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-6800 | Top of Lattice is Unique | Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
Then $S$ has at most one top. | By definition, a top for $S$ is a greatest element.
The result follows from Greatest Element is Unique.
{{qed}} | Let $\struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Lattice (Order Theory)|lattice]].
Then $S$ has at most one [[Definition:Top of Lattice|top]]. | By definition, a [[Definition:Top of Lattice/Definition 1|top]] for $S$ is a [[Definition:Greatest Element|greatest element]].
The result follows from [[Greatest Element is Unique]].
{{qed}} | Top of Lattice is Unique | https://proofwiki.org/wiki/Top_of_Lattice_is_Unique | https://proofwiki.org/wiki/Top_of_Lattice_is_Unique | [
"Lattice Theory"
] | [
"Definition:Lattice (Order Theory)",
"Definition:Top of Lattice"
] | [
"Definition:Top of Lattice/Definition 1",
"Definition:Greatest Element",
"Greatest Element is Unique"
] |
proofwiki-6801 | Bottom of Lattice is Unique | Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
Then $S$ has at most one bottom. | By definition, a bottom for $S$ is a smallest element.
The result follows from Smallest Element is Unique.
{{qed}} | Let $\struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Lattice (Order Theory)|lattice]].
Then $S$ has at most one [[Definition:Bottom of Lattice|bottom]]. | By definition, a [[Definition:Bottom of Lattice/Definition 1|bottom]] for $S$ is a [[Definition:Smallest Element|smallest element]].
The result follows from [[Smallest Element is Unique]].
{{qed}} | Bottom of Lattice is Unique | https://proofwiki.org/wiki/Bottom_of_Lattice_is_Unique | https://proofwiki.org/wiki/Bottom_of_Lattice_is_Unique | [
"Lattice Theory"
] | [
"Definition:Lattice (Order Theory)",
"Definition:Bottom of Lattice"
] | [
"Definition:Bottom of Lattice/Definition 1",
"Definition:Smallest Element",
"Smallest Element is Unique"
] |
proofwiki-6802 | Isometry is Homeomorphism of Induced Topologies | Let $\struct {S_1, d_1}$ and $\struct {S_2, d_2}$ be metric spaces or pseudometric spaces.
Let $f: S_1 \to S_2$ be an isometry from $\struct {S_1, d_1}$ to $\struct {S_2, d_2}$.
Let $\tau_1$ and $\tau_2$ be the topologies induced on $S_1$ and $S_2$ by the metrics $d_1$ and $d_2$, respectively.
Then $f$ is a homeomorphi... | By the definition of an isometry, $f$ is bijective.
By Continuous Mapping is Continuous on Induced Topological Spaces, $f$ is a continuous mapping from $\struct {S_1, \tau_1}$ to $\struct {S_2, \tau_2}$.
{{explain|It needs to be established that $f$ is continuous on the original metric spaces in the first place.}}
By I... | Let $\struct {S_1, d_1}$ and $\struct {S_2, d_2}$ be [[Definition:Metric Space|metric spaces]] or [[Definition:Pseudometric Space|pseudometric spaces]].
Let $f: S_1 \to S_2$ be an [[Definition:Isometry (Metric Spaces)|isometry]] from $\struct {S_1, d_1}$ to $\struct {S_2, d_2}$.
Let $\tau_1$ and $\tau_2$ be the [[Def... | By the definition of an [[Definition:Isometry (Metric Spaces)|isometry]], $f$ is [[Definition:Bijective|bijective]].
By [[Continuous Mapping is Continuous on Induced Topological Spaces]], $f$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] from $\struct {S_1, \tau_1}$ to $\struct {S_2, ... | Isometry is Homeomorphism of Induced Topologies | https://proofwiki.org/wiki/Isometry_is_Homeomorphism_of_Induced_Topologies | https://proofwiki.org/wiki/Isometry_is_Homeomorphism_of_Induced_Topologies | [
"Isometries (Metric Spaces)",
"Homeomorphisms (Topological Spaces)"
] | [
"Definition:Metric Space",
"Definition:Pseudometric/Pseudometric Space",
"Definition:Isometry (Metric Spaces)",
"Definition:Topology Induced by Metric",
"Definition:Metric Space/Metric",
"Definition:Homeomorphism/Topological Spaces"
] | [
"Definition:Isometry (Metric Spaces)",
"Definition:Bijection",
"Continuous Mapping is Continuous on Induced Topological Spaces",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Equivalence of Definitions of Isometry of Metric Spaces",
"Definition:Isometry (Metric Spaces)",
"Continuous Mapping is... |
proofwiki-6803 | Equivalence of Definitions of Distributive Lattice | {{TFAE|def = Distributive Lattice}}
Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice. | In what follows it is shown that if one of the axioms distributive lattice axioms holds then they all hold. | {{TFAE|def = Distributive Lattice}}
Let $\struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Lattice (Order Theory)|lattice]]. | In what follows it is shown that if one of the axioms [[Axiom:Distributive Lattice Axioms|distributive lattice axioms]] holds then they all hold. | Equivalence of Definitions of Distributive Lattice | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Distributive_Lattice | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Distributive_Lattice | [
"Distributive Lattices"
] | [
"Definition:Lattice (Order Theory)"
] | [
"Axiom:Distributive Lattice Axioms"
] |
proofwiki-6804 | Distance-Preserving Surjection is Isometry of Metric Spaces | Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.
Let $\phi: M_1 \to M_2$ be a surjective distance-preserving mapping.
That is:
:$\forall a, b \in M_1: d_1 \tuple {a, b} = d_2 \tuple {\map \phi a, \map \phi b}$
Then $\phi$ is an isometry. | The premises satisfy all elements of the definition of isometry except for bijectivity.
As we presume $\phi$ to be surjective we need only show that it is injective.
By Distance-Preserving Mapping is Injection of Metric Spaces then $\phi$ is injective.
{{qed}}
Category:Isometries (Metric Spaces)
agrybp757xrenqh73ffdlcx... | Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
Let $\phi: M_1 \to M_2$ be a [[Definition:Surjective|surjective]] [[Definition:Distance-Preserving Mapping|distance-preserving mapping]].
That is:
:$\forall a, b \in M_1: d_1 \tuple {a, b} = d_2 \tupl... | The premises satisfy all elements of the definition of [[Definition:Isometry (Metric Spaces)|isometry]] except for [[Definition:Bijective|bijectivity]].
As we presume $\phi$ to be [[Definition:Surjection|surjective]] we need only show that it is [[Definition:Injective|injective]].
By [[Distance-Preserving Mapping is ... | Distance-Preserving Surjection is Isometry of Metric Spaces | https://proofwiki.org/wiki/Distance-Preserving_Surjection_is_Isometry_of_Metric_Spaces | https://proofwiki.org/wiki/Distance-Preserving_Surjection_is_Isometry_of_Metric_Spaces | [
"Isometries (Metric Spaces)"
] | [
"Definition:Metric Space",
"Definition:Surjection",
"Definition:Distance-Preserving Mapping",
"Definition:Isometry (Metric Spaces)"
] | [
"Definition:Isometry (Metric Spaces)",
"Definition:Bijection",
"Definition:Surjection",
"Definition:Injective",
"Distance-Preserving Mapping is Injection of Metric Spaces",
"Definition:Injective",
"Category:Isometries (Metric Spaces)"
] |
proofwiki-6805 | Existence of Positive Root of Positive Real Number | Let $x \in \R$ be a real number such that $x > 0$.
Let $n \in \Z$ be an integer such that $n \ne 0$.
Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$. | === Positive Exponent ===
{{:Existence of Positive Root of Positive Real Number/Positive Exponent|Positive Exponent}} | Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > 0$.
Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n \ne 0$.
Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$. | === [[Existence of Positive Root of Positive Real Number/Positive Exponent|Positive Exponent]] ===
{{:Existence of Positive Root of Positive Real Number/Positive Exponent|Positive Exponent}} | Existence of Positive Root of Positive Real Number | https://proofwiki.org/wiki/Existence_of_Positive_Root_of_Positive_Real_Number | https://proofwiki.org/wiki/Existence_of_Positive_Root_of_Positive_Real_Number | [
"Existence of Positive Root of Positive Real Number",
"Existence and Uniqueness of Positive Root of Positive Real Number",
"Roots of Numbers",
"Real Numbers"
] | [
"Definition:Real Number",
"Definition:Integer"
] | [
"Existence of Positive Root of Positive Real Number/Positive Exponent"
] |
proofwiki-6806 | Existence of Positive Root of Positive Real Number | Let $x \in \R$ be a real number such that $x > 0$.
Let $n \in \Z$ be an integer such that $n \ne 0$.
Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$. | Let $f$ be the real function defined on the unbounded closed interval $\hointr 0 \to$ defined by $\map f y = y^n$.
Consider first the case of $n > 0$.
By Strictly Positive Integer Power Function is Unbounded Above:
:$\exists q \in \R_{>0}: \map f q \ge x$
Since $x \ge 0$:
:$\map f 0 \le x$
By the Intermediate Value The... | Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > 0$.
Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n \ne 0$.
Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$. | Let $f$ be the [[Definition:Real Function|real function]] defined on the [[Definition:Unbounded Closed Real Interval|unbounded closed interval]] $\hointr 0 \to$ defined by $\map f y = y^n$.
Consider first the case of $n > 0$.
By [[Strictly Positive Integer Power Function is Unbounded Above]]:
:$\exists q \in \R_{>0}:... | Existence of Positive Root of Positive Real Number/Positive Exponent/Proof 1 | https://proofwiki.org/wiki/Existence_of_Positive_Root_of_Positive_Real_Number | https://proofwiki.org/wiki/Existence_of_Positive_Root_of_Positive_Real_Number/Positive_Exponent/Proof_1 | [
"Existence of Positive Root of Positive Real Number",
"Existence and Uniqueness of Positive Root of Positive Real Number",
"Roots of Numbers",
"Real Numbers"
] | [
"Definition:Real Number",
"Definition:Integer"
] | [
"Definition:Real Function",
"Definition:Real Interval/Unbounded Closed",
"Strictly Positive Integer Power Function is Unbounded Above",
"Intermediate Value Theorem"
] |
proofwiki-6807 | Existence of Positive Root of Positive Real Number | Let $x \in \R$ be a real number such that $x > 0$.
Let $n \in \Z$ be an integer such that $n \ne 0$.
Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$. | Let $S$ be the set consisting of the positive real numbers $t$ such that $t^n < x$.
Let $t = \dfrac x {1 - x}$.
Then $0 < t < 1$ and so:
:$t^n < t < x$
demonstrating that $S \ne \O$.
Let $t_0 = 1 + x$.
Then $t > t_0$ implies $t^n \ge t > x$.
So $t \notin S$.
Hence $t_0$ is an upper bound of $S$.
By the Continuum Proper... | Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > 0$.
Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n \ne 0$.
Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$. | Let $S$ be the [[Definition:Set|set]] consisting of the [[Definition:Positive Real Number|positive real numbers]] $t$ such that $t^n < x$.
Let $t = \dfrac x {1 - x}$.
Then $0 < t < 1$ and so:
:$t^n < t < x$
demonstrating that $S \ne \O$.
Let $t_0 = 1 + x$.
Then $t > t_0$ implies $t^n \ge t > x$.
So $t \notin S$.
... | Existence of Positive Root of Positive Real Number/Positive Exponent/Proof 2 | https://proofwiki.org/wiki/Existence_of_Positive_Root_of_Positive_Real_Number | https://proofwiki.org/wiki/Existence_of_Positive_Root_of_Positive_Real_Number/Positive_Exponent/Proof_2 | [
"Existence of Positive Root of Positive Real Number",
"Existence and Uniqueness of Positive Root of Positive Real Number",
"Roots of Numbers",
"Real Numbers"
] | [
"Definition:Real Number",
"Definition:Integer"
] | [
"Definition:Set",
"Definition:Positive/Real Number",
"Definition:Upper Bound of Set/Real Numbers",
"Continuum Property",
"Definition:Supremum of Set/Real Numbers",
"Definition:Binomial Coefficient",
"Binomial Theorem",
"Definition:Coefficient of Polynomial",
"Binomial Theorem",
"Binomial Theorem",... |
proofwiki-6808 | Proof by Cases/Formulation 2 | :$\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r}$ | {{BeginTableau|\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \implies \paren {\paren {p \lor q} \implies r} }}
{{Assumption|1|\paren {p \implies r} \land \paren {q \implies r} }}
{{SequentIntro|2|1|\paren {p \lor q} \implies r|1|Proof by Cases: Formulation 1: Forward Implication}}
{{Implication|3||... | :$\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r}$ | {{BeginTableau|\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \implies \paren {\paren {p \lor q} \implies r} }}
{{Assumption|1|\paren {p \implies r} \land \paren {q \implies r} }}
{{SequentIntro|2|1|\paren {p \lor q} \implies r|1|[[Proof by Cases/Formulation 1/Forward Implication|Proof by Cases: For... | Proof by Cases/Formulation 2/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_2 | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_2/Forward_Implication/Proof_1 | [
"Proof by Cases"
] | [] | [
"Proof by Cases/Formulation 1/Forward Implication"
] |
proofwiki-6809 | Proof by Cases/Formulation 2 | :$\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r}$ | 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.
:<nowiki>$\begin{array}{|ccccccc|c|ccccc|} \hline
((p & \implies & r) & \land & (q & \implies & r)) & \implies & ((p & \lor & q) & \implies & r) \\
\... | :$\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r}$ | 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]].
:<now... | Proof by Cases/Formulation 2/Forward Implication/Proof by Truth Table | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_2 | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_2/Forward_Implication/Proof_by_Truth_Table | [
"Proof by Cases"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-6810 | Proof by Cases/Formulation 2 | :$\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r}$ | {{BeginTableau|\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r} }}
{{TheoremIntro|1|\paren {\paren {p \implies r} \land \paren {q \implies r} } \implies \paren {\paren {p \lor q} \implies r}|Proof by Cases: Forward Implication}}
{{TheoremIntro|2|\paren {\pare... | :$\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r}$ | {{BeginTableau|\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r} }}
{{TheoremIntro|1|\paren {\paren {p \implies r} \land \paren {q \implies r} } \implies \paren {\paren {p \lor q} \implies r}|[[Proof by Cases/Formulation 2/Forward Implication|Proof by Cases: F... | Proof by Cases/Formulation 2/Proof | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_2 | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_2/Proof | [
"Proof by Cases"
] | [] | [
"Proof by Cases/Formulation 2/Forward Implication",
"Proof by Cases/Formulation 2/Reverse Implication"
] |
proofwiki-6811 | Proof by Cases/Formulation 2/Forward Implication | :$\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \implies \paren {\paren {p \lor q} \implies r}$ | {{BeginTableau|\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \implies \paren {\paren {p \lor q} \implies r} }}
{{Assumption|1|\paren {p \implies r} \land \paren {q \implies r} }}
{{SequentIntro|2|1|\paren {p \lor q} \implies r|1|Proof by Cases: Formulation 1: Forward Implication}}
{{Implication|3||... | :$\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \implies \paren {\paren {p \lor q} \implies r}$ | {{BeginTableau|\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \implies \paren {\paren {p \lor q} \implies r} }}
{{Assumption|1|\paren {p \implies r} \land \paren {q \implies r} }}
{{SequentIntro|2|1|\paren {p \lor q} \implies r|1|[[Proof by Cases/Formulation 1/Forward Implication|Proof by Cases: For... | Proof by Cases/Formulation 2/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_2/Forward_Implication/Proof_1 | [
"Proof by Cases"
] | [] | [
"Proof by Cases/Formulation 1/Forward Implication"
] |
proofwiki-6812 | Proof by Cases/Formulation 2/Forward Implication | :$\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \implies \paren {\paren {p \lor q} \implies r}$ | 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.
:<nowiki>$\begin{array}{|ccccccc|c|ccccc|} \hline
((p & \implies & r) & \land & (q & \implies & r)) & \implies & ((p & \lor & q) & \implies & r) \\
\... | :$\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \implies \paren {\paren {p \lor q} \implies r}$ | 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]].
:<now... | Proof by Cases/Formulation 2/Forward Implication/Proof by Truth Table | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_2/Forward_Implication/Proof_by_Truth_Table | [
"Proof by Cases"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-6813 | Proof by Cases/Formulation 2/Reverse Implication | :$\vdash \paren {\paren {p \lor q} \implies r} \implies \paren {\paren {p \implies r} \land \paren {q \implies r} }$ | {{BeginTableau|\vdash \paren {\paren {p \lor q} \implies r} \implies \paren {\paren {p \implies r} \land \paren {q \implies r} } }}
{{Assumption|1|\paren {p \lor q} \implies r}}
{{SequentIntro|2|1|\paren {p \implies r} \land \paren {q \implies r}|1|Proof by Cases: Formulation 1: Reverse Implication}}
{{Implication|3||\... | :$\vdash \paren {\paren {p \lor q} \implies r} \implies \paren {\paren {p \implies r} \land \paren {q \implies r} }$ | {{BeginTableau|\vdash \paren {\paren {p \lor q} \implies r} \implies \paren {\paren {p \implies r} \land \paren {q \implies r} } }}
{{Assumption|1|\paren {p \lor q} \implies r}}
{{SequentIntro|2|1|\paren {p \implies r} \land \paren {q \implies r}|1|[[Proof by Cases/Formulation 1/Reverse Implication|Proof by Cases: Form... | Proof by Cases/Formulation 2/Reverse Implication | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_2/Reverse_Implication | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_2/Reverse_Implication | [
"Proof by Cases"
] | [] | [
"Proof by Cases/Formulation 1/Reverse Implication",
"Category:Proof by Cases"
] |
proofwiki-6814 | Set is Subset of Union/Set of Sets | Let $\mathbb S$ be a set of sets.
Then:
:$\ds \forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$ | Let $T$ be any element of $\mathbb S$.
We wish to show that $T \subseteq S$.
Let $x \in T$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = T
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \bigcup \mathbb S
| c = {{Defof|Set Union}}
}}
{{end-eqn}}
Since this holds fo... | Let $\mathbb S$ be a [[Definition:Set of Sets|set of sets]].
