a relation on $S$ which is compatible with $\circ$.
Let $\RR^=$ be the reflexive closure of $\prec$.
That is, $\RR^=$ is defined as the union of $\RR$ with the diagonal relation for $S$.
Then $\RR^=$ is compatible with $\circ$. | By Diagonal Relation is Universally Compatible, the diagonal relation is compatible with $\circ$.
Then by Union of Relations Compatible with Operation is Compatible, $\RR^=$ is compatible with $\circ$.
{{qed}} | Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
Let $\RR$ be a [[Definition:Endorelation|relation]] on $S$ which is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Let $\RR^=$ be the [[Definition:Reflexive Closure|reflexive closure]] of $\prec$.
That is, $\RR^=$ is defined as th... | By [[Diagonal Relation is Universally Compatible]], the [[Definition:Diagonal Relation|diagonal relation]] is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Then by [[Union of Relations Compatible with Operation is Compatible]], $\RR^=$ is compatible with $\circ$.
{{qed}} | Reflexive Closure of Relation Compatible with Operation is Compatible | https://proofwiki.org/wiki/Reflexive_Closure_of_Relation_Compatible_with_Operation_is_Compatible | https://proofwiki.org/wiki/Reflexive_Closure_of_Relation_Compatible_with_Operation_is_Compatible | [
"Compatible Relations",
"Reflexive Closures"
] | [
"Definition:Magma",
"Definition:Endorelation",
"Definition:Relation Compatible with Operation",
"Definition:Reflexive Closure",
"Definition:Diagonal Relation"
] | [
"Diagonal Relation is Universally Compatible",
"Definition:Diagonal Relation",
"Definition:Relation Compatible with Operation",
"Union of Relations Compatible with Operation is Compatible"
] |
proofwiki-6648 | Reflexive Reduction of Relation Compatible with Group Operation is Compatible | Let $\struct {S, \circ}$ be a group.
Let $\RR$ be a relation on $S$ which is compatible with $\circ$.
Let $\RR^\ne$ be the reflexive reduction of $\RR$.
Then $\RR^\ne$ is compatible with $\circ$. | From the Cancellation Laws, $\circ$ is a cancellable operation.
The result then follows directly from Reflexive Reduction of Relation Compatible with Cancellable Operation is Compatible.
{{qed}} | Let $\struct {S, \circ}$ be a [[Definition:Group|group]].
Let $\RR$ be a [[Definition:Endorelation|relation]] on $S$ which is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Let $\RR^\ne$ be the [[Definition:Reflexive Reduction|reflexive reduction]] of $\RR$.
Then $\RR^\ne$ is [[Definitio... | From the [[Cancellation Laws]], $\circ$ is a [[Definition:Cancellable Operation|cancellable operation]].
The result then follows directly from [[Reflexive Reduction of Relation Compatible with Cancellable Operation is Compatible]].
{{qed}} | Reflexive Reduction of Relation Compatible with Group Operation is Compatible/Proof 2 | https://proofwiki.org/wiki/Reflexive_Reduction_of_Relation_Compatible_with_Group_Operation_is_Compatible | https://proofwiki.org/wiki/Reflexive_Reduction_of_Relation_Compatible_with_Group_Operation_is_Compatible/Proof_2 | [
"Compatible Relations",
"Reflexive Reductions",
"Reflexive Reduction of Relation Compatible with Group Operation is Compatible"
] | [
"Definition:Group",
"Definition:Endorelation",
"Definition:Relation Compatible with Operation",
"Definition:Reflexive Reduction",
"Definition:Relation Compatible with Operation"
] | [
"Cancellation Laws",
"Definition:Cancellable Operation",
"Reflexive Reduction of Relation Compatible with Cancellable Operation is Compatible"
] |
proofwiki-6649 | Double Negation Elimination implies Law of Excluded Middle | Let the Law of Double Negation Elimination be supposed to hold:
:$\neg \neg p \vdash p$
Then the Law of Excluded Middle likewise holds:
:$\vdash p \lor \neg p$ | {{BeginTableau|p \lor \neg p}}
{{Assumption |1|\neg \paren {p \lor \neg p}|Assume the contrary}}
{{Assumption |2|p|Assume one disjunct}}
{{Addition |3|2|p \lor \neg p|2|1}}
{{NonContradiction |4|1, 2|3|1}}
{{Contradiction |5|1|\neg p|2|4|demonstrating a contradiction}}
{{Addit... | Let the [[Double Negation Elimination|Law of Double Negation Elimination]] be supposed to hold:
:$\neg \neg p \vdash p$
Then the [[Law of Excluded Middle]] likewise holds:
:$\vdash p \lor \neg p$ | {{BeginTableau|p \lor \neg p}}
{{Assumption |1|\neg \paren {p \lor \neg p}|Assume the contrary}}
{{Assumption |2|p|Assume one disjunct}}
{{Addition |3|2|p \lor \neg p|2|1}}
{{NonContradiction |4|1, 2|3|1}}
{{Contradiction |5|1|\neg p|2|4|demonstrating a contradiction}}
{{Addit... | Double Negation Elimination implies Law of Excluded Middle/Proof 1 | https://proofwiki.org/wiki/Double_Negation_Elimination_implies_Law_of_Excluded_Middle | https://proofwiki.org/wiki/Double_Negation_Elimination_implies_Law_of_Excluded_Middle/Proof_1 | [
"Double Negation Elimination implies Law of Excluded Middle",
"Double Negation Elimination",
"Double Negation"
] | [
"Double Negation/Double Negation Elimination",
"Law of Excluded Middle"
] | [] |
proofwiki-6650 | Double Negation Elimination implies Law of Excluded Middle | Let the Law of Double Negation Elimination be supposed to hold:
:$\neg \neg p \vdash p$
Then the Law of Excluded Middle likewise holds:
:$\vdash p \lor \neg p$ | {{BeginTableau|\vdash p \lor \neg p}}
{{TheoremIntro|1|\neg\neg (p \lor \neg p)|Negation of Excluded Middle is False: Form 2}}
{{DoubleNegElimination|2||p \lor \neg p|1}}
{{EndTableau|qed}} | Let the [[Double Negation Elimination|Law of Double Negation Elimination]] be supposed to hold:
:$\neg \neg p \vdash p$
Then the [[Law of Excluded Middle]] likewise holds:
:$\vdash p \lor \neg p$ | {{BeginTableau|\vdash p \lor \neg p}}
{{TheoremIntro|1|\neg\neg (p \lor \neg p)|[[Negation of Excluded Middle is False/Form 2|Negation of Excluded Middle is False: Form 2]]}}
{{DoubleNegElimination|2||p \lor \neg p|1}}
{{EndTableau|qed}} | Double Negation Elimination implies Law of Excluded Middle/Proof 2 | https://proofwiki.org/wiki/Double_Negation_Elimination_implies_Law_of_Excluded_Middle | https://proofwiki.org/wiki/Double_Negation_Elimination_implies_Law_of_Excluded_Middle/Proof_2 | [
"Double Negation Elimination implies Law of Excluded Middle",
"Double Negation Elimination",
"Double Negation"
] | [
"Double Negation/Double Negation Elimination",
"Law of Excluded Middle"
] | [
"Negation of Excluded Middle is False/Form 2"
] |
proofwiki-6651 | Double Negation/Double Negation Introduction/Sequent Form/Formulation 2 | :$\vdash p \implies \neg \neg p$ | {{BeginTableau|\vdash p \implies \neg \neg p}}
{{Assumption|1|p}}
{{DoubleNegIntro|2|1|\neg \neg p|1}}
{{Implication|3||p \implies \neg \neg p|1|2}}
{{EndTableau|qed}} | :$\vdash p \implies \neg \neg p$ | {{BeginTableau|\vdash p \implies \neg \neg p}}
{{Assumption|1|p}}
{{DoubleNegIntro|2|1|\neg \neg p|1}}
{{Implication|3||p \implies \neg \neg p|1|2}}
{{EndTableau|qed}} | Double Negation/Double Negation Introduction/Sequent Form/Formulation 2 | https://proofwiki.org/wiki/Double_Negation/Double_Negation_Introduction/Sequent_Form/Formulation_2 | https://proofwiki.org/wiki/Double_Negation/Double_Negation_Introduction/Sequent_Form/Formulation_2 | [
"Double Negation Introduction"
] | [] | [] |
proofwiki-6652 | Dual of Dual Ordering | Let $\struct {S, \preceq}$ be an ordered set.
Let $\struct {S, \succeq}$ be its dual.
Then the dual of $\struct {S, \succeq}$ is again $\struct {S, \preceq}$. | Denote with $\preceq'$ the dual of $\succeq$.
By definition of dual ordering, we thus have for all $a, b \in S$:
:$a \preceq b$ {{iff}} $b \succeq a$
:$b \succeq a$ {{iff}} $a \preceq' b$
Hence $a \preceq b$ {{iff}} $a \preceq' b$.
The result follows from Equality of Relations.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $\struct {S, \succeq}$ be its [[Definition:Dual Ordered Set|dual]].
Then the [[Definition:Dual Ordered Set|dual]] of $\struct {S, \succeq}$ is again $\struct {S, \preceq}$. | Denote with $\preceq'$ the [[Definition:Dual Ordering|dual]] of $\succeq$.
By definition of [[Definition:Dual Ordering|dual ordering]], we thus have for all $a, b \in S$:
:$a \preceq b$ {{iff}} $b \succeq a$
:$b \succeq a$ {{iff}} $a \preceq' b$
Hence $a \preceq b$ {{iff}} $a \preceq' b$.
The result follows from [... | Dual of Dual Ordering | https://proofwiki.org/wiki/Dual_of_Dual_Ordering | https://proofwiki.org/wiki/Dual_of_Dual_Ordering | [
"Dual Orderings"
] | [
"Definition:Ordered Set",
"Definition:Dual Ordering/Dual Ordered Set",
"Definition:Dual Ordering/Dual Ordered Set"
] | [
"Definition:Dual Ordering",
"Definition:Dual Ordering",
"Equality of Relations"
] |
proofwiki-6653 | Dual of Dual Statement (Order Theory) | Let $\Sigma$ be a statement about ordered sets.
