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-6700 | Self-Distributive Law for Conditional/Formulation 2 | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | {{BeginTableau |\vdash \paren {\paren {p \implies q} \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} } }}
{{Assumption|1|\paren {p \implies q} \implies \paren {p \implies r} }}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{SequentIntro|4|3|p \implies q|3|True Statement is implied by Every S... | :$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | {{BeginTableau |\vdash \paren {\paren {p \implies q} \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} } }}
{{Assumption|1|\paren {p \implies q} \implies \paren {p \implies r} }}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{SequentIntro|4|3|p \implies q|3|[[True Statement is implied by Every... | Self-Distributive Law for Conditional/Formulation 2/Reverse Implication/Proof 2 | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_2 | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_2/Reverse_Implication/Proof_2 | [
"Self-Distributive Law for Conditional"
] | [] | [
"True Statement is implied by Every Statement"
] |
proofwiki-6701 | Cauchy-Riemann Equations/Necessary Condition | Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$.
Let $f: D \to \C$ be a complex function on $D$.
Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be the two real-valued functions defined as:
{{begin-eqn}}
{{eqn | l = \map u {x, y}
| r = \map \Re {\map f z}
}}
{{eqn | l... | Let $z = x + i y \in D$.
The Epsilon-Function Complex Differentiability Condition shows that there exists $r \in \R_{>0}$ such that for all $t \in \map {B_r} 0 \setminus \set 0$:
:$(\text a): \quad \map f {z + t} = \map f z + t \paren {\map {f'} z + \map \epsilon t}$
where:
:$\map {B_r} 0$ denotes an open ball of $0$
:... | Let $D \subseteq \C$ be an [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$.
Let $f: D \to \C$ be a [[Definition:Complex Function|complex function]] on $D$.
Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y... | Let $z = x + i y \in D$.
The [[Epsilon-Function Complex Differentiability Condition]] shows that there exists $r \in \R_{>0}$ such that for all $t \in \map {B_r} 0 \setminus \set 0$:
:$(\text a): \quad \map f {z + t} = \map f z + t \paren {\map {f'} z + \map \epsilon t}$
where:
:$\map {B_r} 0$ denotes an [[Definitio... | Cauchy-Riemann Equations/Necessary Condition | https://proofwiki.org/wiki/Cauchy-Riemann_Equations/Necessary_Condition | https://proofwiki.org/wiki/Cauchy-Riemann_Equations/Necessary_Condition | [
"Cauchy-Riemann Equations"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Subset",
"Definition:Set",
"Definition:Complex Number",
"Definition:Complex Function",
"Definition:Real-Valued Function",
"Definition:Complex Number/Real Part",
"Definition:Complex Number/Imaginary Part",
"Definition:Differentiable Mapping/Complex ... | [
"Epsilon-Function Differentiability Condition/Complex Case",
"Definition:Open Ball",
"Definition:Complex Function",
"Definition:Complex Number/Real Part",
"Addition of Real and Imaginary Parts",
"Multiplication of Real and Imaginary Parts",
"Definition:Partial Derivative",
"Definition:Complex Number/W... |
proofwiki-6702 | Self-Distributive Law for Conditional/Formulation 1 | :$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$ | {{BeginTableau|p \implies \paren {q \implies r} \vdash \paren {p \implies q} \implies \paren {p \implies r} }}
{{Premise|1|p \implies \paren {q \implies r} }}
{{Assumption|2|p \implies q}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q \implies r|1|3}}
{{ModusPonens|5|2, 3|q|2|3}}
{{ModusPonens|6|1, 2, 3|r|4|5}}
{{Implicati... | :$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$ | {{BeginTableau|p \implies \paren {q \implies r} \vdash \paren {p \implies q} \implies \paren {p \implies r} }}
{{Premise|1|p \implies \paren {q \implies r} }}
{{Assumption|2|p \implies q}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q \implies r|1|3}}
{{ModusPonens|5|2, 3|q|2|3}}
{{ModusPonens|6|1, 2, 3|r|4|5}}
{{Implicati... | Self-Distributive Law for Conditional/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_1 | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_1/Forward_Implication/Proof | [
"Self-Distributive Law for Conditional"
] | [] | [] |
proofwiki-6703 | Self-Distributive Law for Conditional/Formulation 1 | :$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$ | We apply the Method of Truth Tables to the proposition:
$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccccc||ccccccc|} \hline
p & \impl... | :$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$ | We apply the [[Method of Truth Tables]] to the proposition:
$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] mat... | Self-Distributive Law for Conditional/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_1 | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_1/Proof_by_Truth_Table | [
"Self-Distributive Law for Conditional"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6704 | Self-Distributive Law for Conditional/Formulation 1 | :$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$ | {{BeginTableau|\paren {p \implies q} \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r} }}
{{Premise|1|\paren {p \implies q} \implies \paren {p \implies r} }}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{SequentIntro|4|3|p \implies q|3|True Statement is implied by Every Statement}}
{{ModusPonens|5|1, 3|p... | :$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$ | {{BeginTableau|\paren {p \implies q} \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r} }}
{{Premise|1|\paren {p \implies q} \implies \paren {p \implies r} }}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{SequentIntro|4|3|p \implies q|3|[[True Statement is implied by Every Statement]]}}
{{ModusPonens|5|1,... | Self-Distributive Law for Conditional/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_1 | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_1/Reverse_Implication/Proof | [
"Self-Distributive Law for Conditional"
] | [] | [
"True Statement is implied by Every Statement"
] |
proofwiki-6705 | Cauchy-Riemann Equations/Sufficient Condition | Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$.