Then:
:$\ds \forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$ | Let $T$ be any element of $\mathbb S$.
We wish to show that $T \subseteq S$.
Let $x \in T$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = T
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \bigcup \mathbb S
| c = {{Defof|Set Union}}
}}
{{end-eqn}}
Since this hol... | Set is Subset of Union/Set of Sets | https://proofwiki.org/wiki/Set_is_Subset_of_Union/Set_of_Sets | https://proofwiki.org/wiki/Set_is_Subset_of_Union/Set_of_Sets | [
"Set Union",
"Subsets"
] | [
"Definition:Set of Sets"
] | [
"Category:Set Union",
"Category:Subsets"
] |
proofwiki-6815 | Proof by Cases/Formulation 3 | :$\vdash \paren {\paren {p \lor q} \land \paren {p \implies r} \land \paren {q \implies r} } \implies r$ | {{BeginTableau|\paren {\paren {p \lor q} \land \paren {p \implies r} \land \paren {q \implies r} } \implies r}}
{{TheoremIntro|1|\paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}|Constructive Dilemma: Formulation 3}}
{{Substitution|2||\paren {\paren {p \lor q... | :$\vdash \paren {\paren {p \lor q} \land \paren {p \implies r} \land \paren {q \implies r} } \implies r$ | {{BeginTableau|\paren {\paren {p \lor q} \land \paren {p \implies r} \land \paren {q \implies r} } \implies r}}
{{TheoremIntro|1|\paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}|[[Constructive Dilemma/Formulation 3|Constructive Dilemma: Formulation 3]]}}
{{S... | Proof by Cases/Formulation 3 | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_3 | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_3 | [
"Proof by Cases"
] | [] | [
"Constructive Dilemma/Formulation 3"
] |
proofwiki-6816 | Existence of Positive Root of Positive Real Number/Positive Exponent | Let $x \in \R$ be a real number such that $x > 0$.
Let $n \in \Z$ be an integer such that $n > 0$.
Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$. | Let $f$ be the real function defined on the unbounded closed interval $\hointr 0 \to$ defined by $\map f y = y^n$.
Consider first the case of $n > 0$.
By Strictly Positive Integer Power Function is Unbounded Above:
:$\exists q \in \R_{>0}: \map f q \ge x$
Since $x \ge 0$:
:$\map f 0 \le x$
By the Intermediate Value The... | Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > 0$.
Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n > 0$.
Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$. | Let $f$ be the [[Definition:Real Function|real function]] defined on the [[Definition:Unbounded Closed Real Interval|unbounded closed interval]] $\hointr 0 \to$ defined by $\map f y = y^n$.
Consider first the case of $n > 0$.
By [[Strictly Positive Integer Power Function is Unbounded Above]]:
:$\exists q \in \R_{>0}:... | Existence of Positive Root of Positive Real Number/Positive Exponent/Proof 1 | https://proofwiki.org/wiki/Existence_of_Positive_Root_of_Positive_Real_Number/Positive_Exponent | https://proofwiki.org/wiki/Existence_of_Positive_Root_of_Positive_Real_Number/Positive_Exponent/Proof_1 | [
"Existence of Positive Root of Positive Real Number"
] | [
"Definition:Real Number",
"Definition:Integer"
] | [
"Definition:Real Function",
"Definition:Real Interval/Unbounded Closed",
"Strictly Positive Integer Power Function is Unbounded Above",
"Intermediate Value Theorem"
] |
proofwiki-6817 | Existence of Positive Root of Positive Real Number/Positive Exponent | Let $x \in \R$ be a real number such that $x > 0$.
Let $n \in \Z$ be an integer such that $n > 0$.
Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$. | Let $S$ be the set consisting of the positive real numbers $t$ such that $t^n < x$.
Let $t = \dfrac x {1 - x}$.
Then $0 < t < 1$ and so:
:$t^n < t < x$
demonstrating that $S \ne \O$.
Let $t_0 = 1 + x$.
Then $t > t_0$ implies $t^n \ge t > x$.
So $t \notin S$.
Hence $t_0$ is an upper bound of $S$.
By the Continuum Proper... | Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > 0$.
Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n > 0$.
Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$. | Let $S$ be the [[Definition:Set|set]] consisting of the [[Definition:Positive Real Number|positive real numbers]] $t$ such that $t^n < x$.
Let $t = \dfrac x {1 - x}$.
Then $0 < t < 1$ and so:
:$t^n < t < x$
demonstrating that $S \ne \O$.
Let $t_0 = 1 + x$.
Then $t > t_0$ implies $t^n \ge t > x$.
So $t \notin S$.
... | Existence of Positive Root of Positive Real Number/Positive Exponent/Proof 2 | https://proofwiki.org/wiki/Existence_of_Positive_Root_of_Positive_Real_Number/Positive_Exponent | https://proofwiki.org/wiki/Existence_of_Positive_Root_of_Positive_Real_Number/Positive_Exponent/Proof_2 | [
"Existence of Positive Root of Positive Real Number"
] | [
"Definition:Real Number",
"Definition:Integer"
] | [
"Definition:Set",
"Definition:Positive/Real Number",
"Definition:Upper Bound of Set/Real Numbers",
"Continuum Property",
"Definition:Supremum of Set/Real Numbers",
"Definition:Binomial Coefficient",
"Binomial Theorem",
"Definition:Coefficient of Polynomial",
"Binomial Theorem",
"Binomial Theorem",... |
proofwiki-6818 | Existence of Positive Root of Positive Real Number/Negative Exponent | Let $x \in \R$ be a real number such that $x > 0$.
Let $n \in \Z$ be an integer such that $n < 0$.
Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$. | Let $m = -n$.
Then $m > 0$.
Let $g$ be the real function defined on $\hointr 0 \to$ defined by:
:$\map g y = y^m$
Since $x \ge 0$:
:$\dfrac 1 x \ge 0$
By Existence of Positive Root of Positive Real Number: Positive Exponent there is a $y > 0$ such that:
:$\map g y = \dfrac 1 x$
It follows from the definition of power t... | Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > 0$.
Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n < 0$.
Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$. | Let $m = -n$.
Then $m > 0$.
Let $g$ be the [[Definition:Real Function|real function]] defined on $\hointr 0 \to$ defined by:
:$\map g y = y^m$
Since $x \ge 0$:
:$\dfrac 1 x \ge 0$
By [[Existence of Positive Root of Positive Real Number/Positive Exponent|Existence of Positive Root of Positive Real Number: Positive E... | Existence of Positive Root of Positive Real Number/Negative Exponent | https://proofwiki.org/wiki/Existence_of_Positive_Root_of_Positive_Real_Number/Negative_Exponent | https://proofwiki.org/wiki/Existence_of_Positive_Root_of_Positive_Real_Number/Negative_Exponent | [
"Existence of Positive Root of Positive Real Number"
] | [
"Definition:Real Number",
"Definition:Integer"
] | [
"Definition:Real Function",
"Existence of Positive Root of Positive Real Number/Positive Exponent",
"Definition:Power (Algebra)",
"Category:Existence of Positive Root of Positive Real Number"
] |
proofwiki-6819 | Power of Ring Negative | Let $\struct {R, +, \circ}$ be a ring.
Let $n \in \N_{>0}$ be a strictly positive integer.
Let $x \in R$.
Then:
:If $n$ is even:
:::$\map {\circ^n} {-x} = \map {\circ^n} x$
:If $n$ is odd:
:::$\map {\circ^n} {-x} = -\map {\circ^n} x$ | First, suppose that $n$ is even.
Then for some $m \in \N_{>0}$:
:$n = 2 m = m + m$
Thus since $\circ$ is associative:
:$\ds \map {\circ^n} {-x} = \prod_{i \mathop = 1}^m \paren {-x} \circ \paren {-x}$
By Product of Ring Negatives:
:$\paren {-x} \circ \paren {-x} = x \circ x = \map {\circ^2} x$
Thus:
:$\ds \map {\circ^n... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $n \in \N_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Let $x \in R$.
Then:
:If $n$ is [[Definition:Even Integer|even]]:
:::$\map {\circ^n} {-x} = \map {\circ^n} x$
:If $n$ is [[Definition:Odd Intege... | First, suppose that $n$ is [[Definition:Even Integer|even]].
Then for some $m \in \N_{>0}$:
:$n = 2 m = m + m$
Thus since $\circ$ is [[Definition:Associative Operation|associative]]:
:$\ds \map {\circ^n} {-x} = \prod_{i \mathop = 1}^m \paren {-x} \circ \paren {-x}$
By [[Product of Ring Negatives]]:
:$\paren {-x} \ci... | Power of Ring Negative | https://proofwiki.org/wiki/Power_of_Ring_Negative | https://proofwiki.org/wiki/Power_of_Ring_Negative | [
"Ring Theory",
"Powers"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Strictly Positive/Integer",
"Definition:Even Integer",
"Definition:Odd Integer"
] | [
"Definition:Even Integer",
"Definition:Associative Operation",
"Product of Ring Negatives",
"Definition:Associative Operation",
"Definition:Odd Integer",
"Definition:Strictly Positive/Integer",
"Definition:Even Integer",
"Product with Ring Negative",
"Category:Ring Theory",
"Category:Powers"
] |
proofwiki-6820 | Paving Lemma | Let $S$ be an open subset of the Euclidean space $\R^m$ or the set of complex numbers $\C$.
Let $\gamma: \closedint a b \to S$ be a path in $S$.
Then there exists $K \in \R_{>0}$ such that:
:For all $\epsilon \in \openint 0 K$, there exists a normal subdivision $\set {x_0, x_1, \ldots, x_{n - 1}, x_n}$ of the closed in... | === Finding the constant ===
First, suppose that $S$ is a subset of $\R^m$.
From Closed Real Interval is Compact in Metric Space, it follows that $\closedint a b$ is compact.
Then Continuous Image of Compact Space is Compact shows that $\map \gamma {\closedint {x_i} {x_{i + 1} } }$ is compact.
From the Heine-Borel Theo... | Let $S$ be an [[Definition:Open Set (Metric Space)|open]] [[Definition:Subset|subset]] of the [[Definition:Euclidean Space|Euclidean space]] $\R^m$ or the [[Definition:Complex Number|set of complex numbers]] $\C$.
Let $\gamma: \closedint a b \to S$ be a [[Definition:Path (Topology)|path]] in $S$.
Then there exists $... | === Finding the constant ===
First, suppose that $S$ is a [[Definition:Subset|subset]] of $\R^m$.
From [[Closed Real Interval is Compact in Metric Space]], it follows that $\closedint a b$ is [[Definition:Compact (Real Analysis)|compact]].
Then [[Continuous Image of Compact Space is Compact]] shows that $\map \gamma... | Paving Lemma | https://proofwiki.org/wiki/Paving_Lemma | https://proofwiki.org/wiki/Paving_Lemma | [
"Complex Analysis",
"Euclidean Spaces",
"Named Theorems"
] | [
"Definition:Open Set/Metric Space",
"Definition:Subset",
"Definition:Euclidean Space",
"Definition:Complex Number",
"Definition:Path (Topology)",
"Definition:Subdivision of Interval/Normal Subdivision",
"Definition:Real Interval/Closed",
"Definition:Open Ball",
"Definition:Open Ball/Radius"
] | [
"Definition:Subset",
"Closed Real Interval is Compact Space/Metric Space",
"Definition:Compact Space/Real Analysis",
"Continuous Image of Compact Space is Compact",
"Definition:Compact Space/Real Analysis",
"Heine-Borel Theorem/Euclidean Space",
"Definition:Bounded Metric Space",
"Definition:Closed Se... |
proofwiki-6821 | Proof by Cases/Formulation 1/Reverse Implication | :$\paren {p \lor q} \implies r \vdash \paren {p \implies r} \land \paren {q \implies r}$ | {{BeginTableau|\paren {p \lor q} \implies r \vdash \paren {p \implies r} \land \paren {q \implies r} }}
{{Premise|1|\paren {p \lor q} \implies r}}
{{Assumption|2|p}}
{{Addition|3|2|p \lor q|2|1}}
{{ModusPonens|4|1, 2|r|1|3}}
{{Implication|5|1|p \implies r|2|4}}
{{Assumption|6|q}}
{{Addition|7|6|p \lor q|6|2}}
{{ModusPo... | :$\paren {p \lor q} \implies r \vdash \paren {p \implies r} \land \paren {q \implies r}$ | {{BeginTableau|\paren {p \lor q} \implies r \vdash \paren {p \implies r} \land \paren {q \implies r} }}
{{Premise|1|\paren {p \lor q} \implies r}}
{{Assumption|2|p}}
{{Addition|3|2|p \lor q|2|1}}
{{ModusPonens|4|1, 2|r|1|3}}
{{Implication|5|1|p \implies r|2|4}}
{{Assumption|6|q}}
{{Addition|7|6|p \lor q|6|2}}
{{ModusPo... | Proof by Cases/Formulation 1/Reverse Implication | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_1/Reverse_Implication | [
"Proof by Cases"
] | [] | [
"Category:Proof by Cases"
] |
proofwiki-6822 | Praeclarum Theorema/Formulation 1 | :$\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s}$ | {{BeginTableau|\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s} }}
{{Premise|1|\paren {p \implies q} \land \paren {r \implies s} }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|r \implies s|1|2}}
{{Assumption|4|p \land r}}
{{Simplification|5|4|p|4|1... | :$\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s}$ | {{BeginTableau|\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s} }}
{{Premise|1|\paren {p \implies q} \land \paren {r \implies s} }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|r \implies s|1|2}}
{{Assumption|4|p \land r}}
{{Simplification|5|4|p|4|1... | Praeclarum Theorema/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Praeclarum_Theorema/Formulation_1 | https://proofwiki.org/wiki/Praeclarum_Theorema/Formulation_1/Proof_1 | [
"Praeclarum Theorema"
] | [] | [] |
proofwiki-6823 | Praeclarum Theorema/Formulation 1 | :$\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s}$ | {{BeginTableau |\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s} }}
{{Premise |1|\paren {p \implies q} \land \paren {r \implies s} }}
{{Simplification |2|1|p \implies q|1|1}}
{{Simplification |3|1|r \implies s|1|2}}
{{SequentIntro |4|1|p \land r \implies... | :$\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s}$ | {{BeginTableau |\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s} }}
{{Premise |1|\paren {p \implies q} \land \paren {r \implies s} }}
{{Simplification |2|1|p \implies q|1|1}}
{{Simplification |3|1|r \implies s|1|2}}
{{SequentIntro |4|1|p \land r \implies... | Praeclarum Theorema/Formulation 1/Proof 3 | https://proofwiki.org/wiki/Praeclarum_Theorema/Formulation_1 | https://proofwiki.org/wiki/Praeclarum_Theorema/Formulation_1/Proof_3 | [
"Praeclarum Theorema"
] | [] | [
"Factor Principles/Conjunction on Right/Formulation 1/Proof 2",
"Factor Principles/Conjunction on Left/Formulation 1/Proof 2",
"Hypothetical Syllogism"
] |
proofwiki-6824 | Praeclarum Theorema/Formulation 1 | :$\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s}$ | We apply the Method of Truth Tables to the proposition.
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}} is also $T$:
:<nowiki>$\begin{array}{|ccccccc||ccccccc|} \hline
(p & \implies & q) & \land & (r &... | :$\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s}$ | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen for all [[Definition:Boolean Interpretation|boolean interpretations]] by inspection, where the [[Definition:Truth Value|truth value]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] on the {{LHS}} is $T$, that und... | Praeclarum Theorema/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Praeclarum_Theorema/Formulation_1 | https://proofwiki.org/wiki/Praeclarum_Theorema/Formulation_1/Proof_by_Truth_Table | [
"Praeclarum Theorema"
] | [] | [
"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"
] |
proofwiki-6825 | Praeclarum Theorema/Formulation 2 | :$\vdash \paren {\paren {p \implies q} \land \paren {r \implies s} } \implies \paren {\paren {p \land r} \implies \paren {q \land s} }$ | {{BeginTableau|\vdash \paren {\paren {p \implies q} \land \paren {r \implies s} } \implies \paren {\paren {p \land r} \implies \paren {q \land s} } }}
{{Assumption|1|\paren {\paren {p \implies q} \land \paren {r \implies s} } }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|r \implies s|1|2}}
{{Assumptio... | :$\vdash \paren {\paren {p \implies q} \land \paren {r \implies s} } \implies \paren {\paren {p \land r} \implies \paren {q \land s} }$ | {{BeginTableau|\vdash \paren {\paren {p \implies q} \land \paren {r \implies s} } \implies \paren {\paren {p \land r} \implies \paren {q \land s} } }}
{{Assumption|1|\paren {\paren {p \implies q} \land \paren {r \implies s} } }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|r \implies s|1|2}}
{{Assumptio... | Praeclarum Theorema/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Praeclarum_Theorema/Formulation_2 | https://proofwiki.org/wiki/Praeclarum_Theorema/Formulation_2/Proof_1 | [
"Praeclarum Theorema"
] | [] | [] |
proofwiki-6826 | Praeclarum Theorema/Formulation 2 | :$\vdash \paren {\paren {p \implies q} \land \paren {r \implies s} } \implies \paren {\paren {p \land r} \implies \paren {q \land s} }$ | {{BeginTableau|\vdash \paren {\paren {p \implies q} \land \paren {r \implies s} } \implies \paren {\paren {p \land r} \implies \paren {q \land s} } }}
{{Assumption|1|\paren {\paren {p \implies q} \land \paren {r \implies s} } }}
{{SequentIntro|2|1|\paren {p \land r} \implies \paren {q \land s}|1|Praeclarum Theorema: Fo... | :$\vdash \paren {\paren {p \implies q} \land \paren {r \implies s} } \implies \paren {\paren {p \land r} \implies \paren {q \land s} }$ | {{BeginTableau|\vdash \paren {\paren {p \implies q} \land \paren {r \implies s} } \implies \paren {\paren {p \land r} \implies \paren {q \land s} } }}
{{Assumption|1|\paren {\paren {p \implies q} \land \paren {r \implies s} } }}
{{SequentIntro|2|1|\paren {p \land r} \implies \paren {q \land s}|1|[[Praeclarum Theorema/F... | Praeclarum Theorema/Formulation 2/Proof 2 | https://proofwiki.org/wiki/Praeclarum_Theorema/Formulation_2 | https://proofwiki.org/wiki/Praeclarum_Theorema/Formulation_2/Proof_2 | [
"Praeclarum Theorema"
] | [] | [
"Praeclarum Theorema/Formulation 1"
] |
proofwiki-6827 | Strictly Positive Integer Power Function Strictly Succeeds Each Element | Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity.