Let $\Sigma^*$ be its dual statement.
Then $\Sigma$ is also the dual statement of $\Sigma^*$. | By definition, the dual statement $\Sigma^*$ is formed by replacing the ordering $\preceq$ with its dual $\succeq$.
By Dual of Dual Ordering, applying this operation twice results in the original sentence $\Sigma$ again.
{{qed}}
Category:Order Theory
sgenggbsvmbmw7wypp9g6hrj4lesfpe | Let $\Sigma$ be a [[Definition:Statement|statement]] about [[Definition:Ordered Set|ordered sets]].
Let $\Sigma^*$ be its [[Definition:Dual Statement (Order Theory)|dual statement]].
Then $\Sigma$ is also the [[Definition:Dual Statement (Order Theory)|dual statement]] of $\Sigma^*$. | By definition, the [[Definition:Dual Statement (Order Theory)|dual statement]] $\Sigma^*$ is formed by replacing the [[Definition:Ordering|ordering]] $\preceq$ with its [[Definition:Dual Ordering|dual]] $\succeq$.
By [[Dual of Dual Ordering]], applying this operation twice results in the original sentence $\Sigma$ aga... | Dual of Dual Statement (Order Theory) | https://proofwiki.org/wiki/Dual_of_Dual_Statement_(Order_Theory) | https://proofwiki.org/wiki/Dual_of_Dual_Statement_(Order_Theory) | [
"Order Theory"
] | [
"Definition:Statement",
"Definition:Ordered Set",
"Definition:Dual Statement (Order Theory)",
"Definition:Dual Statement (Order Theory)"
] | [
"Definition:Dual Statement (Order Theory)",
"Definition:Ordering",
"Definition:Dual Ordering",
"Dual of Dual Ordering",
"Category:Order Theory"
] |
proofwiki-6654 | Double Negation/Double Negation Elimination/Sequent Form/Formulation 2 | :$\vdash \neg \neg p \implies p$ | {{BeginTableau|\vdash \neg \neg p \implies p}}
{{Assumption|1|\neg \neg p}}
{{DoubleNegElimination|2|1|p|1}}
{{Implication|3||\neg \neg p \implies p|1|2}}
{{EndTableau|qed}} | :$\vdash \neg \neg p \implies p$ | {{BeginTableau|\vdash \neg \neg p \implies p}}
{{Assumption|1|\neg \neg p}}
{{DoubleNegElimination|2|1|p|1}}
{{Implication|3||\neg \neg p \implies p|1|2}}
{{EndTableau|qed}} | Double Negation/Double Negation Elimination/Sequent Form/Formulation 2 | https://proofwiki.org/wiki/Double_Negation/Double_Negation_Elimination/Sequent_Form/Formulation_2 | https://proofwiki.org/wiki/Double_Negation/Double_Negation_Elimination/Sequent_Form/Formulation_2 | [
"Double Negation Elimination"
] | [] | [] |
proofwiki-6655 | Double Negation/Double Negation Elimination | The '''rule of double negation elimination''' is a valid argument in certain types of logic dealing with negation $\neg$.
This includes classical propositional logic and predicate logic, and in particular natural deduction, but for example not intuitionistic propositional logic. | {{BeginTableau|\neg \neg p \vdash p}}
{{Premise|1|\neg \neg p}}
{{ExcludedMiddle|2|p \lor \neg p}}
{{Assumption|3|p}}
{{Assumption|4|\neg p}}
{{NonContradiction|5|1, 4|4|1}}
{{Explosion|6|1, 4|p|5}}
{{ProofByCases|7|1|p|2|3|3|4|6}}
{{EndTableau}}
{{Qed}} | The '''rule of [[Double Negation Elimination|double negation elimination]]''' is a [[Definition:Valid Argument|valid argument]] in certain types of [[Definition:Logic|logic]] dealing with [[Definition:Logical Not|negation]] $\neg$.
This includes [[Definition:Classical Propositional Logic|classical propositional logic]... | {{BeginTableau|\neg \neg p \vdash p}}
{{Premise|1|\neg \neg p}}
{{ExcludedMiddle|2|p \lor \neg p}}
{{Assumption|3|p}}
{{Assumption|4|\neg p}}
{{NonContradiction|5|1, 4|4|1}}
{{Explosion|6|1, 4|p|5}}
{{ProofByCases|7|1|p|2|3|3|4|6}}
{{EndTableau}}
{{Qed}} | Double Negation/Double Negation Elimination/Sequent Form/Formulation 1/Proof | https://proofwiki.org/wiki/Double_Negation/Double_Negation_Elimination | https://proofwiki.org/wiki/Double_Negation/Double_Negation_Elimination/Sequent_Form/Formulation_1/Proof | [
"Double Negation Elimination",
"Double Negation"
] | [
"Double Negation/Double Negation Elimination",
"Definition:Valid Argument",
"Definition:Logic",
"Definition:Logical Not",
"Definition:Classical Propositional Logic",
"Definition:Predicate Logic",
"Definition:Natural Deduction",
"Definition:Intuitionistic Propositional Logic"
] | [] |
proofwiki-6656 | Double Negation/Formulation 1 | :$p \dashv \vdash \neg \neg p$ | === Double Negation Introduction ===
{{:Double Negation/Double Negation Introduction/Sequent Form/Formulation 1/Proof}}
=== Double Negation Elimination ===
{{:Double Negation/Double Negation Elimination/Sequent Form/Formulation 1/Proof}} | :$p \dashv \vdash \neg \neg p$ | === [[Double Negation/Double Negation Introduction/Sequent Form/Formulation 1/Proof|Double Negation Introduction]] ===
{{:Double Negation/Double Negation Introduction/Sequent Form/Formulation 1/Proof}}
=== [[Double Negation/Double Negation Elimination/Sequent Form/Formulation 1/Proof|Double Negation Elimination]] ===
... | Double Negation/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Double_Negation/Formulation_1 | https://proofwiki.org/wiki/Double_Negation/Formulation_1/Proof_1 | [
"Double Negation"
] | [] | [
"Double Negation/Double Negation Introduction/Sequent Form/Formulation 1/Proof",
"Double Negation/Double Negation Elimination/Sequent Form/Formulation 1/Proof"
] |
proofwiki-6657 | Double Negation/Formulation 1 | :$p \dashv \vdash \neg \neg p$ | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, appropriate truth values match for both boolean interpretations.
:<nowiki>$\begin {array} {|c||ccc|} \hline
p & \neg & \neg & p \\
\hline
\F & \F & \T & \F \\
\T & \T & \F & \T \\
\hline
\end {array}$</nowiki>
Hence:
:$p \dashv \vdash... | :$p \dashv \vdash \neg \neg p$ | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen by inspection, appropriate [[Definition:Truth Value|truth values]] match for both [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin {array} {|c||ccc|} \hline
p & \neg & \neg & p \\
\hline
\F & \F & \T & \F \\
\T &... | Double Negation/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Double_Negation/Formulation_1 | https://proofwiki.org/wiki/Double_Negation/Formulation_1/Proof_by_Truth_Table | [
"Double Negation"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Boolean Interpretation"
] |
proofwiki-6658 | Double Negation/Formulation 2 | :$\vdash p \iff \neg \neg p$ | {{BeginTableau|\vdash p \iff \neg \neg p}}
{{TheoremIntro|1|p \implies \neg \neg p|Double Negation Introduction: Formulation 2}}
{{TheoremIntro|2|\neg \neg p \implies p|Double Negation Elimination: Formulation 2}}
{{BiconditionalIntro|3||p \iff \neg \neg p|1|2}}
{{EndTableau}}
{{qed}} | :$\vdash p \iff \neg \neg p$ | {{BeginTableau|\vdash p \iff \neg \neg p}}
{{TheoremIntro|1|p \implies \neg \neg p|[[Double Negation/Double Negation Introduction/Sequent Form/Formulation 2|Double Negation Introduction: Formulation 2]]}}
{{TheoremIntro|2|\neg \neg p \implies p|[[Double Negation/Double Negation Elimination/Sequent Form/Formulation 2|Do... | Double Negation/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Double_Negation/Formulation_2 | https://proofwiki.org/wiki/Double_Negation/Formulation_2/Proof_1 | [
"Double Negation"
] | [] | [
"Double Negation/Double Negation Introduction/Sequent Form/Formulation 2",
"Double Negation/Double Negation Elimination/Sequent Form/Formulation 2"
] |
proofwiki-6659 | Double Negation/Formulation 2 | :$\vdash p \iff \neg \neg p$ | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations.
$\quad \begin{array}{|c|c|ccc|} \hline
p & \iff & \neg & \neg & p \\
\hline
\F & \T & \F & \T & \F \\
\T & \T & \T & \F & \T \\
\hline
\end{array}$
{... | :$\vdash p \iff \neg \neg p$ | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] are [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]].
$\qua... | Double Negation/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Double_Negation/Formulation_2 | https://proofwiki.org/wiki/Double_Negation/Formulation_2/Proof_by_Truth_Table | [
"Double Negation"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-6660 | Rule of Transposition/Formulation 1 | A statement and its contrapositive have the same truth value:
:$p \implies q \dashv \vdash \neg q \implies \neg p$
Its abbreviation in a tableau proof is $\textrm {TP}$. | {{BeginTableau|p \implies q \vdash \neg q \implies \neg p}}
{{Premise|1|p \implies q}}
{{Assumption|2|\neg q}}
{{ModusTollens|3|1, 2|\neg p|1|2}}
{{Implication|4|1|\neg q \implies \neg p|2|3}}
{{EndTableau}}
{{Qed}} | A [[Definition:Statement|statement]] and its [[Definition:Contrapositive Statement|contrapositive]] have the same [[Definition:Truth Value|truth value]]:
:$p \implies q \dashv \vdash \neg q \implies \neg p$
Its abbreviation in a [[Definition:Tableau Proof (Formal Systems)|tableau proof]] is $\textrm {TP}$. | {{BeginTableau|p \implies q \vdash \neg q \implies \neg p}}
{{Premise|1|p \implies q}}
{{Assumption|2|\neg q}}
{{ModusTollens|3|1, 2|\neg p|1|2}}
{{Implication|4|1|\neg q \implies \neg p|2|3}}
{{EndTableau}}
{{Qed}} | Rule of Transposition/Formulation 1/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_1/Forward_Implication/Proof_1 | [
"Rule of Transposition"
] | [
"Definition:Statement",
"Definition:Contrapositive Statement",
"Definition:Truth Value",
"Definition:Tableau Proof (Natural Deduction)"
] | [] |
proofwiki-6661 | Rule of Transposition/Formulation 1 | A statement and its contrapositive have the same truth value:
:$p \implies q \dashv \vdash \neg q \implies \neg p$
Its abbreviation in a tableau proof is $\textrm {TP}$. | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth value under the main connectives match for all boolean interpretations.
:<nowiki>$\begin {array} {|ccc|ccccc|} \hline
p & \implies & q & \neg & q & \implies & \neg & p \\
\hline
\F & \T & \F & \T & \F & \T & \T & \F \\
\F & ... | A [[Definition:Statement|statement]] and its [[Definition:Contrapositive Statement|contrapositive]] have the same [[Definition:Truth Value|truth value]]:
:$p \implies q \dashv \vdash \neg q \implies \neg p$
Its abbreviation in a [[Definition:Tableau Proof (Formal Systems)|tableau proof]] is $\textrm {TP}$. | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen by inspection, the [[Definition:Truth Value|truth value]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin {array} {|c... | Rule of Transposition/Formulation 1/Forward Implication/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_1/Forward_Implication/Proof_by_Truth_Table | [
"Rule of Transposition"
] | [
"Definition:Statement",
"Definition:Contrapositive Statement",
"Definition:Truth Value",
"Definition:Tableau Proof (Natural Deduction)"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6662 | Rule of Transposition/Formulation 1 | A statement and its contrapositive have the same truth value:
:$p \implies q \dashv \vdash \neg q \implies \neg p$
Its abbreviation in a tableau proof is $\textrm {TP}$. | === Proof of Forward Implication ===
{{:Rule of Transposition/Formulation 1/Forward Implication/Proof 1}}
=== Proof of Reverse Implication ===
{{:Rule of Transposition/Formulation 1/Reverse Implication/Proof 1}} | A [[Definition:Statement|statement]] and its [[Definition:Contrapositive Statement|contrapositive]] have the same [[Definition:Truth Value|truth value]]:
:$p \implies q \dashv \vdash \neg q \implies \neg p$
Its abbreviation in a [[Definition:Tableau Proof (Formal Systems)|tableau proof]] is $\textrm {TP}$. | === [[Rule of Transposition/Formulation 1/Forward Implication/Proof 1|Proof of Forward Implication]] ===
{{:Rule of Transposition/Formulation 1/Forward Implication/Proof 1}}
=== [[Rule of Transposition/Formulation 1/Reverse Implication/Proof 1|Proof of Reverse Implication]] ===
{{:Rule of Transposition/Formulation 1/R... | Rule of Transposition/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_1/Proof_1 | [
"Rule of Transposition"
] | [
"Definition:Statement",
"Definition:Contrapositive Statement",
"Definition:Truth Value",
"Definition:Tableau Proof (Natural Deduction)"
] | [
"Rule of Transposition/Formulation 1/Forward Implication/Proof 1",
"Rule of Transposition/Formulation 1/Reverse Implication/Proof 1"
] |
proofwiki-6663 | Rule of Transposition/Formulation 1 | A statement and its contrapositive have the same truth value:
:$p \implies q \dashv \vdash \neg q \implies \neg p$
Its abbreviation in a tableau proof is $\textrm {TP}$. | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin {array} {|ccc||ccccc|} \hline
p & \implies & q & \neg & q & \implies & \neg & p \\
\hline
\F & \T & \F & \T & \F & \T & \T & \F \\
\F ... | A [[Definition:Statement|statement]] and its [[Definition:Contrapositive Statement|contrapositive]] have the same [[Definition:Truth Value|truth value]]:
:$p \implies q \dashv \vdash \neg q \implies \neg p$
Its abbreviation in a [[Definition:Tableau Proof (Formal Systems)|tableau proof]] is $\textrm {TP}$. | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin {array} {|... | Rule of Transposition/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_1/Proof_by_Truth_Table | [
"Rule of Transposition"
] | [
"Definition:Statement",
"Definition:Contrapositive Statement",
"Definition:Truth Value",
"Definition:Tableau Proof (Natural Deduction)"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6664 | Rule of Transposition/Formulation 1 | A statement and its contrapositive have the same truth value:
:$p \implies q \dashv \vdash \neg q \implies \neg p$
Its abbreviation in a tableau proof is $\textrm {TP}$. | {{BeginTableau|\neg q \implies \neg p \vdash p \implies q}}
{{Premise|1|\neg q \implies \neg p}}
{{Assumption|2|p}}
{{DoubleNegIntro|3|2|\neg \neg p|2}}
{{ModusTollens|4|1, 2|\neg \neg q|1|3}}
{{DoubleNegElimination|5|1, 2|q|4}}
{{Implication|6|1|p \implies q|2|5}}
{{EndTableau|Qed}} | A [[Definition:Statement|statement]] and its [[Definition:Contrapositive Statement|contrapositive]] have the same [[Definition:Truth Value|truth value]]:
:$p \implies q \dashv \vdash \neg q \implies \neg p$
Its abbreviation in a [[Definition:Tableau Proof (Formal Systems)|tableau proof]] is $\textrm {TP}$. | {{BeginTableau|\neg q \implies \neg p \vdash p \implies q}}
{{Premise|1|\neg q \implies \neg p}}
{{Assumption|2|p}}
{{DoubleNegIntro|3|2|\neg \neg p|2}}
{{ModusTollens|4|1, 2|\neg \neg q|1|3}}
{{DoubleNegElimination|5|1, 2|q|4}}
{{Implication|6|1|p \implies q|2|5}}
{{EndTableau|Qed}} | Rule of Transposition/Formulation 1/Reverse Implication/Proof 1 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_1/Reverse_Implication/Proof_1 | [
"Rule of Transposition"
] | [
"Definition:Statement",
"Definition:Contrapositive Statement",
"Definition:Truth Value",
"Definition:Tableau Proof (Natural Deduction)"
] | [] |
proofwiki-6665 | Rule of Transposition/Formulation 1 | A statement and its contrapositive have the same truth value:
:$p \implies q \dashv \vdash \neg q \implies \neg p$
Its abbreviation in a tableau proof is $\textrm {TP}$. | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth value under the main connectives match for all boolean interpretations.
:<nowiki>$\begin {array} {|ccccc|ccc|} \hline
\neg & q & \implies & \neg & p & p & \implies & q \\
\hline
\T & \F & \T & \T & \F & \F & \T & \F \\
\F & ... | A [[Definition:Statement|statement]] and its [[Definition:Contrapositive Statement|contrapositive]] have the same [[Definition:Truth Value|truth value]]:
:$p \implies q \dashv \vdash \neg q \implies \neg p$
Its abbreviation in a [[Definition:Tableau Proof (Formal Systems)|tableau proof]] is $\textrm {TP}$. | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen by inspection, the [[Definition:Truth Value|truth value]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin {array} {|c... | Rule of Transposition/Formulation 1/Reverse Implication/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_1/Reverse_Implication/Proof_by_Truth_Table | [
"Rule of Transposition"
] | [
"Definition:Statement",
"Definition:Contrapositive Statement",
"Definition:Truth Value",
"Definition:Tableau Proof (Natural Deduction)"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6666 | Rule of Transposition/Formulation 1/Proof 1 | :$p \implies q \dashv \vdash \neg q \implies \neg p$ | === Proof of Forward Implication ===
{{:Rule of Transposition/Formulation 1/Forward Implication/Proof 1}}
=== Proof of Reverse Implication ===
{{:Rule of Transposition/Formulation 1/Reverse Implication/Proof 1}} | :$p \implies q \dashv \vdash \neg q \implies \neg p$ | === [[Rule of Transposition/Formulation 1/Forward Implication/Proof 1|Proof of Forward Implication]] ===
{{:Rule of Transposition/Formulation 1/Forward Implication/Proof 1}}
=== [[Rule of Transposition/Formulation 1/Reverse Implication/Proof 1|Proof of Reverse Implication]] ===
{{:Rule of Transposition/Formulation 1/R... | Rule of Transposition/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_1/Proof_1 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_1/Proof_1 | [
"Rule of Transposition"
] | [] | [
"Rule of Transposition/Formulation 1/Forward Implication/Proof 1",
"Rule of Transposition/Formulation 1/Reverse Implication/Proof 1"
] |
proofwiki-6667 | Positive Infinity is Maximal | Let $\struct {\overline \R, \le}$ be the extended real numbers with the usual ordering.
Then $+\infty$ is a maximal element of $\overline \R$. | By the definition of the usual ordering on the extended real numbers:
:${\le} = {\le_\R} \cup \set {\tuple {x, +\infty}: x \in \overline \R} \cup \set {\tuple {-\infty, x}: x \in \overline \R}$
Suppose $x \in \overline \R$ and $+\infty \le x$.
That is:
: $\tuple {+\infty, x} \in {\le}$
By the definition of union, $\tup... | Let $\struct {\overline \R, \le}$ be the [[Definition:Extended Real Number Line|extended real numbers]] with the [[Definition:Ordering on Extended Real Numbers|usual ordering]].
Then $+\infty$ is a [[Definition:Maximal Element|maximal element]] of $\overline \R$. | By the definition of the [[Definition:Ordering on Extended Real Numbers|usual ordering]] on the [[Definition:Extended Real Number Line|extended real numbers]]:
:${\le} = {\le_\R} \cup \set {\tuple {x, +\infty}: x \in \overline \R} \cup \set {\tuple {-\infty, x}: x \in \overline \R}$
Suppose $x \in \overline \R$ and $... | Positive Infinity is Maximal | https://proofwiki.org/wiki/Positive_Infinity_is_Maximal | https://proofwiki.org/wiki/Positive_Infinity_is_Maximal | [
"Order Theory",
"Extended Real Numbers"
] | [
"Definition:Extended Real Number Line",
"Definition:Ordering on Extended Real Numbers",
"Definition:Maximal/Element"
] | [
"Definition:Ordering on Extended Real Numbers",
"Definition:Extended Real Number Line",
"Definition:Set Union",
"Definition:Extended Real Number Line",
"Definition:Maximal/Element"
] |
proofwiki-6668 | Negative Infinity is Minimal | Let $\struct {\overline \R, \le}$ be the extended real numbers with the usual ordering.