Let $\struct {R, \le}$ be a directed set with no upper bound.
Let $n \in \N_{>0}$.
Let $f: R \to R$ be defined by:
:$\forall x \in R: \map f x = \circ^n x$
Then the image of $f$ has elements strictly succeeding each elements of $R$. | Let $b \in R$.
By Directed Set has Strict Successors iff Unbounded Above:
:$\exists c \in R: b < c$
:$\exists d \in R: 1 < d$
By the definition of a directed set:
:$\exists e \in R: d \le e, c \le e$
$\struct {R, +, \circ, \le}$ is an ordered ring, so $\le$ is by definition a transitive relation.
Hence by transitivity:... | Let $\struct {R, +, \circ, \le}$ be an [[Definition:Ordered Ring|ordered ring]] [[Definition:Ring with Unity|with unity]].
Let $\struct {R, \le}$ be a [[Definition:Directed Set|directed set]] with no [[Definition:Upper Bound of Set|upper bound]].
Let $n \in \N_{>0}$.
Let $f: R \to R$ be defined by:
:$\forall x \in R... | Let $b \in R$.
By [[Directed Set has Strict Successors iff Unbounded Above]]:
:$\exists c \in R: b < c$
:$\exists d \in R: 1 < d$
By the definition of a [[Definition:Directed Set|directed set]]:
:$\exists e \in R: d \le e, c \le e$
$\struct {R, +, \circ, \le}$ is an [[Definition:Ordered Ring|ordered ring]], so $\le$... | Strictly Positive Integer Power Function Strictly Succeeds Each Element | https://proofwiki.org/wiki/Strictly_Positive_Integer_Power_Function_Strictly_Succeeds_Each_Element | https://proofwiki.org/wiki/Strictly_Positive_Integer_Power_Function_Strictly_Succeeds_Each_Element | [
"Ring Theory"
] | [
"Definition:Ordered Ring",
"Definition:Ring with Unity",
"Definition:Directed Preordering",
"Definition:Upper Bound of Set",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Element",
"Definition:Strictly Succeed",
"Definition:Element"
] | [
"Directed Set has Strict Successors iff Unbounded Above",
"Definition:Directed Preordering",
"Definition:Ordered Ring",
"Definition:Transitive Relation",
"Definition:Transitive Relation",
"Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element",
"Definition:Transitive Rel... |
proofwiki-6828 | Constructive Dilemma/Formulation 3 | :$\vdash \paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}$ | {{BeginTableau |\vdash \paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s} }}
{{Assumption |1|\paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } }}
{{Simplification|2|1|\paren {p \implies q} \land \paren {r \implies s}|1|2|C... | :$\vdash \paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}$ | {{BeginTableau |\vdash \paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s} }}
{{Assumption |1|\paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } }}
{{Simplification|2|1|\paren {p \implies q} \land \paren {r \implies s}|1|2|C... | Constructive Dilemma/Formulation 3 | https://proofwiki.org/wiki/Constructive_Dilemma/Formulation_3 | https://proofwiki.org/wiki/Constructive_Dilemma/Formulation_3 | [
"Constructive Dilemma"
] | [] | [
"Constructive Dilemma/Formulation 1"
] |
proofwiki-6829 | Odd Power Function is Strictly Increasing/Real Numbers | Let $n \in \Z_{> 0}$ be an odd positive integer.
Let $f_n: \R \to \R$ be the real function defined as:
:$\map {f_n} x = x^n$
Then $f_n$ is strictly increasing. | From the Power Rule for Derivatives:
:$\map {D_x} {x^n} = n x^{n - 1}$
As $n$ is odd, $n - 1$ is even.
Thus by Even Power is Non-Negative:
:$\map {D_x} {x^n} \ge 0$
for all $x$.
From Derivative of Monotone Function, it follows that $f_n$ is increasing over the whole of $\R$.
The only place where $\map {D_x} {x^n} = 0$ ... | Let $n \in \Z_{> 0}$ be an [[Definition:Odd Integer|odd]] [[Definition:Positive Integer|positive integer]].
Let $f_n: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\map {f_n} x = x^n$
Then $f_n$ is [[Definition:Strictly Increasing Real Function|strictly increasing]]. | From the [[Power Rule for Derivatives]]:
:$\map {D_x} {x^n} = n x^{n - 1}$
As $n$ is [[Definition:Odd Integer|odd]], $n - 1$ is [[Definition:Even Integer|even]].
Thus by [[Even Power is Non-Negative]]:
:$\map {D_x} {x^n} \ge 0$
for all $x$.
From [[Derivative of Monotone Function]], it follows that $f_n$ is [[Defini... | Odd Power Function is Strictly Increasing/Real Numbers | https://proofwiki.org/wiki/Odd_Power_Function_is_Strictly_Increasing/Real_Numbers | https://proofwiki.org/wiki/Odd_Power_Function_is_Strictly_Increasing/Real_Numbers | [
"Odd Power Function is Strictly Increasing",
"Strictly Increasing Real Functions"
] | [
"Definition:Odd Integer",
"Definition:Positive/Integer",
"Definition:Real Function",
"Definition:Strictly Increasing/Real Function"
] | [
"Power Rule for Derivatives",
"Definition:Odd Integer",
"Definition:Even Integer",
"Even Power is Non-Negative",
"Derivative of Monotone Function",
"Definition:Increasing/Real Function",
"Definition:Strictly Increasing/Real Function",
"Sign of Odd Power",
"Definition:Strictly Increasing/Real Functio... |
proofwiki-6830 | Clavius's Law/Formulation 1 | :$\neg p \implies p \vdash p$ | {{BeginTableau|\neg p \implies p \vdash p}}
{{Premise|1|\neg p \implies p}}
{{ExcludedMiddle|2|p \lor \neg p}}
{{Assumption|3|\neg p|Either $p$ is false ...}}
{{ModusPonens|4|1, 3|p|1|3}}
{{Assumption|5|p|... or $p$ is true}}
{{ProofByCases|6|1|p|2|3|4|5|5}}
{{EndTableau|qed}}
{{LEM||3}} | :$\neg p \implies p \vdash p$ | {{BeginTableau|\neg p \implies p \vdash p}}
{{Premise|1|\neg p \implies p}}
{{ExcludedMiddle|2|p \lor \neg p}}
{{Assumption|3|\neg p|Either $p$ is false ...}}
{{ModusPonens|4|1, 3|p|1|3}}
{{Assumption|5|p|... or $p$ is true}}
{{ProofByCases|6|1|p|2|3|4|5|5}}
{{EndTableau|qed}}
{{LEM||3}} | Clavius's Law/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Clavius's_Law/Formulation_1 | https://proofwiki.org/wiki/Clavius's_Law/Formulation_1/Proof_1 | [
"Clavius's Law"
] | [] | [] |
proofwiki-6831 | Clavius's Law/Formulation 1 | :$\neg p \implies p \vdash p$ | {{BeginTableau|\neg p \implies p \vdash p}}
{{Premise|1|\neg p \implies p}}
{{Assumption|2|p \implies \bot}}
{{SequentIntro|3|2|\neg p|2|Negation as Implication of Bottom}}
{{ModusPonens|4|1,2|p|1|3}}
{{Implication|5|1|(p \implies \bot) \implies p|2|4}}
{{SequentIntro|6|1|p|5|Peirce's Law}}
{{EndTableau|qed}}
{{LEM|Pei... | :$\neg p \implies p \vdash p$ | {{BeginTableau|\neg p \implies p \vdash p}}
{{Premise|1|\neg p \implies p}}
{{Assumption|2|p \implies \bot}}
{{SequentIntro|3|2|\neg p|2|[[Negation as Implication of Bottom]]}}
{{ModusPonens|4|1,2|p|1|3}}
{{Implication|5|1|(p \implies \bot) \implies p|2|4}}
{{SequentIntro|6|1|p|5|[[Peirce's Law/Formulation 1|Peirce's L... | Clavius's Law/Formulation 1/Proof 2 | https://proofwiki.org/wiki/Clavius's_Law/Formulation_1 | https://proofwiki.org/wiki/Clavius's_Law/Formulation_1/Proof_2 | [
"Clavius's Law"
] | [] | [
"Negation as Implication of Bottom",
"Peirce's Law/Formulation 1"
] |
proofwiki-6832 | Clavius's Law/Formulation 2 | :$\vdash \paren {\neg p \implies p} \implies p$ | {{BeginTableau|\vdash \paren {\neg p \implies p} \implies p}}
{{Premise|1|\neg p \implies p}}
{{SequentIntro|2|1|p|1|Clavius's Law: Formulation 1}}
{{Implication|3||\paren {\neg p \implies p} \implies p|1|2}}
{{EndTableau|qed}}
{{LEM|Clavius's Law/Formulation 1|3}} | :$\vdash \paren {\neg p \implies p} \implies p$ | {{BeginTableau|\vdash \paren {\neg p \implies p} \implies p}}
{{Premise|1|\neg p \implies p}}
{{SequentIntro|2|1|p|1|[[Clavius's Law/Formulation 1|Clavius's Law: Formulation 1]]}}
{{Implication|3||\paren {\neg p \implies p} \implies p|1|2}}
{{EndTableau|qed}}
{{LEM|Clavius's Law/Formulation 1|3}} | Clavius's Law/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Clavius's_Law/Formulation_2 | https://proofwiki.org/wiki/Clavius's_Law/Formulation_2/Proof_1 | [
"Clavius's Law"
] | [] | [
"Clavius's Law/Formulation 1"
] |
proofwiki-6833 | Clavius's Law/Formulation 2 | :$\vdash \paren {\neg p \implies p} \implies p$ | {{BeginTableau|\vdash \paren {\neg p \implies p} \implies p}}
{{Premise|1|\neg p \implies p}}
{{ExcludedMiddle|2|p \lor \neg p}}
{{Assumption|3|\neg p|Either $p$ is false ...}}
{{ModusPonens|4|1, 3|p|1|3}}
{{Assumption|5|p|... or $p$ is true}}
{{ProofByCases|6|1|p|2|3|4|5|5}}
{{Implication|7||\paren {\neg p \implies p... | :$\vdash \paren {\neg p \implies p} \implies p$ | {{BeginTableau|\vdash \paren {\neg p \implies p} \implies p}}
{{Premise|1|\neg p \implies p}}
{{ExcludedMiddle|2|p \lor \neg p}}
{{Assumption|3|\neg p|Either $p$ is false ...}}
{{ModusPonens|4|1, 3|p|1|3}}
{{Assumption|5|p|... or $p$ is true}}
{{ProofByCases|6|1|p|2|3|4|5|5}}
{{Implication|7||\paren {\neg p \implies p... | Clavius's Law/Formulation 2/Proof 2 | https://proofwiki.org/wiki/Clavius's_Law/Formulation_2 | https://proofwiki.org/wiki/Clavius's_Law/Formulation_2/Proof_2 | [
"Clavius's Law"
] | [] | [] |
proofwiki-6834 | Clavius's Law/Formulation 2 | :$\vdash \paren {\neg p \implies p} \implies p$ | 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>$\begin{array}{|cccc|c|c|} \hline
(\neg & p & \implies & p) & \implies & p \\
\hline
\T & \F & \F & \F & \T & \F \\
\F & \T & \T & \T & \T & \T \\
\hline
\end{ar... | :$\vdash \paren {\neg p \implies p} \implies p$ | 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... | Clavius's Law/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Clavius's_Law/Formulation_2 | https://proofwiki.org/wiki/Clavius's_Law/Formulation_2/Proof_by_Truth_Table | [
"Clavius's Law"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-6835 | Equivalent Matrices may not be Similar | If two square matrices of order $n > 1$ over a ring with unity $R$ are equivalent, they are not necessarily similar. | Proof by Counterexample:
Let $\mathbf A = \mathbf I_n$ be the unit matrix of order $n > 1$.
Let $\mathbf B$ be an arbitrary nonsingular matrix over $R$ of order $n$ that is different from the unit matrix.
Then:
:$\mathbf I_n^{-1} \mathbf A \mathbf B = \mathbf I_n^{-1} \mathbf I_n \mathbf B = \mathbf B$
showing that $\m... | If two [[Definition:Square Matrix|square matrices of order $n > 1$]] over a [[Definition:Ring with Unity|ring with unity]] $R$ are [[Definition:Matrix Equivalence|equivalent]], they are not necessarily [[Definition:Matrix Similarity|similar]]. | [[Proof by Counterexample]]:
Let $\mathbf A = \mathbf I_n$ be the [[Definition:Unit Matrix|unit matrix]] of order $n > 1$.
Let $\mathbf B$ be an arbitrary [[Definition:Nonsingular Matrix|nonsingular matrix]] over $R$ of [[Definition:Order of Square Matrix|order]] $n$ that is different from the [[Definition:Unit Matri... | Equivalent Matrices may not be Similar | https://proofwiki.org/wiki/Equivalent_Matrices_may_not_be_Similar | https://proofwiki.org/wiki/Equivalent_Matrices_may_not_be_Similar | [
"Matrix Equivalence",
"Matrix Similarity"
] | [
"Definition:Matrix/Square Matrix",
"Definition:Ring with Unity",
"Definition:Matrix Equivalence",
"Definition:Matrix Similarity"
] | [
"Proof by Counterexample",
"Definition:Unit Matrix",
"Definition:Nonsingular Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Unit Matrix",
"Definition:Matrix Equivalence",
"Definition:Nonsingular Matrix",
"Definition:Matrix/Square Matrix",
"Definition:Matrix/Square Matrix/Order",
"De... |
proofwiki-6836 | Odd Power Function is Strictly Increasing/General Result | Let $\struct {R, +, \circ, \le}$ be a totally ordered ring.
Let $n$ be an odd positive integer.
Let $f: R \to R$ be the mapping defined by:
:$\map f x = \map {\circ^n} x$
Then $f$ is strictly increasing on $R$. | {{proofread}}
Let $x, y \in R$ such that $0 < x < y$.
By Power Function is Strictly Increasing on Positive Elements:
:$\map f x < \map f y$
Suppose that $x < y < 0$.
By Properties of Ordered Ring:
:$0 < -y < -x$
By Power Function is Strictly Increasing on Positive Elements (applied to $-y$ and $-x$):
:$0 < \map f {-y} ... | Let $\struct {R, +, \circ, \le}$ be a [[Definition:Totally Ordered Ring|totally ordered ring]].
Let $n$ be an [[Definition:Odd Integer|odd]] [[Definition:Positive Integer|positive integer]].
Let $f: R \to R$ be the [[Definition:Mapping|mapping]] defined by:
:$\map f x = \map {\circ^n} x$
Then $f$ is [[Definition:Str... | {{proofread}}
Let $x, y \in R$ such that $0 < x < y$.
By [[Power Function is Strictly Increasing on Positive Elements]]:
:$\map f x < \map f y$
Suppose that $x < y < 0$.
By [[Properties of Ordered Ring]]:
:$0 < -y < -x$
By [[Power Function is Strictly Increasing on Positive Elements]] (applied to $-y$ and $-x$):
... | Odd Power Function is Strictly Increasing/General Result | https://proofwiki.org/wiki/Odd_Power_Function_is_Strictly_Increasing/General_Result | https://proofwiki.org/wiki/Odd_Power_Function_is_Strictly_Increasing/General_Result | [
"Totally Ordered Rings",
"Odd Power Function is Strictly Increasing"
] | [
"Definition:Totally Ordered Ring",
"Definition:Odd Integer",
"Definition:Positive/Integer",
"Definition:Mapping",
"Definition:Strictly Increasing/Mapping"
] | [
"Power Function is Strictly Increasing on Positive Elements",
"Properties of Ordered Ring",
"Power Function is Strictly Increasing on Positive Elements",
"Power of Ring Negative",
"Properties of Ordered Ring",
"Sign of Odd Power",
"Definition:Strictly Increasing/Mapping",
"Definition:Positivity Proper... |
proofwiki-6837 | Number of Bijective Restrictions | Let $f: S \to T$ be a surjection.
Let $B$ be the set of all bijective restrictions of $f$.
Then the cardinality of $B$ is:
:$\ds \card {\prod_{i \mathop \in I} \family {S / \RR_f}_i}$
where $S / \RR_f$ denotes the quotient set of the induced equivalence of $f$ indexed by $I$. | {{proof wanted}}
Category:Bijections
Category:Restrictions
6kqq43g16lui5yu2a6ryj0y95zgrli4 | Let $f: S \to T$ be a [[Definition:Surjection|surjection]].
Let $B$ be the [[Definition:Set|set]] of all [[Definition:Bijection|bijective]] [[Definition:Restriction of Mapping|restrictions]] of $f$.
Then the [[Definition:Cardinality|cardinality]] of $B$ is:
:$\ds \card {\prod_{i \mathop \in I} \family {S / \RR_f}_i... | {{proof wanted}}
[[Category:Bijections]]
[[Category:Restrictions]]
6kqq43g16lui5yu2a6ryj0y95zgrli4 | Number of Bijective Restrictions | https://proofwiki.org/wiki/Number_of_Bijective_Restrictions | https://proofwiki.org/wiki/Number_of_Bijective_Restrictions | [
"Bijections",
"Restrictions"
] | [
"Definition:Surjection",
"Definition:Set",
"Definition:Bijection",
"Definition:Restriction/Mapping",
"Definition:Cardinality",
"Definition:Quotient Set",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Indexing Set"
] | [
"Category:Bijections",
"Category:Restrictions"
] |
proofwiki-6838 | Number of Injective Restrictions | Let $f: S \to T$ be a mapping.
Let $Q$ be the set of all injective restrictions of $f$.
Then the cardinality of $Q$ is:
:$\ds \card {\prod_{i \mathop \in I} \prod_{j \mathop \in J_i} \family {\family {\powerset {S / \RR_f} }_i}_j}$
where:
:$\powerset \cdot$ denotes power set
:$S / \RR_f$ denotes quotient set of the ind... | {{proof wanted}}
Category:Injections
b3789r3dwtxyt9mvq0dhswfwfa0bkx7 | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $Q$ be the [[Definition:Set|set]] of all [[Definition:Injection|injective]] [[Definition:Restriction of Mapping|restrictions]] of $f$.
Then the [[Definition:Cardinality|cardinality]] of $Q$ is:
:$\ds \card {\prod_{i \mathop \in I} \prod_{j \mathop \in J_i} \... | {{proof wanted}}
[[Category:Injections]]
b3789r3dwtxyt9mvq0dhswfwfa0bkx7 | Number of Injective Restrictions | https://proofwiki.org/wiki/Number_of_Injective_Restrictions | https://proofwiki.org/wiki/Number_of_Injective_Restrictions | [
"Injections"
] | [
"Definition:Mapping",
"Definition:Set",
"Definition:Injection",
"Definition:Restriction/Mapping",
"Definition:Cardinality",
"Definition:Power Set",
"Definition:Quotient Set",
"Definition:Equivalence Relation Induced by Mapping"
] | [
"Category:Injections"
] |
proofwiki-6839 | Box Topology on Finite Product Space is Product Topology | Let $n \in \N$.
For all $k \in \set {1, \ldots, n}$, let $T_k = \struct {X_k, \tau_k}$ be topological spaces.
Let $\ds X = \prod_{k \mathop = 1}^n X_k$ be the cartesian product of $X_1, \ldots, X_n$.
Then the box topology and the product topology on $X$ are identical. | Denote the product topology on $X$ as $\tau$, and the box topology on $X$ as $\tau'$.
Suppose that $U \in \tau'$.
Then there exists an index set $I$ such that:
:$\ds U = \bigcup_{i \mathop \in I} \struct {U_{i, 1} \times U_{i, 2} \times \cdots \times U_{i, n} }$
where $U_{i, k} \in \tau_k$ for all $i \in I, k \in \set ... | Let $n \in \N$.
For all $k \in \set {1, \ldots, n}$, let $T_k = \struct {X_k, \tau_k}$ be [[Definition:Topological Space|topological spaces]].
Let $\ds X = \prod_{k \mathop = 1}^n X_k$ be the [[Definition:Finite Cartesian Product|cartesian product]] of $X_1, \ldots, X_n$.
Then the [[Definition:Box Topology|box topo... | Denote the [[Definition:Product Topology|product topology]] on $X$ as $\tau$, and the [[Definition:Box Topology|box topology]] on $X$ as $\tau'$.
Suppose that $U \in \tau'$.
Then there exists an [[Definition:Indexed Set|index set]] $I$ such that:
:$\ds U = \bigcup_{i \mathop \in I} \struct {U_{i, 1} \times U_{i, 2} ... | Box Topology on Finite Product Space is Product Topology | https://proofwiki.org/wiki/Box_Topology_on_Finite_Product_Space_is_Product_Topology | https://proofwiki.org/wiki/Box_Topology_on_Finite_Product_Space_is_Product_Topology | [
"Box Topology",
"Product Topology"
] | [
"Definition:Topological Space",
"Definition:Cartesian Product/Finite",
"Definition:Box Topology",
"Definition:Product Topology"
] | [
"Definition:Product Topology",
"Definition:Box Topology",
"Definition:Indexing Set/Indexed Set",
"Definition:Projection (Mapping Theory)/Family of Sets",
"Definition:Product Topology",
"Definition:Indexing Set",
"Definition:Topology"
] |
proofwiki-6840 | Strictly Positive Integer Power Function is Unbounded Above/General Case | Let $\struct {R, +, \circ, \le}$ be a totally ordered ring with unity.
Suppose that $R$ has no upper bound.
Let $n \in \N_{>0}$.
Let $f: R \to R$ be defined by:
:$\map f x = \circ^n x$
Then the image of $f$ is unbounded above in $R$. | Let $1_R$ be the unity of $R$.
Let $b \in R$.
We will show that $b$ is not an upper bound of the image of $f$.
Since $R$ is totally ordered and unbounded above, there is an element $c \in R$ such that $b < c$ and $1_R < c$.
By Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element, $c \l... | Let $\struct {R, +, \circ, \le}$ be a [[Definition:Totally Ordered Ring|totally ordered ring]] [[Definition:Ring with Unity|with unity]].
Suppose that $R$ has no [[Definition:Upper Bound of Set|upper bound]].
Let $n \in \N_{>0}$.
Let $f: R \to R$ be defined by:
:$\map f x = \circ^n x$
Then the [[Definition:Image o... | Let $1_R$ be the [[Definition:Unity of Ring|unity]] of $R$.
Let $b \in R$.
We will show that $b$ is not an [[Definition:Upper Bound of Set|upper bound]] of the [[Definition:Image of Mapping|image]] of $f$.
Since $R$ is [[Definition:Totally Ordered Ring|totally ordered]] and [[Definition:Unbounded Above Set|unbounded... | Strictly Positive Integer Power Function is Unbounded Above/General Case | https://proofwiki.org/wiki/Strictly_Positive_Integer_Power_Function_is_Unbounded_Above/General_Case | https://proofwiki.org/wiki/Strictly_Positive_Integer_Power_Function_is_Unbounded_Above/General_Case | [
"Totally Ordered Rings",
"Rings with Unity"
] | [
"Definition:Totally Ordered Ring",
"Definition:Ring with Unity",
"Definition:Upper Bound of Set",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Bounded Above Set/Unbounded"
] | [
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Upper Bound of Set",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Totally Ordered Ring",
"Definition:Bounded Above Set/Unbounded",
"Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element",
"Definition... |
proofwiki-6841 | Strictly Positive Integer Power Function is Unbounded Above | Let $\R$ be the real numbers with the usual ordering.
Let $n \in \N_{>0}$.
Let $f: \R \to \R$ be defined by:
:$\map f x = x^n$
Then $f$ is unbounded above. | If $n = 1$, then $f$ is the identity function.
By the Axiom of Archimedes, the real numbers are unbounded above.
Thus by definition of the identity function:
$f$ is unbounded above.
{{qed|lemma}}
Let $n \ge 2$.
{{AimForCont}} that $f$ is bounded above by $b \in \R$.
{{WLOG}} suppose that $b > 0$.
Then by the definition... | Let $\R$ be the [[Definition:Real Number|real numbers]] with the [[Definition:Usual Ordering|usual ordering]].
Let $n \in \N_{>0}$.
Let $f: \R \to \R$ be defined by:
:$\map f x = x^n$
Then $f$ is [[Definition:Unbounded Above Real-Valued Function|unbounded above]]. | If $n = 1$, then $f$ is the [[Definition:Identity Mapping|identity function]].
By the [[Axiom of Archimedes]], the [[Definition:Real Number|real numbers]] are [[Definition:Unbounded Above Subset of Real Numbers|unbounded above]].
Thus by definition of the [[Definition:Identity Mapping|identity function]]:
$f$ is [[D... | Strictly Positive Integer Power Function is Unbounded Above | https://proofwiki.org/wiki/Strictly_Positive_Integer_Power_Function_is_Unbounded_Above | https://proofwiki.org/wiki/Strictly_Positive_Integer_Power_Function_is_Unbounded_Above | [
"Real Analysis"
] | [
"Definition:Real Number",
"Definition:Usual Ordering",
"Definition:Bounded Above Mapping/Real-Valued/Unbounded"
] | [
"Definition:Identity Mapping",
"Axiom of Archimedes",
"Definition:Real Number",
"Definition:Bounded Above Set/Real Numbers/Unbounded",
"Definition:Identity Mapping",
"Definition:Bounded Above Mapping/Real-Valued/Unbounded",
"Definition:Bounded Above Set/Real Numbers",
"Definition:Upper Bound of Set/Re... |
proofwiki-6842 | Ring Product preserves Inequalities on Positive Elements | Let $\struct {R, +, \circ, \le}$ be an ordered ring.
Let $x, y, z, w \in R$.
Let $0 < x < y$ and $0 < z < w$.
Then:
:$0 < z \circ x < w \circ y$ | By Properties of Ordered Ring $(6)$:
:$z \circ x < z \circ y$
:$z \circ y < w \circ y$
Then by transitivity of $\circ$:
:$z \circ x < w \circ y$
Also by Properties of Ordered Ring $(6)$:
:$z \circ 0 < z \circ x$
Hence by Ring Product with Zero:
:$0 < z \circ x$
{{qed}}
Category:Ordered Rings
Category:Inequalities
e0clz... | Let $\struct {R, +, \circ, \le}$ be an [[Definition:Ordered Ring|ordered ring]].
Let $x, y, z, w \in R$.
Let $0 < x < y$ and $0 < z < w$.
Then:
:$0 < z \circ x < w \circ y$ | By [[Properties of Ordered Ring]] $(6)$:
:$z \circ x < z \circ y$
:$z \circ y < w \circ y$
Then by [[Definition:Transitive Relation|transitivity]] of $\circ$:
:$z \circ x < w \circ y$
Also by [[Properties of Ordered Ring]] $(6)$:
:$z \circ 0 < z \circ x$
Hence by [[Ring Product with Zero]]:
:$0 < z \circ x$
{{qe... | Ring Product preserves Inequalities on Positive Elements | https://proofwiki.org/wiki/Ring_Product_preserves_Inequalities_on_Positive_Elements | https://proofwiki.org/wiki/Ring_Product_preserves_Inequalities_on_Positive_Elements | [
"Ordered Rings",
"Inequalities"
] | [
"Definition:Ordered Ring"
] | [
"Properties of Ordered Ring",
"Definition:Transitive Relation",
"Properties of Ordered Ring",
"Ring Product with Zero",
"Category:Ordered Rings",
"Category:Inequalities"
] |
proofwiki-6843 | Power Function is Strictly Increasing on Positive Elements | Let $\struct {R, +, \circ, \le}$ be an ordered ring.
Let $x, y \in R$.
Let $n \in \N_{>0}$ be a strictly positive integer.
Let $0 < x < y$.
Then:
:$0 < \map {\circ^n} x < \map {\circ^n} y$ | Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$0 < \map {\circ^n} x < \map {\circ^n} y$ | Let $\struct {R, +, \circ, \le}$ be an [[Definition:Ordered Ring|ordered ring]].
Let $x, y \in R$.
Let $n \in \N_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Let $0 < x < y$.
Then:
:$0 < \map {\circ^n} x < \map {\circ^n} y$ | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$0 < \map {\circ^n} x < \map {\circ^n} y$ | Power Function is Strictly Increasing on Positive Elements | https://proofwiki.org/wiki/Power_Function_is_Strictly_Increasing_on_Positive_Elements | https://proofwiki.org/wiki/Power_Function_is_Strictly_Increasing_on_Positive_Elements | [
"Ordered Rings",
"Proofs by Induction"
] | [
"Definition:Ordered Ring",
"Definition:Strictly Positive/Integer"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-6844 | Biconditional is Associative/Formulation 1 | :$p \iff \paren {q \iff r} \dashv \vdash \paren {p \iff q} \iff r$ | Proof of associativity by natural deduction is just too tedious to be considered.
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective match for all boolean interpretations.
$\begin{array}{|ccccc||ccccc|} \hline
p & \iff & (q & \iff & r) & (p & \iff & q) & \iff &... | :$p \iff \paren {q \iff r} \dashv \vdash \paren {p \iff q} \iff r$ | Proof of associativity by [[Definition:Natural Deduction|natural deduction]] is just too tedious to be considered.
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]] match for... | Biconditional is Associative/Formulation 1 | https://proofwiki.org/wiki/Biconditional_is_Associative/Formulation_1 | https://proofwiki.org/wiki/Biconditional_is_Associative/Formulation_1 | [
"Biconditional is Associative"
] | [] | [
"Definition:Natural Deduction",
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6845 | Biconditional is Associative/Formulation 2 | :$\vdash \paren {p \iff \paren {q \iff r} } \iff \paren {\paren {p \iff q} \iff r}$ | {{BeginTableau|\vdash \paren {p \iff \paren {q \iff r} } \iff \paren {\paren {p \iff q} \iff r} }}
{{Assumption|1|p \iff \paren {q \iff r} }}
{{SequentIntro|2|1|\paren {p \iff q} \iff r|1|Biconditional is Associative: Formulation 1}}
{{Implication|3||\paren {p \iff \paren {q \iff r} } \implies \paren {\paren {p \iff q}... | :$\vdash \paren {p \iff \paren {q \iff r} } \iff \paren {\paren {p \iff q} \iff r}$ | {{BeginTableau|\vdash \paren {p \iff \paren {q \iff r} } \iff \paren {\paren {p \iff q} \iff r} }}
{{Assumption|1|p \iff \paren {q \iff r} }}
{{SequentIntro|2|1|\paren {p \iff q} \iff r|1|[[Biconditional is Associative/Formulation 1|Biconditional is Associative: Formulation 1]]}}
{{Implication|3||\paren {p \iff \paren ... | Biconditional is Associative/Formulation 2 | https://proofwiki.org/wiki/Biconditional_is_Associative/Formulation_2 | https://proofwiki.org/wiki/Biconditional_is_Associative/Formulation_2 | [
"Biconditional is Associative"
] | [] | [
"Biconditional is Associative/Formulation 1",
"Biconditional is Associative/Formulation 1"
] |
proofwiki-6846 | Open Ball is Convex Set | Let $V$ be a normed vector space with norm $\norm {\,\cdot\,}$ over $\R$ or $\C$.
An open ball in the metric induced by $\norm {\,\cdot\,}$ is a convex set. | Let $v \in V$ and $\epsilon \in \R_{>0}$.
Denote the open $\epsilon$-ball of $v$ as $\map {B_\epsilon} v$.
Let $x, y \in \map {B_\epsilon} v$.
Then $x + t \paren {y - x}$ lies on line segment joining $x$ and $y$ for all $t \in \closedint 0 1$.
The distance between $x + t \paren {y - x}$ and $v$ is:
{{begin-eqn}}
{{eqn ... | Let $V$ be a [[Definition:Normed Vector Space|normed vector space]] with [[Definition:Norm on Vector Space|norm]] $\norm {\,\cdot\,}$ over $\R$ or $\C$.
An [[Definition:Open Ball|open ball]] in the [[Definition:Metric Induced by Norm|metric induced by $\norm {\,\cdot\,}$]] is a [[Definition:Convex Set (Vector Space)|... | Let $v \in V$ and $\epsilon \in \R_{>0}$.
Denote the [[Definition:Open Ball|open $\epsilon$-ball]] of $v$ as $\map {B_\epsilon} v$.
Let $x, y \in \map {B_\epsilon} v$.
Then $x + t \paren {y - x}$ lies on [[Definition:Convex Set (Vector Space)|line segment]] joining $x$ and $y$ for all $t \in \closedint 0 1$.
The di... | Open Ball is Convex Set | https://proofwiki.org/wiki/Open_Ball_is_Convex_Set | https://proofwiki.org/wiki/Open_Ball_is_Convex_Set | [
"Normed Vector Spaces",
"Open Balls",
"Convex Sets (Vector Spaces)"
] | [
"Definition:Normed Vector Space",
"Definition:Norm/Vector Space",
"Definition:Open Ball",
"Definition:Metric Induced by Norm",
"Definition:Convex Set (Vector Space)"
] | [
"Definition:Open Ball",
"Definition:Convex Set (Vector Space)",
"Definition:Convex Set (Vector Space)"
] |
proofwiki-6847 | Pseudometric Space is Metric Space iff T0 | Let $M = \struct {S, d}$ be a pseudometric space.
Let $T = \struct {S, \tau}$ be the topological space over $S$ induced by $d$.
Then $M$ is a metric space {{iff}} $T$ is a $T_0$ space. | === Necessary Condition ===
Let $M$ be a metric space.
From Metric Space is Hausdorff, $M$ is a Hausdorff space.
From:
:$T_2$ (Hausdorff) Space is $T_1$ Space
and:
:$T_1$ Space is $T_0$ Space
it follows that $M$ is a $T_0$ space.
{{qed|lemma}} | Let $M = \struct {S, d}$ be a [[Definition:Pseudometric Space|pseudometric space]].
Let $T = \struct {S, \tau}$ be the [[Definition:Topological Space|topological space]] over $S$ [[Pseudometric induces Topology|induced by $d$]].
Then $M$ is a [[Definition:Metric Space|metric space]] {{iff}} $T$ is a [[Definition:T0 ... | === Necessary Condition ===
Let $M$ be a [[Definition:Metric Space|metric space]].
From [[Metric Space is Hausdorff]], $M$ is a [[Definition:Hausdorff Space|Hausdorff space]].
From:
:[[T2 Space is T1 Space|$T_2$ (Hausdorff) Space is $T_1$ Space]]
and:
:[[T1 Space is T0 Space|$T_1$ Space is $T_0$ Space]]
it follows t... | Pseudometric Space is Metric Space iff T0 | https://proofwiki.org/wiki/Pseudometric_Space_is_Metric_Space_iff_T0 | https://proofwiki.org/wiki/Pseudometric_Space_is_Metric_Space_iff_T0 | [
"Metric Spaces",
"Pseudometric Spaces",
"T0 Spaces"
] | [
"Definition:Pseudometric/Pseudometric Space",
"Definition:Topological Space",
"Pseudometric induces Topology",
"Definition:Metric Space",
"Definition:T0 Space"
] | [
"Definition:Metric Space",
"Metric Space is T2",
"Definition:T2 Space",
"T2 Space is T1",
"T1 Space is T0",
"Definition:T0 Space",
"Definition:T0 Space",
"Definition:Metric Space"
] |
proofwiki-6848 | Sequence Converges to Point Relative to Metric iff it Converges Relative to Induced Topology | Let $M = \struct {S, d}$ be a metric space or a pseudometric space.