Then $-\infty$ is a minimal element of $\overline \R$. | By the definition of the usual ordering on the extended real numbers:
:${\le} = {\le_\R} \cup \set {\tuple {x, +\infty}: x \in \overline \R} \cup \set {\tuple {-\infty, x}: x \in \overline \R}$
Suppose $x \in \overline \R$ and $x \le -\infty$.
That is:
:$\tuple {x, -\infty} \in {\le}$
By the definition of union, $\tupl... | Let $\struct {\overline \R, \le}$ be the [[Definition:Extended Real Number Line|extended real numbers]] with the [[Definition:Ordering on Extended Real Numbers|usual ordering]].
Then $-\infty$ is a [[Definition:Minimal Element|minimal element]] of $\overline \R$. | By the definition of the [[Definition:Ordering on Extended Real Numbers|usual ordering]] on the [[Definition:Extended Real Number Line|extended real numbers]]:
:${\le} = {\le_\R} \cup \set {\tuple {x, +\infty}: x \in \overline \R} \cup \set {\tuple {-\infty, x}: x \in \overline \R}$
Suppose $x \in \overline \R$ and $... | Negative Infinity is Minimal | https://proofwiki.org/wiki/Negative_Infinity_is_Minimal | https://proofwiki.org/wiki/Negative_Infinity_is_Minimal | [
"Order Theory",
"Extended Real Numbers"
] | [
"Definition:Extended Real Number Line",
"Definition:Ordering on Extended Real Numbers",
"Definition:Minimal/Element"
] | [
"Definition:Ordering on Extended Real Numbers",
"Definition:Extended Real Number Line",
"Definition:Set Union",
"Definition:Extended Real Number Line",
"Definition:Minimal/Element"
] |
proofwiki-6669 | Inversion Mapping Reverses Ordering in Ordered Group/Corollary | {{begin-eqn}}
{{eqn | q = \forall x \in G
| l = x \preccurlyeq e
| o = \iff
| r = e \preccurlyeq x^{-1}
}}
{{eqn | l = e \preccurlyeq x
| o = \iff
| r = x^{-1} \preccurlyeq e
}}
{{eqn | l = x \prec e
| o = \iff
| r = e \prec x^{-1}
}}
{{eqn | l = e \prec x
| o = \iff
... | By Inversion Mapping Reverses Ordering in Ordered Group:
{{begin-eqn}}
{{eqn | q = \forall x \in G
| l = x \preccurlyeq e
| o = \iff
| r = e^{-1} \preccurlyeq x^{-1}
}}
{{eqn | l = e \preccurlyeq x
| o = \iff
| r = x^{-1} \preccurlyeq e^{-1}
}}
{{eqn | l = x \prec e
| o = \iff
... | {{begin-eqn}}
{{eqn | q = \forall x \in G
| l = x \preccurlyeq e
| o = \iff
| r = e \preccurlyeq x^{-1}
}}
{{eqn | l = e \preccurlyeq x
| o = \iff
| r = x^{-1} \preccurlyeq e
}}
{{eqn | l = x \prec e
| o = \iff
| r = e \prec x^{-1}
}}
{{eqn | l = e \prec x
| o = \iff
... | By [[Inversion Mapping Reverses Ordering in Ordered Group]]:
{{begin-eqn}}
{{eqn | q = \forall x \in G
| l = x \preccurlyeq e
| o = \iff
| r = e^{-1} \preccurlyeq x^{-1}
}}
{{eqn | l = e \preccurlyeq x
| o = \iff
| r = x^{-1} \preccurlyeq e^{-1}
}}
{{eqn | l = x \prec e
| o = \iff
... | Inversion Mapping Reverses Ordering in Ordered Group/Corollary/Proof 1 | https://proofwiki.org/wiki/Inversion_Mapping_Reverses_Ordering_in_Ordered_Group/Corollary | https://proofwiki.org/wiki/Inversion_Mapping_Reverses_Ordering_in_Ordered_Group/Corollary/Proof_1 | [
"Inversion Mapping Reverses Ordering in Ordered Group"
] | [] | [
"Inversion Mapping Reverses Ordering in Ordered Group"
] |
proofwiki-6670 | Inversion Mapping Reverses Ordering in Ordered Group/Corollary | {{begin-eqn}}
{{eqn | q = \forall x \in G
| l = x \preccurlyeq e
| o = \iff
| r = e \preccurlyeq x^{-1}
}}
{{eqn | l = e \preccurlyeq x
| o = \iff
| r = x^{-1} \preccurlyeq e
}}
{{eqn | l = x \prec e
| o = \iff
| r = e \prec x^{-1}
}}
{{eqn | l = e \prec x
| o = \iff
... | By the definition of an ordered group, $\preccurlyeq$ is a relation compatible with $\circ$.
Thus by Inverses of Elements Related by Compatible Relation: Corollary:
{{begin-eqn}}
{{eqn | q = \forall x \in G
| l = x \preccurlyeq e
| o = \iff
| r = e \preccurlyeq x^{-1}
}}
{{eqn | l = e \preccurlyeq x
... | {{begin-eqn}}
{{eqn | q = \forall x \in G
| l = x \preccurlyeq e
| o = \iff
| r = e \preccurlyeq x^{-1}
}}
{{eqn | l = e \preccurlyeq x
| o = \iff
| r = x^{-1} \preccurlyeq e
}}
{{eqn | l = x \prec e
| o = \iff
| r = e \prec x^{-1}
}}
{{eqn | l = e \prec x
| o = \iff
... | By the definition of an [[Definition:Ordered Group|ordered group]], $\preccurlyeq$ is a relation [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Thus by [[Inverses of Elements Related by Compatible Relation/Corollary|Inverses of Elements Related by Compatible Relation: Corollary]]:
{{begin-... | Inversion Mapping Reverses Ordering in Ordered Group/Corollary/Proof 2 | https://proofwiki.org/wiki/Inversion_Mapping_Reverses_Ordering_in_Ordered_Group/Corollary | https://proofwiki.org/wiki/Inversion_Mapping_Reverses_Ordering_in_Ordered_Group/Corollary/Proof_2 | [
"Inversion Mapping Reverses Ordering in Ordered Group"
] | [] | [
"Definition:Ordered Group",
"Definition:Relation Compatible with Operation",
"Inverses of Elements Related by Compatible Relation/Corollary",
"Reflexive Reduction of Relation Compatible with Group Operation is Compatible",
"Inverses of Elements Related by Compatible Relation/Corollary"
] |
proofwiki-6671 | Inversion Mapping Reverses Ordering in Ordered Group | {{begin-eqn}}
{{eqn | q = \forall x, y \in G
| l = x \preccurlyeq y
| o = \iff
| r = e \prec y^{-1} \preccurlyeq x^{-1}
}}
{{eqn | q = \forall x, y \in S
| l = x \prec y
| o = \iff
| r = y^{-1} \prec x^{-1}
}}
{{end-eqn}} | By the definition of an ordered group, $\preceq$ is a relation compatible with $\circ$.
Thus by Inverses of Elements Related by Compatible Relation, we obtain the first result:
:$x \preccurlyeq y \iff y^{-1} \preccurlyeq x^{-1}$
By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $\prec$ i... | {{begin-eqn}}
{{eqn | q = \forall x, y \in G
| l = x \preccurlyeq y
| o = \iff
| r = e \prec y^{-1} \preccurlyeq x^{-1}
}}
{{eqn | q = \forall x, y \in S
| l = x \prec y
| o = \iff
| r = y^{-1} \prec x^{-1}
}}
{{end-eqn}} | By the definition of an [[Definition:Ordered Group|ordered group]], $\preceq$ is a relation [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Thus by [[Inverses of Elements Related by Compatible Relation]], we obtain the first result:
:$x \preccurlyeq y \iff y^{-1} \preccurlyeq x^{-1}$
By [... | Inversion Mapping Reverses Ordering in Ordered Group | https://proofwiki.org/wiki/Inversion_Mapping_Reverses_Ordering_in_Ordered_Group | https://proofwiki.org/wiki/Inversion_Mapping_Reverses_Ordering_in_Ordered_Group | [
"Ordered Groups",
"Inversion Mappings",
"Inversion Mapping Reverses Ordering in Ordered Group"
] | [] | [
"Definition:Ordered Group",
"Definition:Relation Compatible with Operation",
"Inverses of Elements Related by Compatible Relation",
"Reflexive Reduction of Relation Compatible with Group Operation is Compatible",
"Inverses of Elements Related by Compatible Relation"
] |
proofwiki-6672 | Inversion Mapping Reverses Ordering in Ordered Group | {{begin-eqn}}
{{eqn | q = \forall x, y \in G
| l = x \preccurlyeq y
| o = \iff
| r = e \prec y^{-1} \preccurlyeq x^{-1}
}}
{{eqn | q = \forall x, y \in S
| l = x \prec y
| o = \iff
| r = y^{-1} \prec x^{-1}
}}
{{end-eqn}} | By Inversion Mapping Reverses Ordering in Ordered Group:
{{begin-eqn}}
{{eqn | q = \forall x \in G
| l = x \preccurlyeq e
| o = \iff
| r = e^{-1} \preccurlyeq x^{-1}
}}
{{eqn | l = e \preccurlyeq x
| o = \iff
| r = x^{-1} \preccurlyeq e^{-1}
}}
{{eqn | l = x \prec e
| o = \iff
... | {{begin-eqn}}
{{eqn | q = \forall x, y \in G
| l = x \preccurlyeq y
| o = \iff
| r = e \prec y^{-1} \preccurlyeq x^{-1}
}}
{{eqn | q = \forall x, y \in S
| l = x \prec y
| o = \iff
| r = y^{-1} \prec x^{-1}
}}
{{end-eqn}} | By [[Inversion Mapping Reverses Ordering in Ordered Group]]:
{{begin-eqn}}
{{eqn | q = \forall x \in G
| l = x \preccurlyeq e
| o = \iff
| r = e^{-1} \preccurlyeq x^{-1}
}}
{{eqn | l = e \preccurlyeq x
| o = \iff
| r = x^{-1} \preccurlyeq e^{-1}
}}
{{eqn | l = x \prec e
| o = \iff
... | Inversion Mapping Reverses Ordering in Ordered Group/Corollary/Proof 1 | https://proofwiki.org/wiki/Inversion_Mapping_Reverses_Ordering_in_Ordered_Group | https://proofwiki.org/wiki/Inversion_Mapping_Reverses_Ordering_in_Ordered_Group/Corollary/Proof_1 | [
"Ordered Groups",
"Inversion Mappings",
"Inversion Mapping Reverses Ordering in Ordered Group"
] | [] | [
"Inversion Mapping Reverses Ordering in Ordered Group"
] |
proofwiki-6673 | Inversion Mapping Reverses Ordering in Ordered Group | {{begin-eqn}}
{{eqn | q = \forall x, y \in G
| l = x \preccurlyeq y
| o = \iff
| r = e \prec y^{-1} \preccurlyeq x^{-1}
}}
{{eqn | q = \forall x, y \in S
| l = x \prec y
| o = \iff
| r = y^{-1} \prec x^{-1}
}}
{{end-eqn}} | By the definition of an ordered group, $\preccurlyeq$ is a relation compatible with $\circ$.
Thus by Inverses of Elements Related by Compatible Relation: Corollary:
{{begin-eqn}}
{{eqn | q = \forall x \in G
| l = x \preccurlyeq e
| o = \iff
| r = e \preccurlyeq x^{-1}
}}
{{eqn | l = e \preccurlyeq x
... | {{begin-eqn}}
{{eqn | q = \forall x, y \in G
| l = x \preccurlyeq y
| o = \iff
| r = e \prec y^{-1} \preccurlyeq x^{-1}
}}
{{eqn | q = \forall x, y \in S
| l = x \prec y
| o = \iff
| r = y^{-1} \prec x^{-1}
}}
{{end-eqn}} | By the definition of an [[Definition:Ordered Group|ordered group]], $\preccurlyeq$ is a relation [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Thus by [[Inverses of Elements Related by Compatible Relation/Corollary|Inverses of Elements Related by Compatible Relation: Corollary]]:
{{begin-... | Inversion Mapping Reverses Ordering in Ordered Group/Corollary/Proof 2 | https://proofwiki.org/wiki/Inversion_Mapping_Reverses_Ordering_in_Ordered_Group | https://proofwiki.org/wiki/Inversion_Mapping_Reverses_Ordering_in_Ordered_Group/Corollary/Proof_2 | [
"Ordered Groups",
"Inversion Mappings",
"Inversion Mapping Reverses Ordering in Ordered Group"
] | [] | [
"Definition:Ordered Group",
"Definition:Relation Compatible with Operation",
"Inverses of Elements Related by Compatible Relation/Corollary",
"Reflexive Reduction of Relation Compatible with Group Operation is Compatible",
"Inverses of Elements Related by Compatible Relation/Corollary"
] |
proofwiki-6674 | Operating on Transitive Relationships Compatible with Operation | Let $\struct {S, \circ}$ be a magma.
Let $\RR$ be a transitive relation on $S$ which is compatible with $\circ$.
Let $\RR^=$ be the reflexive closure of $\RR$.
Let $x, y, z, w \in S$.
Then the following implications hold:
$(1): \quad$ If $x \mathrel \RR y$ and $z \mathrel \RR w$, then $x \circ z \mathrel \RR y \circ w$... | We will prove $(1)$ and $(2)$. $(3)$ can be proven using precisely the same argument as $(2)$ and $(4)$ follows from $(1)$.
Suppose that $x \mathrel \RR y$ and $z \mathrel \RR w$.
By the definition of compatibility:
:$x \mathrel \RR y \implies x \circ z \mathrel \RR y \circ z$
:$z \mathrel \RR w \implies y \circ z \mat... | Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
Let $\RR$ be a [[Definition:Transitive Relation|transitive relation]] on $S$ which is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Let $\RR^=$ be the [[Definition:Reflexive Closure|reflexive closure]] of $\RR$.
Let $x, y, z, w \i... | We will prove $(1)$ and $(2)$. $(3)$ can be proven using precisely the same argument as $(2)$ and $(4)$ follows from $(1)$.
Suppose that $x \mathrel \RR y$ and $z \mathrel \RR w$.
By the definition of [[Definition:Relation Compatible with Operation|compatibility]]:
:$x \mathrel \RR y \implies x \circ z \mathrel \RR ... | Operating on Transitive Relationships Compatible with Operation | https://proofwiki.org/wiki/Operating_on_Transitive_Relationships_Compatible_with_Operation | https://proofwiki.org/wiki/Operating_on_Transitive_Relationships_Compatible_with_Operation | [
"Compatible Relations",
"Transitive Relations"
] | [
"Definition:Magma",
"Definition:Transitive Relation",
"Definition:Relation Compatible with Operation",
"Definition:Reflexive Closure"
] | [
"Definition:Relation Compatible with Operation",
"Definition:Transitive Relation",
"Reflexive Closure of Relation Compatible with Operation is Compatible",
"Definition:Relation Compatible with Operation",
"Definition:Relation Compatible with Operation",
"Extended Transitivity",
"Reflexive Closure of Rel... |
proofwiki-6675 | Operating on Ordered Group Inequalities | If $x \prec y$ and $z \prec w$, then $x \circ z \prec y \circ w$.
If $x \prec y$ and $z \preccurlyeq w$, then $x \circ z \prec y \circ w$.
If $x \preccurlyeq y$ and $z \prec w$, then $x \circ z \prec y \circ w$.
If $x \preccurlyeq y$ and $z \preccurlyeq w$, then $x \circ z \preccurlyeq y \circ w$. | Because:
:$\struct {G, \circ, \preccurlyeq}$ is a group
:$\preccurlyeq$ is compatible with $\circ$
it follows from Reflexive Reduction of Relation Compatible with Group Operation is Compatible that $\prec$ is compatible with $\circ$ .
By the definition of an ordering, $\preccurlyeq$ is transitive and antisymmetric.
The... | If $x \prec y$ and $z \prec w$, then $x \circ z \prec y \circ w$.
If $x \prec y$ and $z \preccurlyeq w$, then $x \circ z \prec y \circ w$.
If $x \preccurlyeq y$ and $z \prec w$, then $x \circ z \prec y \circ w$.
If $x \preccurlyeq y$ and $z \preccurlyeq w$, then $x \circ z \preccurlyeq y \circ w$. | Because:
:$\struct {G, \circ, \preccurlyeq}$ is a [[Definition:Ordered Group|group]]
:$\preccurlyeq$ is compatible with $\circ$
it follows from [[Reflexive Reduction of Relation Compatible with Group Operation is Compatible]] that $\prec$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$ .
... | Operating on Ordered Group Inequalities | https://proofwiki.org/wiki/Operating_on_Ordered_Group_Inequalities | https://proofwiki.org/wiki/Operating_on_Ordered_Group_Inequalities | [
"Ordered Groups"
] | [] | [
"Definition:Ordered Group",
"Reflexive Reduction of Relation Compatible with Group Operation is Compatible",
"Definition:Relation Compatible with Operation",
"Definition:Ordering",
"Definition:Transitive Relation",
"Definition:Antisymmetric Relation",
"Reflexive Reduction of Transitive Antisymmetric Rel... |
proofwiki-6676 | Rule of Transposition/Formulation 2 | :$\vdash \paren {p \implies q} \iff \paren {\neg q \implies \neg p}$ | {{BeginTableau |\vdash \paren {p \implies q} \implies \paren {\neg q \implies \neg p} }}
{{Assumption |1|p \implies q}}
{{Assumption |2|\neg q}}
{{ModusTollens |3|1, 2|\neg p|1|2}}
{{Implication |4|1|\neg q \implies \neg p|2|3}}
{{Implication |5||\paren {p \implies q} \implies \paren {\neg q \implies \neg p}|1|4}... | :$\vdash \paren {p \implies q} \iff \paren {\neg q \implies \neg p}$ | {{BeginTableau |\vdash \paren {p \implies q} \implies \paren {\neg q \implies \neg p} }}
{{Assumption |1|p \implies q}}
{{Assumption |2|\neg q}}
{{ModusTollens |3|1, 2|\neg p|1|2}}
{{Implication |4|1|\neg q \implies \neg p|2|3}}
{{Implication |5||\paren {p \implies q} \implies \paren {\neg q \implies \neg p}|1|4}... | Rule of Transposition/Formulation 2/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_2/Forward_Implication/Proof_1 | [
"Rule of Transposition"
] | [] | [] |
proofwiki-6677 | Rule of Transposition/Formulation 2 | :$\vdash \paren {p \implies q} \iff \paren {\neg q \implies \neg p}$ | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth value under the main connective is true for all boolean interpretations.