Let $T = \struct {S, \tau}$ be the topological space induced by $d$.
Let $\sequence {x_n}$ be a infinite sequence in $S$.
Let $l \in S$.
Then $\sequence {x_n}$ converges to $l$ relative to $d$ {{iff}} $\sequence {x_n}$ converges to $l$ relative to $\ta... | === Necessary Condition ===
Suppose that $\sequence {x_n}$ converges to $l$ relative to $d$.
When $\map {B_\epsilon} l$ denotes the open $\epsilon$-ball of $l$, this means:
:$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies x_n \in \map {B_\epsilon} l$
Let $U \in \tau$ with $l \in U$.
By definition of ind... | Let $M = \struct {S, d}$ be a [[Definition:Metric Space|metric space]] or a [[Definition:Pseudometric Space|pseudometric space]].
Let $T = \struct {S, \tau}$ be the [[Definition:Topology Induced by Metric|topological space induced by $d$]].
Let $\sequence {x_n}$ be a [[Definition:Infinite Sequence|infinite sequence]]... | === Necessary Condition ===
Suppose that $\sequence {x_n}$ [[Definition:Convergent Sequence (Metric Space)|converges]] to $l$ relative to $d$.
When $\map {B_\epsilon} l$ denotes the [[Definition:Open Ball|open $\epsilon$-ball]] of $l$, this means:
:$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies x_n ... | Sequence Converges to Point Relative to Metric iff it Converges Relative to Induced Topology | https://proofwiki.org/wiki/Sequence_Converges_to_Point_Relative_to_Metric_iff_it_Converges_Relative_to_Induced_Topology | https://proofwiki.org/wiki/Sequence_Converges_to_Point_Relative_to_Metric_iff_it_Converges_Relative_to_Induced_Topology | [
"Metric Spaces"
] | [
"Definition:Metric Space",
"Definition:Pseudometric/Pseudometric Space",
"Definition:Topology Induced by Metric",
"Definition:Sequence/Infinite Sequence",
"Definition:Convergent Sequence/Metric Space",
"Definition:Convergent Sequence/Topology"
] | [
"Definition:Convergent Sequence/Metric Space",
"Definition:Open Ball",
"Definition:Topology Induced by Metric",
"Definition:Convergent Sequence/Topology",
"Definition:Topology Induced by Metric",
"Definition:Convergent Sequence/Topology",
"Definition:Convergent Sequence/Metric Space"
] |
proofwiki-6849 | Isometry Preserves Sequence Convergence | Let $M_1 = \struct {S_1, d_1}$ and $M_2 = \struct {S_2, d_2}$ both be metric spaces or pseudometric spaces.
Let $\phi: S_1 \to S_2$ be an isometry.
Let $\sequence {x_n}$ be an infinite sequence in $S_1$.
Suppose that $\sequence {x_n}$ converges to a point $p \in S_1$.
Then $\sequence {\map \phi {x_n}}$ converges to $\m... | {{begin-eqn}}
{{eqn | o =
| r = \lim_{n \mathop \to \infty} \map {d_2} {\map \phi {x_n}, \map \phi p}
}}
{{eqn | r = \lim_{n \mathop \to \infty} \map {d_1} {x_n, p}
| c = {{Defof|Isometry (Metric Spaces)|index = 1}}
}}
{{eqn | r = 0
| c = {{Defof|Convergent Sequence|subdef = Metric Space|index = 3}}
}... | Let $M_1 = \struct {S_1, d_1}$ and $M_2 = \struct {S_2, d_2}$ both be [[Definition:Metric Space|metric spaces]] or [[Definition:Pseudometric Space|pseudometric spaces]].
Let $\phi: S_1 \to S_2$ be an [[Definition:Isometry (Metric Spaces)|isometry]].
Let $\sequence {x_n}$ be an [[Definition:Infinite Sequence|infinite ... | {{begin-eqn}}
{{eqn | o =
| r = \lim_{n \mathop \to \infty} \map {d_2} {\map \phi {x_n}, \map \phi p}
}}
{{eqn | r = \lim_{n \mathop \to \infty} \map {d_1} {x_n, p}
| c = {{Defof|Isometry (Metric Spaces)|index = 1}}
}}
{{eqn | r = 0
| c = {{Defof|Convergent Sequence|subdef = Metric Space|index = 3}}
}... | Isometry Preserves Sequence Convergence/Proof 1 | https://proofwiki.org/wiki/Isometry_Preserves_Sequence_Convergence | https://proofwiki.org/wiki/Isometry_Preserves_Sequence_Convergence/Proof_1 | [
"Isometry Preserves Sequence Convergence",
"Isometries (Metric Spaces)",
"Sequences"
] | [
"Definition:Metric Space",
"Definition:Pseudometric/Pseudometric Space",
"Definition:Isometry (Metric Spaces)",
"Definition:Sequence/Infinite Sequence",
"Definition:Convergent Sequence/Metric Space",
"Definition:Convergent Sequence/Metric Space"
] | [
"Definition:Convergent Sequence/Metric Space/Definition 3"
] |
proofwiki-6850 | Isometry Preserves Sequence Convergence | Let $M_1 = \struct {S_1, d_1}$ and $M_2 = \struct {S_2, d_2}$ both be metric spaces or pseudometric spaces.
Let $\phi: S_1 \to S_2$ be an isometry.
Let $\sequence {x_n}$ be an infinite sequence in $S_1$.
Suppose that $\sequence {x_n}$ converges to a point $p \in S_1$.
Then $\sequence {\map \phi {x_n}}$ converges to $\m... | {{proof wanted|proof from the fact that an isometry is a homeomorphism}} | Let $M_1 = \struct {S_1, d_1}$ and $M_2 = \struct {S_2, d_2}$ both be [[Definition:Metric Space|metric spaces]] or [[Definition:Pseudometric Space|pseudometric spaces]].
Let $\phi: S_1 \to S_2$ be an [[Definition:Isometry (Metric Spaces)|isometry]].
Let $\sequence {x_n}$ be an [[Definition:Infinite Sequence|infinite ... | {{proof wanted|proof from the fact that an isometry is a homeomorphism}} | Isometry Preserves Sequence Convergence/Proof 2 | https://proofwiki.org/wiki/Isometry_Preserves_Sequence_Convergence | https://proofwiki.org/wiki/Isometry_Preserves_Sequence_Convergence/Proof_2 | [
"Isometry Preserves Sequence Convergence",
"Isometries (Metric Spaces)",
"Sequences"
] | [
"Definition:Metric Space",
"Definition:Pseudometric/Pseudometric Space",
"Definition:Isometry (Metric Spaces)",
"Definition:Sequence/Infinite Sequence",
"Definition:Convergent Sequence/Metric Space",
"Definition:Convergent Sequence/Metric Space"
] | [] |
proofwiki-6851 | Connected Domain is Connected by Staircase Contours | Let $D \subseteq \C$ be an open set.
Then $D$ is a connected domain {{iff}}:
:for all $z, w \in \C$, there exists a staircase contour in $D$ with start point $z$ and end point $w$. | === Necessary Condition ===
Suppose $D$ is a connected domain.
If $z, w \in D$, there exists a path $\gamma: \closedint 0 1 \to D$ with $\map \gamma 0 = z$ and $\map \gamma 0 = w$.
From the Paving Lemma, it follows that there exist $\epsilon \in \R_{>0}$ and a subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint 0... | Let $D \subseteq \C$ be an [[Definition:Open Set (Complex Analysis)|open set]].
Then $D$ is a [[Definition:Connected Domain (Complex Analysis)|connected domain]] {{iff}}:
:for all $z, w \in \C$, there exists a [[Definition:Staircase Contour|staircase contour]] in $D$ with [[Definition:Start Point of Contour (Complex... | === Necessary Condition ===
Suppose $D$ is a [[Definition:Connected Domain (Complex Analysis)|connected domain]].
If $z, w \in D$, there exists a [[Definition:Path (Topology)|path]] $\gamma: \closedint 0 1 \to D$ with $\map \gamma 0 = z$ and $\map \gamma 0 = w$.
From the [[Paving Lemma]], it follows that there exist... | Connected Domain is Connected by Staircase Contours | https://proofwiki.org/wiki/Connected_Domain_is_Connected_by_Staircase_Contours | https://proofwiki.org/wiki/Connected_Domain_is_Connected_by_Staircase_Contours | [
"Complex Analysis"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Connected Domain (Complex Analysis)",
"Definition:Staircase Contour",
"Definition:Contour/Endpoints/Complex Plane",
"Definition:Contour/Endpoints/Complex Plane"
] | [
"Definition:Connected Domain (Complex Analysis)",
"Definition:Path (Topology)",
"Paving Lemma",
"Definition:Subdivision of Interval",
"Definition:Open Ball",
"Definition:Complex Number/Real Part",
"Definition:Complex Number/Imaginary Part",
"Definition:Smooth Path/Complex",
"Definition:Convex Set (V... |
proofwiki-6852 | Dedekind Completeness is Self-Dual | Let $\struct {S, \preceq}$ be an ordered set.
Then $\struct {S, \preceq}$ is Dedekind complete {{iff}} every non-empty subset of $S$ that is bounded below admits an infimum in $S$.
That is, an ordered set is Dedekind complete {{iff}} its dual is Dedekind complete. | === Necessary Condition ===
Let $\struct {S, \preceq}$ be Dedekind complete.
Let $A \subseteq S$ be non-empty and bounded below.
Let $B \subseteq S$ be set of all lower bounds for $A$.
Then every element of $A$ is an upper bound for $B$.
Therefore, $B$ is non-empty and bounded above.
By the definition of Dedekind compl... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Then $\struct {S, \preceq}$ is [[Definition:Dedekind Complete|Dedekind complete]] {{iff}} every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$ that is [[Definition:Bounded Below Set|bounded below]] admits an [[Definit... | === Necessary Condition ===
Let $\struct {S, \preceq}$ be [[Definition:Dedekind Complete|Dedekind complete]].
Let $A \subseteq S$ be [[Definition:Non-Empty Set|non-empty]] and [[Definition:Bounded Below Set|bounded below]].
Let $B \subseteq S$ be [[Definition:Set|set]] of all [[Definition:Lower Bound of Set|lower b... | Dedekind Completeness is Self-Dual | https://proofwiki.org/wiki/Dedekind_Completeness_is_Self-Dual | https://proofwiki.org/wiki/Dedekind_Completeness_is_Self-Dual | [
"Order Theory"
] | [
"Definition:Ordered Set",
"Definition:Dedekind Completeness Property",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Bounded Below Set",
"Definition:Infimum of Set",
"Definition:Ordered Set",
"Definition:Dedekind Completeness Property",
"Definition:Dual Ordering/Dual Ordered Set",
"... | [
"Definition:Dedekind Completeness Property",
"Definition:Non-Empty Set",
"Definition:Bounded Below Set",
"Definition:Set",
"Definition:Lower Bound of Set",
"Definition:Element",
"Definition:Upper Bound of Set",
"Definition:Non-Empty Set",
"Definition:Bounded Above Set",
"Definition:Dedekind Comple... |
proofwiki-6853 | Biconditional is Transitive/Formulation 1 | :$p \iff q, q \iff r \vdash p \iff r$ | {{BeginTableau|p \iff q, q \iff r \vdash p \iff r}}
{{Premise|1|p \iff q}}
{{Premise|2|q \iff r}}
{{BiconditionalElimination|3|1|p \implies q|1|1}}
{{BiconditionalElimination|4|2|q \implies r|2|1}}
{{SequentIntro|5|1, 2|p \implies r|1, 2|Hypothetical Syllogism: Formulation 1}}
{{BiconditionalElimination|6|1|q \implies ... | :$p \iff q, q \iff r \vdash p \iff r$ | {{BeginTableau|p \iff q, q \iff r \vdash p \iff r}}
{{Premise|1|p \iff q}}
{{Premise|2|q \iff r}}
{{BiconditionalElimination|3|1|p \implies q|1|1}}
{{BiconditionalElimination|4|2|q \implies r|2|1}}
{{SequentIntro|5|1, 2|p \implies r|1, 2|[[Hypothetical Syllogism/Formulation 1|Hypothetical Syllogism: Formulation 1]]}}
{... | Biconditional is Transitive/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Biconditional_is_Transitive/Formulation_1 | https://proofwiki.org/wiki/Biconditional_is_Transitive/Formulation_1/Proof_1 | [
"Biconditional is Transitive"
] | [] | [
"Hypothetical Syllogism/Formulation 1",
"Hypothetical Syllogism/Formulation 1"
] |
proofwiki-6854 | Biconditional is Transitive/Formulation 1 | :$p \iff q, q \iff r \vdash p \iff r$ | We apply the Method of Truth Tables.
As can be seen for all boolean interpretations by inspection, where the truth values under the main connective on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$:
:<nowiki>$\begin {array} {|ccccccc||ccc|} \hline
(p & \iff & q) & \land & (q & \iff & r) & p & \iff & ... | :$p \iff q, q \iff r \vdash p \iff r$ | We apply the [[Method of Truth Tables]].
As can be seen for all [[Definition:Boolean Interpretation|boolean interpretations]] by inspection, where the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] on the {{LHS}} is $T$, that under the one on the ... | Biconditional is Transitive/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Biconditional_is_Transitive/Formulation_1 | https://proofwiki.org/wiki/Biconditional_is_Transitive/Formulation_1/Proof_by_Truth_Table | [
"Biconditional is Transitive"
] | [] | [
"Method of Truth Tables",
"Definition:Boolean Interpretation",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic"
] |
proofwiki-6855 | Biconditional is Transitive/Formulation 2 | :$\vdash \paren {\paren {p \iff q} \land \paren {q \iff r} } \implies \paren {p \iff r}$ | {{BeginTableau|\vdash \paren {\paren {p \iff q} \land \paren {q \iff r} } \implies \paren {p \iff r} }}
{{Assumption|1|\paren {p \iff q} \land \paren {q \iff r} }}
{{Simplification|2|1|p \iff q|1|1}}
{{Simplification|3|1|q \iff r|1|2}}
{{SequentIntro|4|1|p \iff r|2, 3|Biconditional is Transitive: Formulation 1}}
{{Impl... | :$\vdash \paren {\paren {p \iff q} \land \paren {q \iff r} } \implies \paren {p \iff r}$ | {{BeginTableau|\vdash \paren {\paren {p \iff q} \land \paren {q \iff r} } \implies \paren {p \iff r} }}
{{Assumption|1|\paren {p \iff q} \land \paren {q \iff r} }}
{{Simplification|2|1|p \iff q|1|1}}
{{Simplification|3|1|q \iff r|1|2}}
{{SequentIntro|4|1|p \iff r|2, 3|[[Biconditional is Transitive/Formulation 1|Bicondi... | Biconditional is Transitive/Formulation 2 | https://proofwiki.org/wiki/Biconditional_is_Transitive/Formulation_2 | https://proofwiki.org/wiki/Biconditional_is_Transitive/Formulation_2 | [
"Biconditional is Transitive"
] | [] | [
"Biconditional is Transitive/Formulation 1"
] |
proofwiki-6856 | Biconditional as Disjunction of Conjunctions/Formulation 1/Forward Implication | :$p \iff q \vdash \paren {p \land q} \lor \paren {\neg p \land \neg q}$ | {{BeginTableau|p \iff q \vdash \paren {p \land q} \lor \paren {\neg p \land \neg q} }}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|p \implies q|1|1}}
{{BiconditionalElimination|3|1|q \implies p|1|2}}
{{ExcludedMiddle|4|p \lor \neg p}}
{{Assumption|5|p}}
{{ModusPonens|6|1, 5|q|2|5}}
{{Conjunction|7|1, 5|p \lan... | :$p \iff q \vdash \paren {p \land q} \lor \paren {\neg p \land \neg q}$ | {{BeginTableau|p \iff q \vdash \paren {p \land q} \lor \paren {\neg p \land \neg q} }}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|p \implies q|1|1}}
{{BiconditionalElimination|3|1|q \implies p|1|2}}
{{ExcludedMiddle|4|p \lor \neg p}}
{{Assumption|5|p}}
{{ModusPonens|6|1, 5|q|2|5}}
{{Conjunction|7|1, 5|p \lan... | Biconditional as Disjunction of Conjunctions/Formulation 1/Forward Implication | https://proofwiki.org/wiki/Biconditional_as_Disjunction_of_Conjunctions/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Biconditional_as_Disjunction_of_Conjunctions/Formulation_1/Forward_Implication | [
"Biconditional as Disjunction of Conjunctions"
] | [] | [
"Category:Biconditional as Disjunction of Conjunctions"
] |
proofwiki-6857 | Biconditional as Disjunction of Conjunctions/Formulation 1/Reverse Implication | : $\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right) \vdash p \iff q$ | {{BeginTableau|\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right) \vdash p \iff q}}
{{Premise|1|\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)}}
{{Assumption|2|p \land q}}
{{Assumption|3|p}}
{{Simplification|4|2|q|2|2}}
{{Implication|5|2|p \implies q|3|4}}
{{Assumption|6|q}}
{{Simplifica... | : $\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right) \vdash p \iff q$ | {{BeginTableau|\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right) \vdash p \iff q}}
{{Premise|1|\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)}}
{{Assumption|2|p \land q}}
{{Assumption|3|p}}
{{Simplification|4|2|q|2|2}}
{{Implication|5|2|p \implies q|3|4}}
{{Assumption|6|q}}
{{Simplifica... | Biconditional as Disjunction of Conjunctions/Formulation 1/Reverse Implication | https://proofwiki.org/wiki/Biconditional_as_Disjunction_of_Conjunctions/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Biconditional_as_Disjunction_of_Conjunctions/Formulation_1/Reverse_Implication | [
"Biconditional as Disjunction of Conjunctions"
] | [] | [
"Rule of Transposition",
"Rule of Transposition",
"Category:Biconditional as Disjunction of Conjunctions"
] |
proofwiki-6858 | Biconditional as Disjunction of Conjunctions/Formulation 1 | :$p \iff q \dashv \vdash \paren {p \land q} \lor \paren {\neg p \land \neg q}$ | We apply the Method of Truth Tables.