:<nowiki>$\begin {array} {|ccc|c|ccccc|} \hline
(p & \implies & q) & \implies & (\neg & q & \implies & \neg & p) \\
\hline
\F & \T & \F & \T & \T & \F ... | :$\vdash \paren {p \implies q} \iff \paren {\neg q \implies \neg p}$ | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen by inspection, the [[Definition:Truth Value|truth value]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowik... | Rule of Transposition/Formulation 2/Forward Implication/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_2/Forward_Implication/Proof_by_Truth_Table | [
"Rule of Transposition"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-6678 | Rule of Transposition/Formulation 2 | :$\vdash \paren {p \implies q} \iff \paren {\neg q \implies \neg p}$ | === Proof of Forward Implication ===
{{:Rule of Transposition/Formulation 2/Forward Implication/Proof 1}}
=== Proof of Reverse Implication ===
{{:Rule of Transposition/Formulation 2/Reverse Implication/Proof 1}} | :$\vdash \paren {p \implies q} \iff \paren {\neg q \implies \neg p}$ | === [[Rule of Transposition/Formulation 2/Forward Implication/Proof 1|Proof of Forward Implication]] ===
{{:Rule of Transposition/Formulation 2/Forward Implication/Proof 1}}
=== [[Rule of Transposition/Formulation 2/Reverse Implication/Proof 1|Proof of Reverse Implication]] ===
{{:Rule of Transposition/Formulation 2/R... | Rule of Transposition/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_2/Proof_1 | [
"Rule of Transposition"
] | [] | [
"Rule of Transposition/Formulation 2/Forward Implication/Proof 1",
"Rule of Transposition/Formulation 2/Reverse Implication/Proof 1"
] |
proofwiki-6679 | Rule of Transposition/Formulation 2 | :$\vdash \paren {p \implies q} \iff \paren {\neg q \implies \neg p}$ | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin{array}{|ccc|c|ccccc|} \hline
(p & \implies & q) & \iff & (\neg & q & \implies & \neg & p) \\
\hline
\F & \T & \F & \T & \T & \F & \T &... | :$\vdash \paren {p \implies q} \iff \paren {\neg q \implies \neg p}$ | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|cc... | Rule of Transposition/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_2/Proof_by_Truth_Table | [
"Rule of Transposition"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6680 | Rule of Transposition/Formulation 2 | :$\vdash \paren {p \implies q} \iff \paren {\neg q \implies \neg p}$ | {{BeginTableau|\vdash \paren {\neg q \implies \neg p} \implies \paren {p \implies q} }}
{{Assumption|1|\neg q \implies \neg p}}
{{Assumption|2|p}}
{{DoubleNegIntro|3|2|\neg \neg p|2}}
{{ModusTollens|4|1, 2|\neg \neg q|1|3}}
{{DoubleNegElimination|5|1, 2|q|4}}
{{Implication|6|1|p \implies q|2|5}}
{{Implication|7||\paren... | :$\vdash \paren {p \implies q} \iff \paren {\neg q \implies \neg p}$ | {{BeginTableau|\vdash \paren {\neg q \implies \neg p} \implies \paren {p \implies q} }}
{{Assumption|1|\neg q \implies \neg p}}
{{Assumption|2|p}}
{{DoubleNegIntro|3|2|\neg \neg p|2}}
{{ModusTollens|4|1, 2|\neg \neg q|1|3}}
{{DoubleNegElimination|5|1, 2|q|4}}
{{Implication|6|1|p \implies q|2|5}}
{{Implication|7||\paren... | Rule of Transposition/Formulation 2/Reverse Implication/Proof 1 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_2/Reverse_Implication/Proof_1 | [
"Rule of Transposition"
] | [] | [] |
proofwiki-6681 | Rule of Transposition/Formulation 2 | :$\vdash \paren {p \implies q} \iff \paren {\neg q \implies \neg p}$ | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth value under the main connective is is true for all boolean interpretations.
:<nowiki>$\begin {array} {|ccccc|c|ccc|} \hline
(\neg & q & \implies & \neg & p) & \implies & (p & \implies & q) \\
\hline
\T & \F & \T & \T & \F & ... | :$\vdash \paren {p \implies q} \iff \paren {\neg q \implies \neg p}$ | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen by inspection, the [[Definition:Truth Value|truth value]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<no... | Rule of Transposition/Formulation 2/Reverse Implication/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Transposition/Formulation_2/Reverse_Implication/Proof_by_Truth_Table | [
"Rule of Transposition"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-6682 | Complex Modulus is Norm | The complex modulus is a norm on the set of complex numbers $\C$. | We prove the norm axioms. | The [[Definition:Complex Modulus|complex modulus]] is a [[Definition:Norm on Division Ring|norm]] on the [[Definition:Complex Number|set of complex numbers]] $\C$. | We prove the [[Definition:Norm on Division Ring|norm axioms]]. | Complex Modulus is Norm | https://proofwiki.org/wiki/Complex_Modulus_is_Norm | https://proofwiki.org/wiki/Complex_Modulus_is_Norm | [
"Examples of Norms",
"Complex Modulus"
] | [
"Definition:Complex Modulus",
"Definition:Norm/Division Ring",
"Definition:Complex Number"
] | [
"Definition:Norm/Division Ring"
] |
proofwiki-6683 | Set Difference of Relations Compatible with Group Operation is Compatible | Let $\struct {G, \circ}$ be a group.
Let $\RR, \QQ$ be relations on $G$ which are compatible with $\circ$.
Then the difference $\RR \setminus \QQ$ is compatible with $\circ$. | By Complement of Relation Compatible with Group is Compatible, $\relcomp {G \times G} \QQ$ is compatible with $\circ$.
Thus by Intersection of Relations Compatible with Operation is Compatible, $\RR \cap \relcomp {G \times G} \QQ$ is compatible with $\circ$.
But:
:$\RR \cap \relcomp {G \times G} \QQ = \RR \setminus \QQ... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\RR, \QQ$ be [[Definition:Endorelation|relations]] on $G$ which are [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Then the [[Definition:Set Difference|difference]] $\RR \setminus \QQ$ is [[Definition:Relation Compatible with ... | By [[Complement of Relation Compatible with Group is Compatible]], $\relcomp {G \times G} \QQ$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Thus by [[Intersection of Relations Compatible with Operation is Compatible]], $\RR \cap \relcomp {G \times G} \QQ$ is [[Definition:Relation Compa... | Set Difference of Relations Compatible with Group Operation is Compatible | https://proofwiki.org/wiki/Set_Difference_of_Relations_Compatible_with_Group_Operation_is_Compatible | https://proofwiki.org/wiki/Set_Difference_of_Relations_Compatible_with_Group_Operation_is_Compatible | [
"Compatible Relations"
] | [
"Definition:Group",
"Definition:Endorelation",
"Definition:Relation Compatible with Operation",
"Definition:Set Difference",
"Definition:Relation Compatible with Operation"
] | [
"Complement of Relation Compatible with Group is Compatible",
"Definition:Relation Compatible with Operation",
"Intersection of Relations Compatible with Operation is Compatible",
"Definition:Relation Compatible with Operation",
"Category:Compatible Relations"
] |
proofwiki-6684 | Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements | Let $\struct {S, \circ}$ be a semigroup.
Let $\RR$ be a transitive relation on $S$ which is compatible with $\circ$.
Let $x, y \in S$ such that $x \mathrel \RR y$.
Let $n \in \N_{>0}$ be a strictly positive integer.
Then:
:$x^n \mathrel \RR y^n$
where $x^n$ is the $n$th power of $x$. | We proceed by mathematical induction.
By definition of power:
:$x^1 = x$
:$y^1 = y$
Hence, by assumption:
:$x^1 \mathrel \RR y^1$
Suppose now that for $n \ge 1$:
:$x^n \mathrel \RR y^n$
Recall the assumption that $x \mathrel \RR y$.
Applying Operating on Transitive Relationships Compatible with Operation to these relat... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $\RR$ be a [[Definition:Transitive Relation|transitive relation]] on $S$ which is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Let $x, y \in S$ such that $x \mathrel \RR y$.
Let $n \in \N_{>0}$ be a [[Definition:Stric... | We proceed by [[Principle of Mathematical Induction|mathematical induction]].
By definition of [[Definition:Power of Element of Semigroup|power]]:
:$x^1 = x$
:$y^1 = y$
Hence, by assumption:
:$x^1 \mathrel \RR y^1$
Suppose now that for $n \ge 1$:
:$x^n \mathrel \RR y^n$
Recall the assumption that $x \mathrel \RR... | Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements | https://proofwiki.org/wiki/Transitive_Relation_Compatible_with_Semigroup_Operation_Relates_Powers_of_Related_Elements | https://proofwiki.org/wiki/Transitive_Relation_Compatible_with_Semigroup_Operation_Relates_Powers_of_Related_Elements | [
"Compatible Relations",
"Semigroups"
] | [
"Definition:Semigroup",
"Definition:Transitive Relation",
"Definition:Relation Compatible with Operation",
"Definition:Strictly Positive/Integer",
"Definition:Power of Element/Semigroup"
] | [
"Principle of Mathematical Induction",
"Definition:Power of Element/Semigroup",
"Operating on Transitive Relationships Compatible with Operation",
"Definition:Power of Element/Semigroup",
"Principle of Mathematical Induction",
"Category:Compatible Relations",
"Category:Semigroups"
] |
proofwiki-6685 | Reflexive Closure of Transitive Antisymmetric Relation is Ordering | Let $S$ be a set.
Let $\prec$ be a transitive, antisymmetric relation on $S$.
Let $\preceq$ be the reflexive closure of $\prec$.
Then $\preceq$ is an ordering on $S$. | === Reflexive ===
Follows from Reflexive Closure is Reflexive.
{{qed|lemma}} | Let $S$ be a set.
Let $\prec$ be a [[Definition:Transitive Relation|transitive]], [[Definition:Antisymmetric Relation|antisymmetric]] [[Definition:Endorelation|relation]] on $S$.
Let $\preceq$ be the [[Definition:Reflexive Closure|reflexive closure]] of $\prec$.
Then $\preceq$ is an [[Definition:Ordering|ordering]]... | === Reflexive ===
Follows from [[Reflexive Closure is Reflexive]].
{{qed|lemma}} | Reflexive Closure of Transitive Antisymmetric Relation is Ordering | https://proofwiki.org/wiki/Reflexive_Closure_of_Transitive_Antisymmetric_Relation_is_Ordering | https://proofwiki.org/wiki/Reflexive_Closure_of_Transitive_Antisymmetric_Relation_is_Ordering | [
"Order Theory",
"Reflexive Closures",
"Transitive Relations",
"Antisymmetric Relations"
] | [
"Definition:Transitive Relation",
"Definition:Antisymmetric Relation",
"Definition:Endorelation",
"Definition:Reflexive Closure",
"Definition:Ordering"
] | [
"Reflexive Closure is Reflexive"
] |
proofwiki-6686 | Reflexive Closure of Transitive Relation is Transitive | Let $S$ be a set.