As can be seen by inspection, in all cases the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin {array} {|ccc||ccccccccc|} \hline
p & \iff & q & (p & \land & q) & \lor & (\neg & p & \land & \neg & q) \\
\hline
\F & \T & \F & \F & \F & \... | :$p \iff q \dashv \vdash \paren {p \land q} \lor \paren {\neg p \land \neg q}$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, in all cases the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin {array} {|ccc||c... | Biconditional as Disjunction of Conjunctions/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Biconditional_as_Disjunction_of_Conjunctions/Formulation_1 | https://proofwiki.org/wiki/Biconditional_as_Disjunction_of_Conjunctions/Formulation_1/Proof_by_Truth_Table | [
"Biconditional as Disjunction of Conjunctions"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6859 | Biconditional as Disjunction of Conjunctions/Formulation 2 | :$\vdash \paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} }$ | {{BeginTableau|\vdash \paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} } }}
{{Assumption |1|p \iff q}}
{{SequentIntro |2|1|\paren {p \land q} \lor \paren {\neg p \land \neg q}|1|Biconditional as Disjunction of Conjunctions: Formulation 1}}
{{Implication |3| |\paren {... | :$\vdash \paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} }$ | {{BeginTableau|\vdash \paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} } }}
{{Assumption |1|p \iff q}}
{{SequentIntro |2|1|\paren {p \land q} \lor \paren {\neg p \land \neg q}|1|[[Biconditional as Disjunction of Conjunctions/Formulation 1|Biconditional as Disjunction of Co... | Biconditional as Disjunction of Conjunctions/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Biconditional_as_Disjunction_of_Conjunctions/Formulation_2 | https://proofwiki.org/wiki/Biconditional_as_Disjunction_of_Conjunctions/Formulation_2/Proof_1 | [
"Biconditional as Disjunction of Conjunctions"
] | [] | [
"Biconditional as Disjunction of Conjunctions/Formulation 1",
"Biconditional as Disjunction of Conjunctions/Formulation 1"
] |
proofwiki-6860 | Biconditional as Disjunction of Conjunctions/Formulation 2 | :$\vdash \paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} }$ | We apply the Method of Truth Tables.
As can be seen by inspection, in all cases the truth values under the main connective is true for all boolean interpretations.
$\begin{array}{|ccc|c|ccccccccc|} \hline
(p & \iff & q) & \iff & ((p & \land & q) & \lor & (\neg & p & \land & \neg & q)) \\
\hline
\F & \T & \F & \T & \F &... | :$\vdash \paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} }$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, in all cases 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]].
$\begin{arra... | Biconditional as Disjunction of Conjunctions/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Biconditional_as_Disjunction_of_Conjunctions/Formulation_2 | https://proofwiki.org/wiki/Biconditional_as_Disjunction_of_Conjunctions/Formulation_2/Proof_by_Truth_Table | [
"Biconditional as Disjunction of Conjunctions"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-6861 | Biconditional Equivalent to Biconditional of Negations/Formulation 1/Forward Implication | :$p \iff q \vdash \neg p \iff \neg q$ | {{BeginTableau|p \iff q \vdash \neg p \iff \neg q}}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|p \implies q|1|1}}
{{SequentIntro|3|1|\neg q \implies \neg p|2|Rule of Transposition}}
{{BiconditionalElimination|4|1|q \implies p|1|2}}
{{SequentIntro|5|1|\neg p \implies \neg q|4|Rule of Transposition}}
{{Bicondi... | :$p \iff q \vdash \neg p \iff \neg q$ | {{BeginTableau|p \iff q \vdash \neg p \iff \neg q}}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|p \implies q|1|1}}
{{SequentIntro|3|1|\neg q \implies \neg p|2|[[Rule of Transposition]]}}
{{BiconditionalElimination|4|1|q \implies p|1|2}}
{{SequentIntro|5|1|\neg p \implies \neg q|4|[[Rule of Transposition]]}}
{... | Biconditional Equivalent to Biconditional of Negations/Formulation 1/Forward Implication | https://proofwiki.org/wiki/Biconditional_Equivalent_to_Biconditional_of_Negations/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Biconditional_Equivalent_to_Biconditional_of_Negations/Formulation_1/Forward_Implication | [
"Biconditional Equivalent to Biconditional of Negations"
] | [] | [
"Rule of Transposition",
"Rule of Transposition"
] |
proofwiki-6862 | Biconditional Equivalent to Biconditional of Negations/Formulation 1/Reverse Implication | :$\neg p \iff \neg q \vdash p \iff q$ | {{BeginTableau|\neg p \iff \neg q \vdash p \iff q}}
{{Premise|1|\neg p \iff \neg q}}
{{BiconditionalElimination|2|1|\neg p \implies \neg q|1|1}}
{{SequentIntro|3|1|\neg \neg q \implies \neg \neg p|2|Rule of Transposition}}
{{DoubleNegElimination|4|1|q \implies p|3|(twice)}}
{{BiconditionalElimination|5|1|\neg q \implie... | :$\neg p \iff \neg q \vdash p \iff q$ | {{BeginTableau|\neg p \iff \neg q \vdash p \iff q}}
{{Premise|1|\neg p \iff \neg q}}
{{BiconditionalElimination|2|1|\neg p \implies \neg q|1|1}}
{{SequentIntro|3|1|\neg \neg q \implies \neg \neg p|2|[[Rule of Transposition]]}}
{{DoubleNegElimination|4|1|q \implies p|3|(twice)}}
{{BiconditionalElimination|5|1|\neg q \im... | Biconditional Equivalent to Biconditional of Negations/Formulation 1/Reverse Implication | https://proofwiki.org/wiki/Biconditional_Equivalent_to_Biconditional_of_Negations/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Biconditional_Equivalent_to_Biconditional_of_Negations/Formulation_1/Reverse_Implication | [
"Biconditional Equivalent to Biconditional of Negations"
] | [] | [
"Rule of Transposition",
"Rule of Transposition"
] |
proofwiki-6863 | Biconditional iff Disjunction implies Conjunction/Formulation 1/Forward Implication | :$p \iff q \vdash \paren {p \lor q} \implies \paren {p \land q}$ | {{BeginTableau|p \iff q \vdash \paren {p \lor q} \implies \paren {p \land q} }}
{{Premise |1|p \iff q}}
{{SequentIntro|2|1|\paren {p \land q} \lor \paren {\neg p \land \neg q}|1|Biconditional as Disjunction of Conjunctions}}
{{DeMorgan |3|1|\paren {p \land q} \lor \neg \paren {p \lor q}|2|Conjunction of Negation... | :$p \iff q \vdash \paren {p \lor q} \implies \paren {p \land q}$ | {{BeginTableau|p \iff q \vdash \paren {p \lor q} \implies \paren {p \land q} }}
{{Premise |1|p \iff q}}
{{SequentIntro|2|1|\paren {p \land q} \lor \paren {\neg p \land \neg q}|1|[[Biconditional as Disjunction of Conjunctions]]}}
{{DeMorgan |3|1|\paren {p \land q} \lor \neg \paren {p \lor q}|2|Conjunction of Nega... | Biconditional iff Disjunction implies Conjunction/Formulation 1/Forward Implication | https://proofwiki.org/wiki/Biconditional_iff_Disjunction_implies_Conjunction/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Biconditional_iff_Disjunction_implies_Conjunction/Formulation_1/Forward_Implication | [
"Biconditional iff Disjunction implies Conjunction"
] | [] | [
"Biconditional as Disjunction of Conjunctions",
"Rule of Material Implication",
"Category:Biconditional iff Disjunction implies Conjunction"
] |
proofwiki-6864 | Biconditional iff Disjunction implies Conjunction/Formulation 1/Reverse Implication | :$\paren {p \lor q} \implies \paren {p \land q} \vdash p \iff q$ | {{BeginTableau|\paren {p \lor q} \implies \paren {p \land q} \vdash p \iff q}}
{{Premise|1|\paren {p \lor q} \implies \paren {p \land q} }}
{{SequentIntro|2|1|\neg \paren {p \lor q} \lor \paren {p \land q}|1|Rule of Material Implication}}
{{Commutation|3|1|\paren {p \land q} \lor \neg \paren {p \lor q}|2|Disjunction}}
... | :$\paren {p \lor q} \implies \paren {p \land q} \vdash p \iff q$ | {{BeginTableau|\paren {p \lor q} \implies \paren {p \land q} \vdash p \iff q}}
{{Premise|1|\paren {p \lor q} \implies \paren {p \land q} }}
{{SequentIntro|2|1|\neg \paren {p \lor q} \lor \paren {p \land q}|1|[[Rule of Material Implication]]}}
{{Commutation|3|1|\paren {p \land q} \lor \neg \paren {p \lor q}|2|Disjunctio... | Biconditional iff Disjunction implies Conjunction/Formulation 1/Reverse Implication | https://proofwiki.org/wiki/Biconditional_iff_Disjunction_implies_Conjunction/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Biconditional_iff_Disjunction_implies_Conjunction/Formulation_1/Reverse_Implication | [
"Biconditional iff Disjunction implies Conjunction"
] | [] | [
"Rule of Material Implication",
"Biconditional as Disjunction of Conjunctions",
"Rule of Material Equivalence",
"Category:Biconditional iff Disjunction implies Conjunction"
] |
proofwiki-6865 | Biconditional iff Disjunction implies Conjunction/Formulation 1 | :$p \iff q \dashv \vdash \paren {p \lor q} \implies \paren {p \land q}$ | We apply the Method of Truth Tables.
As can be seen by inspection, in all cases the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin {array} {|ccc||ccccccc|} \hline
p & \iff & q & (p & \lor & q) & \implies & (p & \land & q) \\
\hline
\F & \T & \F & \F & \F & \F & \T & \F &... | :$p \iff q \dashv \vdash \paren {p \lor q} \implies \paren {p \land q}$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, in all cases the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin {array} {|ccc||c... | Biconditional iff Disjunction implies Conjunction/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Biconditional_iff_Disjunction_implies_Conjunction/Formulation_1 | https://proofwiki.org/wiki/Biconditional_iff_Disjunction_implies_Conjunction/Formulation_1/Proof_by_Truth_Table | [
"Biconditional iff Disjunction implies Conjunction"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6866 | Biconditional iff Disjunction implies Conjunction/Formulation 2 | :$\vdash \paren {p \iff q} \iff \paren {\paren {p \lor q} \implies \paren {p \land q} }$ | {{BeginTableau|\vdash \paren {p \iff q} \iff \paren {\paren {p \lor q} \implies \paren {p \land q} } }}
{{Assumption|1|p \iff q}}
{{SequentIntro|2|1|\paren {p \lor q} \implies \paren {p \land q}|1|Biconditional iff Disjunction implies Conjunction: Formulation 1}}
{{Implication|3||\paren {p \iff q} \implies \paren {\par... | :$\vdash \paren {p \iff q} \iff \paren {\paren {p \lor q} \implies \paren {p \land q} }$ | {{BeginTableau|\vdash \paren {p \iff q} \iff \paren {\paren {p \lor q} \implies \paren {p \land q} } }}
{{Assumption|1|p \iff q}}
{{SequentIntro|2|1|\paren {p \lor q} \implies \paren {p \land q}|1|[[Biconditional iff Disjunction implies Conjunction/Formulation 1|Biconditional iff Disjunction implies Conjunction: Formul... | Biconditional iff Disjunction implies Conjunction/Formulation 2 | https://proofwiki.org/wiki/Biconditional_iff_Disjunction_implies_Conjunction/Formulation_2 | https://proofwiki.org/wiki/Biconditional_iff_Disjunction_implies_Conjunction/Formulation_2 | [
"Biconditional iff Disjunction implies Conjunction"
] | [] | [
"Biconditional iff Disjunction implies Conjunction/Formulation 1",
"Biconditional iff Disjunction implies Conjunction/Formulation 1",
"Category:Biconditional iff Disjunction implies Conjunction"
] |
proofwiki-6867 | Order of Squares in Totally Ordered Ring without Proper Zero Divisors | Let $\struct {R, +, \circ, \le}$ be a totally ordered ring without proper zero divisors whose zero is $0_R$.
Let $x, y \in R$ be positive, that is, $0_R \le x, y$.
Then $x \le y \iff x \circ x \le y \circ y$.
That is, the square mapping is an order embedding of $\struct {R_{\ge 0}, \le}$ into itself.
When $R$ is one of... | From Order of Squares in Ordered Ring, we have:
:$x \le y \implies x \circ x \le y \circ y$
To prove the opposite implication, we use a Proof by Contradiction.
{{AimForCont}} $x \circ x \le y \circ y$ but $x \not\le y$.
Since $\le$ is a total ordering, this means $y < x$.
Since $0_R \le y$, Extended Transitivity shows ... | Let $\struct {R, +, \circ, \le}$ be a [[Definition:Totally Ordered Ring|totally ordered ring]] without [[Definition:Proper Zero Divisor|proper zero divisors]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Let $x, y \in R$ be [[Definition:Positive|positive]], that is, $0_R \le x, y$.
Then $x \le y \iff x \circ x \le ... | From [[Order of Squares in Ordered Ring]], we have:
:$x \le y \implies x \circ x \le y \circ y$
To prove the opposite [[Definition:Implication|implication]], we use a [[Proof by Contradiction]].
{{AimForCont}} $x \circ x \le y \circ y$ but $x \not\le y$.
Since $\le$ is a [[Definition:Total Ordering|total ordering]... | Order of Squares in Totally Ordered Ring without Proper Zero Divisors | https://proofwiki.org/wiki/Order_of_Squares_in_Totally_Ordered_Ring_without_Proper_Zero_Divisors | https://proofwiki.org/wiki/Order_of_Squares_in_Totally_Ordered_Ring_without_Proper_Zero_Divisors | [
"Totally Ordered Rings"
] | [
"Definition:Totally Ordered Ring",
"Definition:Proper Zero Divisor",
"Definition:Ring Zero",
"Definition:Positive",
"Definition:Square/Mapping",
"Definition:Order Embedding",
"Definition:Number",
"Definition:Positive/Number"
] | [
"Order of Squares in Ordered Ring",
"Definition:Conditional",
"Proof by Contradiction",
"Definition:Total Ordering",
"Extended Transitivity",
"Definition:Ordering Compatible with Ring Structure",
"Definition:Ring (Abstract Algebra)",
"Definition:Positive",
"Ordering Cycle implies Equality",
"Produ... |
proofwiki-6868 | Order of Squares in Ordered Field | Let $\struct {R, +, \circ, \le}$ be an ordered field whose zero is $0_R$ and whose unity is $1_R$.
Suppose that $\forall a \in R: 0 < a \implies 0 < a^{-1}$.
Let $x, y \in \struct {R, +, \circ, \le}$ such that $0_R \le x, y$.
Then $x \le y \iff x \circ x \le y \circ y$.
That is, the square function is an order embeddin... | From Order of Squares in Ordered Ring, we have:
:$x \le y \implies x \circ x \le y \circ y$
To prove the reverse implication, suppose that $x \circ x \le y \circ y$.
Thus:
{{begin-eqn}}
{{eqn | l = x \circ x
| o = \le
| r = y \circ y
| c =
}}
{{eqn | ll=\leadsto
| l = x \circ x + \paren {-\pare... | Let $\struct {R, +, \circ, \le}$ be an [[Definition:Ordered Field|ordered field]] whose [[Definition:Field Zero|zero]] is $0_R$ and whose [[Definition:Unity of Field|unity]] is $1_R$.
Suppose that $\forall a \in R: 0 < a \implies 0 < a^{-1}$.
Let $x, y \in \struct {R, +, \circ, \le}$ such that $0_R \le x, y$.
Then ... | From [[Order of Squares in Ordered Ring]], we have:
:$x \le y \implies x \circ x \le y \circ y$
To prove the reverse implication, suppose that $x \circ x \le y \circ y$.
Thus:
{{begin-eqn}}
{{eqn | l = x \circ x
| o = \le
| r = y \circ y
| c =
}}
{{eqn | ll=\leadsto
| l = x \circ x + \paren... | Order of Squares in Ordered Field | https://proofwiki.org/wiki/Order_of_Squares_in_Ordered_Field | https://proofwiki.org/wiki/Order_of_Squares_in_Ordered_Field | [
"Ordered Fields"
] | [
"Definition:Ordered Field",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Order Embedding",
"Definition:Standard Number Field",
"Definition:Positive"
] | [
"Order of Squares in Ordered Ring",
"Difference of Two Squares",
"Definition:Commutative Ring"
] |
proofwiki-6869 | Law of Identity/Formulation 1 | :$p \vdash p$ | {{BeginTableau|p \vdash p}}
{{Premise|1|p}}
{{EndTableau|qed}}
This is the shortest tableau proof possible. | :$p \vdash p$ | {{BeginTableau|p \vdash p}}
{{Premise|1|p}}
{{EndTableau|qed}}
This is the shortest [[Definition:Tableau Proof (Formal Systems)|tableau proof]] possible. | Law of Identity/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Law_of_Identity/Formulation_1 | https://proofwiki.org/wiki/Law_of_Identity/Formulation_1/Proof_1 | [
"Law of Identity"
] | [] | [
"Definition:Tableau Proof (Natural Deduction)"
] |
proofwiki-6870 | Law of Identity/Formulation 1 | :$p \vdash p$ | We apply the Method of Truth Tables (trivially) to the proposition.
:<nowiki>$\begin{array}{|c|c|} \hline
p & p \\
\hline
\F & \F \\
\T & \T \\
\hline
\end{array}$</nowiki>
{{qed}} | :$p \vdash p$ | We apply the [[Method of Truth Tables]] (trivially) to the proposition.
:<nowiki>$\begin{array}{|c|c|} \hline
p & p \\
\hline
\F & \F \\
\T & \T \\
\hline
\end{array}$</nowiki>
{{qed}} | Law of Identity/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Law_of_Identity/Formulation_1 | https://proofwiki.org/wiki/Law_of_Identity/Formulation_1/Proof_by_Truth_Table | [
"Law of Identity"
] | [] | [
"Method of Truth Tables"
] |
proofwiki-6871 | Law of Identity/Formulation 2 | Every proposition entails itself:
:$\vdash p \implies p$ | {{BeginTableau|\vdash p \implies p}}
{{Premise|1|p}}
{{Implication|2||p \implies p|1|1}}
{{EndTableau|qed}} | Every [[Definition:Proposition|proposition]] entails itself:
:$\vdash p \implies p$ | {{BeginTableau|\vdash p \implies p}}
{{Premise|1|p}}
{{Implication|2||p \implies p|1|1}}
{{EndTableau|qed}} | Law of Identity/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Law_of_Identity/Formulation_2 | https://proofwiki.org/wiki/Law_of_Identity/Formulation_2/Proof_1 | [
"Law of Identity"
] | [
"Definition:Proposition"
] | [] |
proofwiki-6872 | Law of Identity/Formulation 2 | Every proposition entails itself:
:$\vdash p \implies p$ | Using a tableau proof for instance 1 of a Hilbert proof system:
{{BeginTableau|p \implies p|nohead = 1}}
{{TableauLine|n = 1
| f = \paren {p \implies \paren {\paren {p \implies p} \implies p} } \implies \paren {\paren {p \implies \paren {p \implies p} } \implies \paren {p \implies p} }
| rtxt = Axiom 2
| c = $\ma... | Every [[Definition:Proposition|proposition]] entails itself:
:$\vdash p \implies p$ | Using a [[Definition:Tableau Proof (Formal Systems)|tableau proof]] for [[Definition:Hilbert Proof System/Instance 1|instance 1 of a Hilbert proof system]]:
{{BeginTableau|p \implies p|nohead = 1}}
{{TableauLine|n = 1
| f = \paren {p \implies \paren {\paren {p \implies p} \implies p} } \implies \paren {\paren {p \imp... | Law of Identity/Formulation 2/Proof 2 | https://proofwiki.org/wiki/Law_of_Identity/Formulation_2 | https://proofwiki.org/wiki/Law_of_Identity/Formulation_2/Proof_2 | [
"Law of Identity"
] | [
"Definition:Proposition"
] | [
"Definition:Tableau Proof (Natural Deduction)",
"Definition:Hilbert Proof System/Instance 1"
] |
proofwiki-6873 | Law of Identity/Formulation 2 | Every proposition entails itself:
:$\vdash p \implies p$ | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth value under the main connective is $\T$ throughout.
:<nowiki>$\begin{array}{|ccc|} \hline
p & \implies & p \\
\hline
\F & \T & \F \\
\T & \T & \T \\
\hline
\end{array}$</nowiki>
{{qed}} | Every [[Definition:Proposition|proposition]] entails itself:
:$\vdash p \implies p$ | 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 $\T$ throughout.
:<nowiki>$\begin{array}{|ccc|} \hline
p & \implies & p \\
\hline
\F & \T & \F \\
\T & ... | Law of Identity/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Law_of_Identity/Formulation_2 | https://proofwiki.org/wiki/Law_of_Identity/Formulation_2/Proof_by_Truth_Table | [
"Law of Identity"
] | [
"Definition:Proposition"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic"
] |
proofwiki-6874 | Neighborhood Condition for Coarser Topology | Let $S$ be a set.
Let $\tau_1$ and $\tau_2$ be two topologies on $S$.
Suppose that for all $z \in S$ and for all open neighborhoods $N_z$ of $z$ with respect to $\tau_1$, there exists $U \in \tau_2$ such that $U \subseteq N_z$.
Then $\tau_1$ is coarser than $\tau_2$. | Let $V \in \tau_1$.
For all $z \in V$, we have that $V$ is an open neighborhood of $z$ with respect to $\tau_1$.
Then for all $z \in V$, we can find $U_z \in \tau_2$ such that $z \in U_z \subseteq V$.
Then:
:$\ds V = \bigcup_{z \mathop \in V } \set z \subseteq \bigcup_{z \mathop \in V } U_z \subseteq V$
By definition o... | Let $S$ be a [[Definition:Set|set]].
Let $\tau_1$ and $\tau_2$ be two [[Definition:Topology|topologies]] on $S$.
Suppose that for all $z \in S$ and for all [[Definition:Open Neighborhood of Point|open neighborhoods]] $N_z$ of $z$ with respect to $\tau_1$, there exists $U \in \tau_2$ such that $U \subseteq N_z$.
Th... | Let $V \in \tau_1$.
For all $z \in V$, we have that $V$ is an [[Definition:Open Neighborhood of Point|open neighborhood]] of $z$ with respect to $\tau_1$.
Then for all $z \in V$, we can find $U_z \in \tau_2$ such that $z \in U_z \subseteq V$.
Then:
:$\ds V = \bigcup_{z \mathop \in V } \set z \subseteq \bigcup_{z \m... | Neighborhood Condition for Coarser Topology | https://proofwiki.org/wiki/Neighborhood_Condition_for_Coarser_Topology | https://proofwiki.org/wiki/Neighborhood_Condition_for_Coarser_Topology | [
"Topology"
] | [
"Definition:Set",
"Definition:Topology",
"Definition:Open Neighborhood/Point",
"Definition:Coarser Topology"
] | [
"Definition:Open Neighborhood/Point",
"Definition:Set Equality/Definition 2",
"Definition:Coarser Topology",
"Category:Topology"
] |
proofwiki-6875 | Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Proof 2 | :$p \land \paren {p \lor q} \dashv \vdash p$ | By calculation:
{{begin-eqn}}
{{eqn | l = p \land \paren {p \lor q}
| r = \paren {p \lor \bot} \land \paren {p \lor q}
| c = Disjunction with Contradiction
}}
{{eqn | r = p \lor \paren {\bot \land q}
| c = Disjunction is Left Distributive over Conjunction
}}
{{eqn | r = p \lor \bot
| c = Conjunc... | :$p \land \paren {p \lor q} \dashv \vdash p$ | By calculation:
{{begin-eqn}}
{{eqn | l = p \land \paren {p \lor q}
| r = \paren {p \lor \bot} \land \paren {p \lor q}
| c = [[Disjunction with Contradiction]]
}}
{{eqn | r = p \lor \paren {\bot \land q}
| c = [[Disjunction is Left Distributive over Conjunction]]
}}
{{eqn | r = p \lor \bot
| c ... | Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Proof 2 | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Conjunction_Absorbs_Disjunction/Proof_2 | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Conjunction_Absorbs_Disjunction/Proof_2 | [
"Absorption Laws (Logic)"
] | [] | [
"Disjunction with Contradiction",
"Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive",
"Conjunction with Contradiction",
"Disjunction with Contradiction"
] |
proofwiki-6876 | Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Forward Implication | :$p \land \paren {p \lor q} \vdash p$ | {{BeginTableau|p \land \paren {p \lor q} \vdash p}}
{{Premise|1|p \land \paren {p \lor q} }}
{{Simplification|2|1|p|1|1}}
{{EndTableau|qed}} | :$p \land \paren {p \lor q} \vdash p$ | {{BeginTableau|p \land \paren {p \lor q} \vdash p}}
{{Premise|1|p \land \paren {p \lor q} }}
{{Simplification|2|1|p|1|1}}
{{EndTableau|qed}} | Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Forward Implication | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Conjunction_Absorbs_Disjunction/Forward_Implication | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Conjunction_Absorbs_Disjunction/Forward_Implication | [
"Absorption Laws (Logic)"
] | [] | [] |
proofwiki-6877 | Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Reverse Implication | :$p \vdash p \land \paren {p \lor q}$ | {{BeginTableau|p \vdash p \land \paren {p \lor q} }}
{{Premise|1|p}}
{{Addition|2|1|p \lor q|1|1}}
{{Conjunction|3|1|p \land \paren {p \lor q}|1|2}}
{{EndTableau|qed}} | :$p \vdash p \land \paren {p \lor q}$ | {{BeginTableau|p \vdash p \land \paren {p \lor q} }}
{{Premise|1|p}}
{{Addition|2|1|p \lor q|1|1}}
{{Conjunction|3|1|p \land \paren {p \lor q}|1|2}}
{{EndTableau|qed}} | Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Reverse Implication | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Conjunction_Absorbs_Disjunction/Reverse_Implication | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Conjunction_Absorbs_Disjunction/Reverse_Implication | [
"Absorption Laws (Logic)"
] | [] | [] |
proofwiki-6878 | Absorption Laws (Logic)/Conjunction Absorbs Disjunction | :$p \land \paren {p \lor q} \dashv \vdash p$ | By calculation:
{{begin-eqn}}
{{eqn | l = p \land \paren {p \lor q}
| r = \paren {p \lor \bot} \land \paren {p \lor q}
| c = Disjunction with Contradiction
}}
{{eqn | r = p \lor \paren {\bot \land q}
| c = Disjunction is Left Distributive over Conjunction
}}
{{eqn | r = p \lor \bot
| c = Conjunc... | :$p \land \paren {p \lor q} \dashv \vdash p$ | By calculation:
{{begin-eqn}}
{{eqn | l = p \land \paren {p \lor q}
| r = \paren {p \lor \bot} \land \paren {p \lor q}
| c = [[Disjunction with Contradiction]]
}}
{{eqn | r = p \lor \paren {\bot \land q}
| c = [[Disjunction is Left Distributive over Conjunction]]
}}
{{eqn | r = p \lor \bot
| c ... | Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Proof 2 | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Conjunction_Absorbs_Disjunction | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Conjunction_Absorbs_Disjunction/Proof_2 | [
"Absorption Laws (Logic)"
] | [] | [
"Disjunction with Contradiction",
"Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive",
"Conjunction with Contradiction",
"Disjunction with Contradiction"
] |
proofwiki-6879 | Absorption Laws (Logic)/Conjunction Absorbs Disjunction | :$p \land \paren {p \lor q} \dashv \vdash p$ | We apply the Method of Truth Tables.
As can be seen by inspection, the appropriate truth values match for all boolean interpretations.
$\begin{array}{|ccccc||c|} \hline
p & \land & (p & \lor & q) & p \\
\hline
\F & \F & \F & \F & \F & \F \\
\F & \F & \F & \T & \T & \F \\
\T & \T & \T & \T & \F & \T \\
\T & \T & \T & \T... | :$p \land \paren {p \lor q} \dashv \vdash p$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the appropriate [[Definition:Truth Value|truth values]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|ccccc||c|} \hline
p & \land & (p & \lor & q) & p \\
\hline
\F & \F & \F & \F & \F & \F \\
\F & \F ... | Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Proof by Truth Table | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Conjunction_Absorbs_Disjunction | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Conjunction_Absorbs_Disjunction/Proof_by_Truth_Table | [
"Absorption Laws (Logic)"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Boolean Interpretation"
] |
proofwiki-6880 | Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Forward Implication | :$p \lor \paren {p \land q} \vdash p$ | {{BeginTableau|p \lor \paren {p \land q} \vdash p}}
{{Premise|1|p \lor \paren {p \land q} }}
{{Assumption|2|p}}
{{Assumption|3|p \land q}}
{{Simplification|4|3|p|3|1}}
{{ProofByCases|5|1|p|1|2|2|3|4}}
{{EndTableau|qed}} | :$p \lor \paren {p \land q} \vdash p$ | {{BeginTableau|p \lor \paren {p \land q} \vdash p}}
{{Premise|1|p \lor \paren {p \land q} }}
{{Assumption|2|p}}
{{Assumption|3|p \land q}}
{{Simplification|4|3|p|3|1}}
{{ProofByCases|5|1|p|1|2|2|3|4}}
{{EndTableau|qed}} | Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Forward Implication | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Disjunction_Absorbs_Conjunction/Forward_Implication | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Disjunction_Absorbs_Conjunction/Forward_Implication | [
"Absorption Laws (Logic)"
] | [] | [] |
proofwiki-6881 | Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Reverse Implication | :$p \vdash p \lor \paren {p \land q}$ | {{BeginTableau|p \vdash p \lor \paren {p \land q} }}
{{Premise|1|p}}
{{Addition|2|1|p \lor \paren {p \land q}|1|1}}
{{EndTableau|qed}} | :$p \vdash p \lor \paren {p \land q}$ | {{BeginTableau|p \vdash p \lor \paren {p \land q} }}
{{Premise|1|p}}
{{Addition|2|1|p \lor \paren {p \land q}|1|1}}
{{EndTableau|qed}} | Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Reverse Implication | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Disjunction_Absorbs_Conjunction/Reverse_Implication | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Disjunction_Absorbs_Conjunction/Reverse_Implication | [
"Absorption Laws (Logic)"
] | [] | [] |
proofwiki-6882 | Absorption Laws (Logic)/Disjunction Absorbs Conjunction | :$p \lor \paren {p \land q} \dashv \vdash p$ | {{begin-eqn}}
{{eqn | l = p \lor \paren {p \land q}
| r = \paren {p \land \top} \lor \paren {p \land q}
| c = Conjunction with Tautology
}}
{{eqn | r = p \land \paren {\top \lor q}
| c = Conjunction is Left Distributive over Disjunction
}}
{{eqn | r = p \land \top
| c = Disjunction with Tautolog... | :$p \lor \paren {p \land q} \dashv \vdash p$ | {{begin-eqn}}
{{eqn | l = p \lor \paren {p \land q}
| r = \paren {p \land \top} \lor \paren {p \land q}
| c = [[Conjunction with Tautology]]
}}
{{eqn | r = p \land \paren {\top \lor q}
| c = [[Conjunction is Left Distributive over Disjunction]]
}}
{{eqn | r = p \land \top
| c = [[Disjunction wit... | Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Proof 2 | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Disjunction_Absorbs_Conjunction | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Disjunction_Absorbs_Conjunction/Proof_2 | [
"Absorption Laws (Logic)",
"Disjunction",
"Conjunction"
] | [] | [
"Conjunction with Tautology",
"Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive",
"Disjunction with Tautology",
"Conjunction with Tautology"
] |
proofwiki-6883 | Absorption Laws (Logic)/Disjunction Absorbs Conjunction | :$p \lor \paren {p \land q} \dashv \vdash p$ | We apply the Method of Truth Tables.
As can be seen by inspection, the appropriate truth values match for all boolean interpretations.
$\begin{array}{|ccccc||c|} \hline
p & \lor & (p & \land & q) & p \\
\hline
\F & \F & \F & \F & \F & \F \\
\F & \F & \F & \F & \T & \F \\
\T & \T & \T & \F & \F & \T \\
\T & \T & \T & \T... | :$p \lor \paren {p \land q} \dashv \vdash p$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the appropriate [[Definition:Truth Value|truth values]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|ccccc||c|} \hline
p & \lor & (p & \land & q) & p \\
\hline
\F & \F & \F & \F & \F & \F \\
\F & \F ... | Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Proof by Truth Table | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Disjunction_Absorbs_Conjunction | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Disjunction_Absorbs_Conjunction/Proof_by_Truth_Table | [
"Absorption Laws (Logic)",
"Disjunction",
"Conjunction"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Boolean Interpretation"
] |
proofwiki-6884 | Extended Transitivity | Let $S$ be a set.
Let $\RR$ be a transitive relation on $S$.
Let $\RR^=$ be the reflexive closure of $\RR$.
Let $a, b, c \in S$.
Then:
{{begin-eqn}}
{{eqn | n = 1
| l = \paren {a \mathrel \RR b} \land \paren {b \mathrel \RR c}
| o = \implies
| r = a \mathrel \RR c
}}
{{eqn | n = 2
| l = \paren {... | $(1)$ follows from the definition of a transitive relation.
$(4)$ follows from Reflexive Closure of Transitive Relation is Transitive.
Suppose that:
:$\paren {a \mathrel \RR b} \land \paren {b \mathrel {\RR^=} c}$
By the definition of reflexive closure:
:$b \mathrel \RR c$ or $b = c$
If $b = c$, then since $a \mathrel ... | Let $S$ be a [[Definition:Set|set]].
Let $\RR$ be a [[Definition:Transitive Relation|transitive relation]] on $S$.
Let $\RR^=$ be the [[Definition:Reflexive Closure|reflexive closure]] of $\RR$.
Let $a, b, c \in S$.
Then:
{{begin-eqn}}
{{eqn | n = 1
| l = \paren {a \mathrel \RR b} \land \paren {b \mathrel \... | $(1)$ follows from the definition of a [[Definition:Transitive Relation|transitive relation]].
$(4)$ follows from [[Reflexive Closure of Transitive Relation is Transitive]].
Suppose that:
:$\paren {a \mathrel \RR b} \land \paren {b \mathrel {\RR^=} c}$
By the definition of [[Definition:Reflexive Closure|reflexive c... | Extended Transitivity | https://proofwiki.org/wiki/Extended_Transitivity | https://proofwiki.org/wiki/Extended_Transitivity | [
"Transitive Relations",
"Reflexive Closures"
] | [
"Definition:Set",
"Definition:Transitive Relation",
"Definition:Reflexive Closure"
] | [
"Definition:Transitive Relation",
"Reflexive Closure of Transitive Relation is Transitive",
"Definition:Reflexive Closure",
"Definition:Transitive Relation",
"Definition:Reflexive Closure",
"Definition:Transitive Relation",
"Category:Transitive Relations",
"Category:Reflexive Closures"
] |
proofwiki-6885 | Rule of Idempotence/Conjunction | {{:Rule of Idempotence/Conjunction/Formulation 1}} | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for each boolean interpretations.
$\begin{array} {|c||ccc|} \hline
p & p & \land & p \\
\hline
\T & \T & \T & \T \\
\F & \F & \F & \F \\
\hline
\end{array}$
{{qed}} | {{:Rule of Idempotence/Conjunction/Formulation 1}} | 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 connectives]] match for each [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array} {|c||ccc|} \hline
p & p & \lan... | Rule of Idempotence/Conjunction/Formulation 1/Proof | https://proofwiki.org/wiki/Rule_of_Idempotence/Conjunction | https://proofwiki.org/wiki/Rule_of_Idempotence/Conjunction/Formulation_1/Proof | [
"Rule of Idempotence"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6886 | Rule of Idempotence/Disjunction | {{:Rule of Idempotence/Disjunction/Formulation 1}} | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for each boolean interpretations.