Let $\RR$ be a transitive relation.
Let $\RR^=$ be the reflexive closure of $\RR$.
Then $\RR^=$ is also transitive. | Let $a, b, c \in S$.
Suppose that $a \mathrel {\RR^=} b$ and $b \mathrel {\RR^=} c$.
If $a = b$, then since $b \mathrel {\RR^=} c$, also $a \mathrel {\RR^=} c$.
If $b = c$, then since $a \mathrel {\RR^=} b$, also $a \mathrel {\RR^=} c$.
The only case that remains is that $a \ne b$ and $b \ne c$.
Then by the definition ... | Let $S$ be a [[Definition:Set|set]].
Let $\RR$ be a [[Definition:Transitive Relation|transitive relation]].
Let $\RR^=$ be the [[Definition:Reflexive Closure|reflexive closure]] of $\RR$.
Then $\RR^=$ is also [[Definition:Transitive Relation|transitive]]. | Let $a, b, c \in S$.
Suppose that $a \mathrel {\RR^=} b$ and $b \mathrel {\RR^=} c$.
If $a = b$, then since $b \mathrel {\RR^=} c$, also $a \mathrel {\RR^=} c$.
If $b = c$, then since $a \mathrel {\RR^=} b$, also $a \mathrel {\RR^=} c$.
The only case that remains is that $a \ne b$ and $b \ne c$.
Then by the [[Def... | Reflexive Closure of Transitive Relation is Transitive | https://proofwiki.org/wiki/Reflexive_Closure_of_Transitive_Relation_is_Transitive | https://proofwiki.org/wiki/Reflexive_Closure_of_Transitive_Relation_is_Transitive | [
"Reflexive Closures"
] | [
"Definition:Set",
"Definition:Transitive Relation",
"Definition:Reflexive Closure",
"Definition:Transitive Relation"
] | [
"Definition:Reflexive Closure",
"Definition:Transitive Relation",
"Definition:Transitive Relation"
] |
proofwiki-6687 | Reflexive Closure of Antisymmetric Relation is Antisymmetric | Let $S$ be a set.
Let $RR$ be an antisymmetric relation on $S$.
Let $\RR^=$ be the reflexive closure of $\RR$.
Then $\RR^=$ is also antisymmetric. | Let $a, b \in S$.
Suppose that $a \mathrel {\RR^=} b$ and $b \mathrel {\RR^=} a$.
By definition of $\RR^=$, this means:
:$a \mathrel \RR b$ or $a = b$
:$b \mathrel \RR a$ or $b = a$
If $a = b$ or $b = a$ we are done, by definition of an antisymmetric relation.
So suppose $a \mathrel \RR b$ and $b \mathrel \RR a$.
Becau... | Let $S$ be a [[Definition:Set|set]].
Let $RR$ be an [[Definition:Antisymmetric Relation|antisymmetric relation]] on $S$.
Let $\RR^=$ be the [[Definition:Reflexive Closure|reflexive closure]] of $\RR$.
Then $\RR^=$ is also [[Definition:Antisymmetric Relation|antisymmetric]]. | Let $a, b \in S$.
Suppose that $a \mathrel {\RR^=} b$ and $b \mathrel {\RR^=} a$.
By [[Definition:Reflexive Closure|definition of $\RR^=$]], this means:
:$a \mathrel \RR b$ or $a = b$
:$b \mathrel \RR a$ or $b = a$
If $a = b$ or $b = a$ we are done, by definition of an [[Definition:Antisymmetric Relation|antisymmet... | Reflexive Closure of Antisymmetric Relation is Antisymmetric | https://proofwiki.org/wiki/Reflexive_Closure_of_Antisymmetric_Relation_is_Antisymmetric | https://proofwiki.org/wiki/Reflexive_Closure_of_Antisymmetric_Relation_is_Antisymmetric | [
"Reflexive Closures"
] | [
"Definition:Set",
"Definition:Antisymmetric Relation",
"Definition:Reflexive Closure",
"Definition:Antisymmetric Relation"
] | [
"Definition:Reflexive Closure",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation",
"Category:Reflexive Closures"
] |
proofwiki-6688 | Positive Infinity is Greatest Element | Let $\left({\overline \R, \le}\right)$ be the extended real numbers with their usual ordering.
Then $+\infty$ is the greatest element of $\overline \R$. | We have, by definition of the usual ordering on $\overline \R$:
:$\forall x \in \overline \R: x \le +\infty$
That is, $+\infty$ is the greatest element of $\overline \R$.
{{qed}} | Let $\left({\overline \R, \le}\right)$ be the [[Definition:Extended Real Number Line|extended real numbers]] with their [[Definition:Ordering on Extended Real Numbers|usual ordering]].
Then $+\infty$ is the [[Definition:Greatest Element|greatest element]] of $\overline \R$. | We have, by definition of the [[Definition:Ordering on Extended Real Numbers|usual ordering]] on $\overline \R$:
:$\forall x \in \overline \R: x \le +\infty$
That is, $+\infty$ is the [[Definition:Greatest Element|greatest element]] of $\overline \R$.
{{qed}} | Positive Infinity is Greatest Element | https://proofwiki.org/wiki/Positive_Infinity_is_Greatest_Element | https://proofwiki.org/wiki/Positive_Infinity_is_Greatest_Element | [
"Order Theory",
"Extended Real Numbers"
] | [
"Definition:Extended Real Number Line",
"Definition:Ordering on Extended Real Numbers",
"Definition:Greatest Element"
] | [
"Definition:Ordering on Extended Real Numbers",
"Definition:Greatest Element"
] |
proofwiki-6689 | Negative Infinity is Smallest Element | Let $\left({\overline \R, \le}\right)$ be the extended real numbers with their usual ordering.
Then $-\infty$ is the smallest element of $\overline \R$. | We have, by definition of the usual ordering on $\overline \R$:
:$\forall x \in \overline \R: -\infty \le x$
That is, $-\infty$ is the smallest element of $\overline \R$.
{{qed}} | Let $\left({\overline \R, \le}\right)$ be the [[Definition:Extended Real Number Line|extended real numbers]] with their [[Definition:Ordering on Extended Real Numbers|usual ordering]].
Then $-\infty$ is the [[Definition:Smallest Element|smallest element]] of $\overline \R$. | We have, by definition of the [[Definition:Ordering on Extended Real Numbers|usual ordering]] on $\overline \R$:
:$\forall x \in \overline \R: -\infty \le x$
That is, $-\infty$ is the [[Definition:Smallest Element|smallest element]] of $\overline \R$.
{{qed}} | Negative Infinity is Smallest Element | https://proofwiki.org/wiki/Negative_Infinity_is_Smallest_Element | https://proofwiki.org/wiki/Negative_Infinity_is_Smallest_Element | [
"Order Theory",
"Extended Real Numbers"
] | [
"Definition:Extended Real Number Line",
"Definition:Ordering on Extended Real Numbers",
"Definition:Smallest Element"
] | [
"Definition:Ordering on Extended Real Numbers",
"Definition:Smallest Element"
] |
proofwiki-6690 | Relation Compatible with Group Operation is Strongly Compatible/Corollary | :$\forall x, y \in G:$
::$(1): \quad x \mathrel \RR y \iff e \mathrel \RR y \circ x^{-1}$
::$(2): \quad x \mathrel \RR y \iff e \mathrel \RR x^{-1} \circ y$
::$(3): \quad x \mathrel \RR y \iff x \circ y^{-1} \mathrel \RR e$
::$(4): \quad x \mathrel \RR y \iff y^{-1} \circ x \mathrel \RR e$ | Applying Relation Compatible with Group Operation is Strongly Compatible to $x$, $y$, and $x^{-1}$ we obtain:
:$x \mathrel \RR y \iff x \circ x^{-1} \mathrel \RR y \circ x^{-1}$
:$x \mathrel \RR y \iff x^{-1} \circ x \mathrel \RR x^{-1} \circ y$
Applying Relation Compatible with Group Operation is Strongly Compatible t... | :$\forall x, y \in G:$
::$(1): \quad x \mathrel \RR y \iff e \mathrel \RR y \circ x^{-1}$
::$(2): \quad x \mathrel \RR y \iff e \mathrel \RR x^{-1} \circ y$
::$(3): \quad x \mathrel \RR y \iff x \circ y^{-1} \mathrel \RR e$
::$(4): \quad x \mathrel \RR y \iff y^{-1} \circ x \mathrel \RR e$ | Applying [[Relation Compatible with Group Operation is Strongly Compatible]] to $x$, $y$, and $x^{-1}$ we obtain:
:$x \mathrel \RR y \iff x \circ x^{-1} \mathrel \RR y \circ x^{-1}$
:$x \mathrel \RR y \iff x^{-1} \circ x \mathrel \RR x^{-1} \circ y$
Applying [[Relation Compatible with Group Operation is Strongly Comp... | Relation Compatible with Group Operation is Strongly Compatible/Corollary | https://proofwiki.org/wiki/Relation_Compatible_with_Group_Operation_is_Strongly_Compatible/Corollary | https://proofwiki.org/wiki/Relation_Compatible_with_Group_Operation_is_Strongly_Compatible/Corollary | [
"Relations Compatible with Group Operation"
] | [] | [
"Relation Compatible with Group Operation is Strongly Compatible",
"Relation Compatible with Group Operation is Strongly Compatible",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Category:Relations Compatible with Group Operation"
] |
proofwiki-6691 | Dual Pairs (Order Theory) | Let $\left({S, \preceq}\right)$ be an ordered set.
Let $a, b \in S$, and let $T \subseteq S$.
Then the following phrases about, and concepts pertaining to $\left({S, \preceq}\right)$ are dual to one another:
:{| style="text-align:center"
| $b \preceq a$
| width = "20px" |
| $a \preceq b$
|-
| $a$ succeeds $b$
|
| $a$ p... | Let $\succeq$ be the dual ordering of $\preceq$.
By definition of dual statement:
:$b \preceq a$
is dual to:
:$b \succeq a$
and by definition of dual ordering, this is equivalent to:
:$a \preceq b$
By virtue of Dual of Dual Statement (Order Theory), the converse follows.