$\begin{array}{|c||ccc|} \hline
p & p & \lor & p \\
\hline
\T & \T & \T & \T \\
\F & \F & \F & \F \\
\hline
\end{array}$
{{qed}} | {{:Rule of Idempotence/Disjunction/Formulation 1}} | 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 connectives]] match for each [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|c||ccc|} \hline
p & p & \lor ... | Rule of Idempotence/Disjunction/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Idempotence/Disjunction | https://proofwiki.org/wiki/Rule_of_Idempotence/Disjunction/Formulation_1/Proof_by_Truth_Table | [
"Rule of Idempotence"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6887 | Rule of Idempotence/Disjunction/Formulation 1/Forward Implication | :$p \vdash p \lor p$ | {{BeginTableau|p \vdash p \lor p}}
{{Premise|1|p}}
{{Addition|2|1|p \lor p|1|1}}
{{EndTableau|qed}} | :$p \vdash p \lor p$ | {{BeginTableau|p \vdash p \lor p}}
{{Premise|1|p}}
{{Addition|2|1|p \lor p|1|1}}
{{EndTableau|qed}} | Rule of Idempotence/Disjunction/Formulation 1/Forward Implication | https://proofwiki.org/wiki/Rule_of_Idempotence/Disjunction/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Rule_of_Idempotence/Disjunction/Formulation_1/Forward_Implication | [
"Rule of Idempotence"
] | [] | [] |
proofwiki-6888 | Rule of Idempotence/Disjunction/Formulation 1/Reverse Implication | :$p \lor p \vdash p$ | {{BeginTableau|p \lor p \vdash p}}
{{Premise|1|p \lor p}}
{{Assumption|2|p}}
{{ProofByCases|3|1|p|1|2|2|2|2}}
{{EndTableau|qed}} | :$p \lor p \vdash p$ | {{BeginTableau|p \lor p \vdash p}}
{{Premise|1|p \lor p}}
{{Assumption|2|p}}
{{ProofByCases|3|1|p|1|2|2|2|2}}
{{EndTableau|qed}} | Rule of Idempotence/Disjunction/Formulation 1/Reverse Implication | https://proofwiki.org/wiki/Rule_of_Idempotence/Disjunction/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Rule_of_Idempotence/Disjunction/Formulation_1/Reverse_Implication | [
"Rule of Idempotence"
] | [] | [] |
proofwiki-6889 | Rule of Idempotence/Disjunction/Formulation 1 | :$p \dashv \vdash p \lor p$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for each boolean interpretations.
$\begin{array}{|c||ccc|} \hline
p & p & \lor & p \\
\hline
\T & \T & \T & \T \\
\F & \F & \F & \F \\
\hline
\end{array}$
{{qed}} | :$p \dashv \vdash p \lor p$ | 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 connectives]] match for each [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|c||ccc|} \hline
p & p & \lor ... | Rule of Idempotence/Disjunction/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Idempotence/Disjunction/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Idempotence/Disjunction/Formulation_1/Proof_by_Truth_Table | [
"Rule of Idempotence"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6890 | Rule of Idempotence/Disjunction/Formulation 2 | :$\vdash p \iff \paren {p \lor p}$ | {{BeginTableau |p \iff \paren {p \lor p} }}
{{Assumption | 1| p}}
{{Addition | 2| 1|p \lor p |1|1}}
{{Implication | 3| |p \implies \paren {p \lor p}|1|2}}
{{Assumption | 4| p \lor p}}
{{Assumption | 5| \neg p}}
{{ModusTollendoPonens| 6|4... | :$\vdash p \iff \paren {p \lor p}$ | {{BeginTableau |p \iff \paren {p \lor p} }}
{{Assumption | 1| p}}
{{Addition | 2| 1|p \lor p |1|1}}
{{Implication | 3| |p \implies \paren {p \lor p}|1|2}}
{{Assumption | 4| p \lor p}}
{{Assumption | 5| \neg p}}
{{ModusTollendoPonens| 6|4... | Rule of Idempotence/Disjunction/Formulation 2 | https://proofwiki.org/wiki/Rule_of_Idempotence/Disjunction/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Idempotence/Disjunction/Formulation_2 | [
"Rule of Idempotence"
] | [] | [] |
proofwiki-6891 | Rule of Idempotence/Conjunction/Formulation 1/Forward Implication | :$p \vdash p \land p$ | {{BeginTableau|p \vdash p \land p}}
{{Premise|1|p}}
{{Conjunction|2|1|p \land q|1|1}}
{{EndTableau|qed}}
Category:Rule of Idempotence
dpjnyjtco9tniusxt10j76s1lzcf52l | :$p \vdash p \land p$ | {{BeginTableau|p \vdash p \land p}}
{{Premise|1|p}}
{{Conjunction|2|1|p \land q|1|1}}
{{EndTableau|qed}}
[[Category:Rule of Idempotence]]
dpjnyjtco9tniusxt10j76s1lzcf52l | Rule of Idempotence/Conjunction/Formulation 1/Forward Implication | https://proofwiki.org/wiki/Rule_of_Idempotence/Conjunction/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Rule_of_Idempotence/Conjunction/Formulation_1/Forward_Implication | [
"Rule of Idempotence"
] | [] | [
"Category:Rule of Idempotence"
] |
proofwiki-6892 | Rule of Idempotence/Conjunction/Formulation 1/Reverse Implication | :$p \land p \vdash p$ | {{BeginTableau|p \land p \vdash p}}
{{Premise|1|p \land p}}
{{Simplification|2|1|p|1|1}}
{{EndTableau|qed}}
Category:Rule of Idempotence
6ejt9d025qxqrds4cn8tiwy441h4fcc | :$p \land p \vdash p$ | {{BeginTableau|p \land p \vdash p}}
{{Premise|1|p \land p}}
{{Simplification|2|1|p|1|1}}
{{EndTableau|qed}}
[[Category:Rule of Idempotence]]
6ejt9d025qxqrds4cn8tiwy441h4fcc | Rule of Idempotence/Conjunction/Formulation 1/Reverse Implication | https://proofwiki.org/wiki/Rule_of_Idempotence/Conjunction/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Rule_of_Idempotence/Conjunction/Formulation_1/Reverse_Implication | [
"Rule of Idempotence"
] | [] | [
"Category:Rule of Idempotence"
] |
proofwiki-6893 | Rule of Idempotence/Conjunction/Formulation 1 | :$p \dashv \vdash p \land p$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for each boolean interpretations.
$\begin{array} {|c||ccc|} \hline
p & p & \land & p \\
\hline
\T & \T & \T & \T \\
\F & \F & \F & \F \\
\hline
\end{array}$
{{qed}} | :$p \dashv \vdash p \land p$ | 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 connectives]] match for each [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array} {|c||ccc|} \hline
p & p & \lan... | Rule of Idempotence/Conjunction/Formulation 1/Proof | https://proofwiki.org/wiki/Rule_of_Idempotence/Conjunction/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Idempotence/Conjunction/Formulation_1/Proof | [
"Rule of Idempotence"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6894 | Rule of Idempotence/Conjunction/Formulation 2 | :$\vdash p \iff \paren {p \land p}$ | {{BeginTableau|\vdash p \iff \paren {p \land p} }}
{{Assumption|1|p}}
{{Conjunction|2|1|p \land p|1|1}}
{{Implication|3||p \implies \paren {p \land p}|1|2}}
{{Assumption|4|p \land p}}
{{Simplification|5|4|p|4|1}}
{{Implication|6||\paren {p \land p} \implies p|4|5}}
{{BiconditionalIntro|7||p \iff \paren {p \land p}|3|6}... | :$\vdash p \iff \paren {p \land p}$ | {{BeginTableau|\vdash p \iff \paren {p \land p} }}
{{Assumption|1|p}}
{{Conjunction|2|1|p \land p|1|1}}
{{Implication|3||p \implies \paren {p \land p}|1|2}}
{{Assumption|4|p \land p}}
{{Simplification|5|4|p|4|1}}
{{Implication|6||\paren {p \land p} \implies p|4|5}}
{{BiconditionalIntro|7||p \iff \paren {p \land p}|3|6}... | Rule of Idempotence/Conjunction/Formulation 2 | https://proofwiki.org/wiki/Rule_of_Idempotence/Conjunction/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Idempotence/Conjunction/Formulation_2 | [
"Rule of Idempotence"
] | [] | [] |
proofwiki-6895 | Ordering Cycle implies Equality | Let $\struct {S, \preceq}$ be an ordered set.
Let $x_1$, $x_2$, and $x_3$ be elements of $S$.
Suppose that
{{begin-eqn}}
{{eqn | l = x_1
| o = \preceq
| r = x_2
}}
{{eqn | l = x_2
| o = \preceq
| r = x_3
}}
{{eqn | l = x_3
| o = \preceq
| r = x_1
}}
{{end-eqn}}
Then $x_1 = x_2 = x_3$... | Because $\preceq$ is an ordering, it is transitive and antisymmetric.
By transitivity, $x_1 \preceq x_3$.
Because $x_1 \preceq x_3$ and $x_3 \preceq x_1$, antisymmetry allows us to conclude that $x_1 = x_3$.
Because $x_1 = x_3$ and $x_1 \preceq x_2$, we must have $x_3 \preceq x_2$.
Because $x_3 \preceq x_2$ and $x_2 \p... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $x_1$, $x_2$, and $x_3$ be elements of $S$.
Suppose that
{{begin-eqn}}
{{eqn | l = x_1
| o = \preceq
| r = x_2
}}
{{eqn | l = x_2
| o = \preceq
| r = x_3
}}
{{eqn | l = x_3
| o = \preceq
| r = x_1
}}
{{end... | Because $\preceq$ is an [[Definition:Ordering|ordering]], it is [[Definition:Transitive Relation|transitive]] and [[Definition:Antisymmetric Relation|antisymmetric]].
By [[Definition:Transitive Relation|transitivity]], $x_1 \preceq x_3$.
Because $x_1 \preceq x_3$ and $x_3 \preceq x_1$, [[Definition:Antisymmetric Rela... | Ordering Cycle implies Equality | https://proofwiki.org/wiki/Ordering_Cycle_implies_Equality | https://proofwiki.org/wiki/Ordering_Cycle_implies_Equality | [
"Order Theory"
] | [
"Definition:Ordered Set"
] | [
"Definition:Ordering",
"Definition:Transitive Relation",
"Definition:Antisymmetric Relation",
"Definition:Transitive Relation",
"Definition:Antisymmetric Relation",
"Category:Order Theory"
] |
proofwiki-6896 | Euclidean Topology is Product Topology | Let $T_1 = \struct {\R, \tau_1}$ be the topological space such that $\tau_1$ is the Euclidean topology on $\R$.
Let $T_n = \struct {\R^n, \tau_n}$ be the topological space such that $\tau_n$ is the product topology on the cartesian product $\ds \R_n = \prod_{i \mathop = 1}^n \R$.
Then the Euclidean topology on $\R^n$ a... | Denote the Euclidean topology on $\R^n$ as $\tau$, and denote the product topology on $\R^n$ as $\tau'$.
Let $U \in \tau$, and let $x = \tuple{x_1, \ldots, x_n} \in U$.
Then there exists $\epsilon \in \R_{>0}$ such that the open ball $\map {B_\epsilon} x \subseteq U$.
We show that:
:$\ds B' = \prod_{i \mathop = 1}^n \o... | Let $T_1 = \struct {\R, \tau_1}$ be the [[Definition:Topological Space|topological space]] such that $\tau_1$ is the [[Definition:Euclidean Topology|Euclidean topology]] on $\R$.
Let $T_n = \struct {\R^n, \tau_n}$ be the [[Definition:Topological Space|topological space]] such that $\tau_n$ is the [[Definition:Product ... | Denote the [[Definition:Euclidean Topology|Euclidean topology]] on $\R^n$ as $\tau$, and denote the [[Definition:Product Topology|product topology]] on $\R^n$ as $\tau'$.
Let $U \in \tau$, and let $x = \tuple{x_1, \ldots, x_n} \in U$.
Then there exists $\epsilon \in \R_{>0}$ such that the [[Definition:Open Ball of Me... | Euclidean Topology is Product Topology/Proof 1 | https://proofwiki.org/wiki/Euclidean_Topology_is_Product_Topology | https://proofwiki.org/wiki/Euclidean_Topology_is_Product_Topology/Proof_1 | [
"Euclidean Topology is Product Topology",
"Euclidean Topology",
"Euclidean Spaces",
"Product Topology"
] | [
"Definition:Topological Space",
"Definition:Euclidean Space/Euclidean Topology",
"Definition:Topological Space",
"Definition:Product Topology",
"Definition:Cartesian Product",
"Definition:Euclidean Space/Euclidean Topology",
"Definition:Product Topology"
] | [
"Definition:Euclidean Space/Euclidean Topology",
"Definition:Product Topology",
"Definition:Open Ball",
"Definition:Euclidean Metric/Real Vector Space",
"Minkowski's Inequality for Sums",
"Natural Basis of Product Topology/Finite Product",
"Neighborhood Condition for Coarser Topology",
"Natural Basis ... |
proofwiki-6897 | Euclidean Topology is Product Topology | Let $T_1 = \struct {\R, \tau_1}$ be the topological space such that $\tau_1$ is the Euclidean topology on $\R$.
Let $T_n = \struct {\R^n, \tau_n}$ be the topological space such that $\tau_n$ is the product topology on the cartesian product $\ds \R_n = \prod_{i \mathop = 1}^n \R$.
Then the Euclidean topology on $\R^n$ a... | The proof proceeds by induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:the Euclidean topology on $\R^n$ and the product topology on $\R^n$ are the same.
=== Basis for the Induction ===
$\map P 2$ is the case:
:the Euclidean topology on $\R^n$ and the product topology on $\R^n$ are the same.
Th... | Let $T_1 = \struct {\R, \tau_1}$ be the [[Definition:Topological Space|topological space]] such that $\tau_1$ is the [[Definition:Euclidean Topology|Euclidean topology]] on $\R$.
Let $T_n = \struct {\R^n, \tau_n}$ be the [[Definition:Topological Space|topological space]] such that $\tau_n$ is the [[Definition:Product ... | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:the [[Definition:Euclidean Topology|Euclidean topology]] on $\R^n$ and the [[Definition:Product Topology|product topology]] on $\R^n$ are the same.
=== B... | Euclidean Topology is Product Topology/Proof 2 | https://proofwiki.org/wiki/Euclidean_Topology_is_Product_Topology | https://proofwiki.org/wiki/Euclidean_Topology_is_Product_Topology/Proof_2 | [
"Euclidean Topology is Product Topology",
"Euclidean Topology",
"Euclidean Spaces",
"Product Topology"
] | [
"Definition:Topological Space",
"Definition:Euclidean Space/Euclidean Topology",
"Definition:Topological Space",
"Definition:Product Topology",
"Definition:Cartesian Product",
"Definition:Euclidean Space/Euclidean Topology",
"Definition:Product Topology"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Euclidean Space/Euclidean Topology",
"Definition:Product Topology",
"Definition:Euclidean Space/Euclidean Topology",
"Definition:Product Topology",
"Euclidean Topology on Cartesian Plane is Product Topology",
"Definition:Basi... |
proofwiki-6898 | Modus Ponendo Tollens/Variant/Formulation 1/Proof | :$\neg \paren {p \land q} \dashv \vdash p \implies \neg q$ | As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|cccc||cccc|} \hline
\neg & (p & \land & q) & p & \implies & \neg & q \\
\hline
T & F & F & F & F & T & T & F \\
T & F & F & T & F & T & F & T \\
T & T & F & F & T & T & T & F \\
F & T & T & T... | :$\neg \paren {p \land q} \dashv \vdash p \implies \neg q$ | 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]].
$\begin{array}{|cccc||cccc|} \hline
\neg & (p & \land & q) & p & \implies & \neg & q \\
... | Modus Ponendo Tollens/Variant/Formulation 1/Proof | https://proofwiki.org/wiki/Modus_Ponendo_Tollens/Variant/Formulation_1/Proof | https://proofwiki.org/wiki/Modus_Ponendo_Tollens/Variant/Formulation_1/Proof | [
"Modus Ponendo Tollens",
"Truth Table Proofs"
] | [] | [
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6899 | Modus Ponendo Tollens/Variant/Formulation 1/Forward Implication | :$\neg \paren {p \land q} \vdash p \implies \neg q$ | {{BeginTableau|\neg \paren {p \land q} \vdash p \implies \neg q}}
{{Premise|1|\neg \paren {p \land q} }}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{Conjunction|4|2, 3|p \land q|2|3}}
{{NonContradiction|5|1, 2, 5|4|1}}
{{Contradiction|6|1, 2|\neg q|3|5}}
{{Implication|7|1|p \implies \neg q|2|6}}
{{EndTableau|qed}}
Category... | :$\neg \paren {p \land q} \vdash p \implies \neg q$ | {{BeginTableau|\neg \paren {p \land q} \vdash p \implies \neg q}}
{{Premise|1|\neg \paren {p \land q} }}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{Conjunction|4|2, 3|p \land q|2|3}}
{{NonContradiction|5|1, 2, 5|4|1}}
{{Contradiction|6|1, 2|\neg q|3|5}}
{{Implication|7|1|p \implies \neg q|2|6}}
{{EndTableau|qed}}
[[Categ... | Modus Ponendo Tollens/Variant/Formulation 1/Forward Implication | https://proofwiki.org/wiki/Modus_Ponendo_Tollens/Variant/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Modus_Ponendo_Tollens/Variant/Formulation_1/Forward_Implication | [
"Modus Ponendo Tollens"
] | [] | [
"Category:Modus Ponendo Tollens"
] |
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