The other claims are proved on the following pag... | Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
Let $a, b \in S$, and let $T \subseteq S$.
Then the following phrases about, and concepts pertaining to $\left({S, \preceq}\right)$ are [[Definition:Dual Statement (Order Theory)|dual]] to one another:
:{| style="text-align:center"
| $b \... | Let $\succeq$ be the [[Definition:Dual Ordering|dual ordering]] of $\preceq$.
By definition of [[Definition:Dual Statement (Order Theory)|dual statement]]:
:$b \preceq a$
is dual to:
:$b \succeq a$
and by definition of [[Definition:Dual Ordering|dual ordering]], this is [[Definition:Logical Equivalence|equivalent]... | Dual Pairs (Order Theory) | https://proofwiki.org/wiki/Dual_Pairs_(Order_Theory) | https://proofwiki.org/wiki/Dual_Pairs_(Order_Theory) | [
"Order Theory",
"Dual Pairs (Order Theory)"
] | [
"Definition:Ordered Set",
"Definition:Dual Statement (Order Theory)",
"Definition:Succeed",
"Definition:Precede",
"Definition:Strictly Succeed",
"Definition:Strictly Precede",
"Definition:Upper Bound of Set",
"Definition:Lower Bound of Set",
"Definition:Supremum of Set",
"Definition:Infimum of Set... | [
"Definition:Dual Ordering",
"Definition:Dual Statement (Order Theory)",
"Definition:Dual Ordering",
"Definition:Logical Equivalence",
"Dual of Dual Statement (Order Theory)",
"Succeed is Dual to Precede",
"Strictly Succeed is Dual to Strictly Precede",
"Upper Bound is Dual to Lower Bound",
"Supremum... |
proofwiki-6692 | Conditional is not Right Self-Distributive/Formulation 1 | While this holds:
:$\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$
its converse does not:
:$\paren {p \implies r} \implies \paren {q \implies r} \not \vdash \paren {p \implies q} \implies r$ | We apply the Method of Truth Tables to the proposition:
:$\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$
As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the {{LHS}} is $\T$, that under the one on the {{RHS}} ... | While this holds:
:$\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$
its converse does not:
:$\paren {p \implies r} \implies \paren {q \implies r} \not \vdash \paren {p \implies q} \implies r$ | We apply the [[Method of Truth Tables]] to the proposition:
:$\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$
As can be seen for all [[Definition:Boolean Interpretation|boolean interpretations]] by inspection, where the [[Definition:Truth Value|truth value]] under the [[D... | Conditional is not Right Self-Distributive/Formulation 1 | https://proofwiki.org/wiki/Conditional_is_not_Right_Self-Distributive/Formulation_1 | https://proofwiki.org/wiki/Conditional_is_not_Right_Self-Distributive/Formulation_1 | [
"Truth Table Proofs",
"Conditional is not Right Self-Distributive"
] | [] | [
"Method of Truth Tables",
"Definition:Boolean Interpretation",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Language of Propositional Logic/Formal Grammar/WFF",
"Definition:Logical Equivalence",
"Category:Truth Table Proofs",
"Category:Conditional is not Righ... |
proofwiki-6693 | Conditional is not Right Self-Distributive/Formulation 2 | While this holds:
:$\vdash \paren {\paren {p \implies q} \implies r} \implies \paren {\paren {p \implies r} \implies \paren {q \implies r} }$
its converse does not:
:$\not \vdash \paren {\paren {p \implies r} \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies r}$ | {{ProofWanted|Can be seen to be logically equivalent to Conditional is not Right Self-Distributive/Formulation 1 by application of the Rule of Implication and Modus Ponendo Ponens.}} | While this holds:
:$\vdash \paren {\paren {p \implies q} \implies r} \implies \paren {\paren {p \implies r} \implies \paren {q \implies r} }$
its converse does not:
:$\not \vdash \paren {\paren {p \implies r} \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies r}$ | {{ProofWanted|Can be seen to be [[Definition:Logical Equivalence|logically equivalent]] to [[Conditional is not Right Self-Distributive/Formulation 1]] by application of the [[Rule of Implication]] and [[Modus Ponendo Ponens]].}} | Conditional is not Right Self-Distributive/Formulation 2 | https://proofwiki.org/wiki/Conditional_is_not_Right_Self-Distributive/Formulation_2 | https://proofwiki.org/wiki/Conditional_is_not_Right_Self-Distributive/Formulation_2 | [
"Conditional is not Right Self-Distributive"
] | [] | [
"Definition:Logical Equivalence",
"Conditional is not Right Self-Distributive/Formulation 1",
"Rule of Implication",
"Modus Ponendo Ponens"
] |
proofwiki-6694 | Cauchy-Riemann Equations | Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$.
Let $f: D \to \C$ be a complex function on $D$.
Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be the two real-valued functions defined as:
{{begin-eqn}}
{{eqn | l = \map u {x, y}
| r = \map \Re {\map f z}
}}
{{eqn | l... | === Necessary Condition ===
{{:Cauchy-Riemann Equations/Necessary Condition}}{{qed|lemma}} | Let $D \subseteq \C$ be an [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$.
Let $f: D \to \C$ be a [[Definition:Complex Function|complex function]] on $D$.
Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y... | === [[Cauchy-Riemann Equations/Necessary Condition|Necessary Condition]] ===
{{:Cauchy-Riemann Equations/Necessary Condition}}{{qed|lemma}} | Cauchy-Riemann Equations | https://proofwiki.org/wiki/Cauchy-Riemann_Equations | https://proofwiki.org/wiki/Cauchy-Riemann_Equations | [
"Cauchy-Riemann Equations",
"Complex Analysis"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Subset",
"Definition:Set",
"Definition:Complex Number",
"Definition:Complex Function",
"Definition:Real-Valued Function",
"Definition:Complex Number/Real Part",
"Definition:Complex Number/Imaginary Part",
"Definition:Differentiable Mapping/Complex ... | [
"Cauchy-Riemann Equations/Necessary Condition"
] |
proofwiki-6695 | Self-Distributive Law for Conditional/Formulation 2 | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} } }}
{{TheoremIntro|1|\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }|Self-Distributive Law for Conditional: Formulation 2... | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} } }}
{{TheoremIntro|1|\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }|[[Self-Distributive Law for Conditional/Formulation ... | Self-Distributive Law for Conditional/Formulation 2 | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_2 | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_2 | [
"Self-Distributive Law for Conditional"
] | [] | [
"Self-Distributive Law for Conditional/Formulation 2/Forward Implication",
"Self-Distributive Law for Conditional/Formulation 2/Reverse Implication"
] |
proofwiki-6696 | Self-Distributive Law for Conditional/Formulation 2 | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | {{BeginTableau|\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} } }}
{{Assumption |1|p \implies \paren {q \implies r} }}
{{SequentIntro|2|1|\paren {p \implies q} \implies \paren {p \implies r}|1|Self-Distributive Law for Conditional: Formulation 1}}
{{Imp... | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | {{BeginTableau|\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} } }}
{{Assumption |1|p \implies \paren {q \implies r} }}
{{SequentIntro|2|1|\paren {p \implies q} \implies \paren {p \implies r}|1|[[Self-Distributive Law for Conditional/Formulation 1/Forwar... | Self-Distributive Law for Conditional/Formulation 2/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_2 | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_2/Forward_Implication/Proof_1 | [
"Self-Distributive Law for Conditional"
] | [] | [
"Self-Distributive Law for Conditional/Formulation 1/Forward Implication"
] |
proofwiki-6697 | Self-Distributive Law for Conditional/Formulation 2 | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | {{BeginTableau|\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} } }}
{{Assumption |1|p \implies \paren {q \implies r} }}
{{Assumption |2|p \implies q}}
{{Assumption |3|p}}
{{ModusPonens |4|1,3|q \implies r|1|3}}
{{ModusPonens |5|2,3|q|2|3}}
{{ModusPonen... | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | {{BeginTableau|\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} } }}
{{Assumption |1|p \implies \paren {q \implies r} }}
{{Assumption |2|p \implies q}}
{{Assumption |3|p}}
{{ModusPonens |4|1,3|q \implies r|1|3}}
{{ModusPonens |5|2,3|q|2|3}}
{{ModusPonen... | Self-Distributive Law for Conditional/Formulation 2/Forward Implication/Proof 2 | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_2 | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_2/Forward_Implication/Proof_2 | [
"Self-Distributive Law for Conditional"
] | [] | [] |
proofwiki-6698 | Self-Distributive Law for Conditional/Formulation 2 | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth value under the main connective is true for all boolean interpretations.
:<nowiki>$\begin{array}{|ccccc|c|ccccccc|}
\hline
(p & \implies & (q & \implies & r)) & \implies & ((p & \implies & q) & \implies & (p & \implies & r)) \\
\hline
\F & \T ... | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth value]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|c... | Self-Distributive Law for Conditional/Formulation 2/Forward Implication/Proof by Truth Table | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_2 | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_2/Forward_Implication/Proof_by_Truth_Table | [
"Self-Distributive Law for Conditional"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-6699 | Self-Distributive Law for Conditional/Formulation 2 | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | {{BeginTableau |\vdash \paren {\paren {p \implies q} \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} } }}
{{Assumption |1|\paren {p \implies q} \implies \paren {p \implies r} }}
{{SequentIntro |2|1|p \implies \paren {q \implies r}|1|Self-Distributive Law for Conditional: Formulation... | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | {{BeginTableau |\vdash \paren {\paren {p \implies q} \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} } }}
{{Assumption |1|\paren {p \implies q} \implies \paren {p \implies r} }}
{{SequentIntro |2|1|p \implies \paren {q \implies r}|1|[[Self-Distributive Law for Conditional/Formulatio... | Self-Distributive Law for Conditional/Formulation 2/Reverse Implication/Proof 1 | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_2 | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_2/Reverse_Implication/Proof_1 | [
"Self-Distributive Law for Conditional"
] | [] | [
"Self-Distributive Law for Conditional/Formulation 1/Reverse Implication"
] |
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