Let $f: D \to \C$ be a complex function on $D$.
Let $u, v: \set {\paren {x, y} \in \R^2: x + i y = z \in D} \to \R$ be two real-valued functions defined as:
:$\map u {x, y} = \map \Re {\map f z}$
:$\map v {x, y} = \map \Im {\map f z}$
where:
:$\... | Suppose that the '''Cauchy-Riemann equations''' hold for $u$ and $v$ in their entire domain.
Let $h, k \in \R \setminus \set 0$, and put $t = h + i k \in \C$.
Let $\tuple {x, y} \in \R^2$ be a point in the domain of $u$ and $v$.
Put:
:$a = \map {\dfrac {\partial u} {\partial x} } {x, y} = \map {\dfrac {\partial v} {\pa... | Let $D \subseteq \C$ be an [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$.
Let $f: D \to \C$ be a [[Definition:Complex Function|complex function]] on $D$.
Let $u, v: \set {\paren {x, y} \in \R^2: x + i y... | Suppose that the '''Cauchy-Riemann equations''' hold for $u$ and $v$ in their entire [[Definition:Domain of Mapping|domain]].
Let $h, k \in \R \setminus \set 0$, and put $t = h + i k \in \C$.
Let $\tuple {x, y} \in \R^2$ be a [[Definition:Point|point]] in the [[Definition:Domain of Mapping|domain]] of $u$ and $v$.
P... | Cauchy-Riemann Equations/Sufficient Condition | https://proofwiki.org/wiki/Cauchy-Riemann_Equations/Sufficient_Condition | https://proofwiki.org/wiki/Cauchy-Riemann_Equations/Sufficient_Condition | [
"Cauchy-Riemann Equations"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Subset",
"Definition:Set",
"Definition:Complex Number",
"Definition:Complex Function",
"Definition:Real-Valued Function",
"Definition:Complex Number/Real Part",
"Definition:Complex Number/Imaginary Part",
"Definition:Differentiable Mapping/Real-Val... | [
"Definition:Domain (Set Theory)/Mapping",
"Definition:Point",
"Definition:Domain (Set Theory)/Mapping",
"Epsilon-Function Differentiability Condition",
"Definition:Real Function",
"Definition:Convergent Mapping",
"Definition:Partial Derivative",
"Definition:Continuous Real Function",
"Epsilon-Functi... |
proofwiki-6706 | Cauchy-Riemann Equations/Expression of Derivative | Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$.
Let $f: D \to \C$ be a complex function on $D$.
Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be two real-valued functions defined as:
{{begin-eqn}}
{{eqn | l = \map u {x, y}
| r = \map \Re {\map f z}
}}
{{eqn | l = \... | Let $z = x + i y$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\dfrac {\partial f} {\partial x} } z
| r = \map {\dfrac {\partial u} {\partial x} } {x, y} + i \map {\dfrac {\partial v} {\partial x} } {x, y}
}}
{{eqn | r = \map \Re {\map {f'} z} + i \, \map \Im {\map {f'} z}
| c = from the last part of the proof fo... | Let $D \subseteq \C$ be an [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$.
Let $f: D \to \C$ be a [[Definition:Complex Function|complex function]] on $D$.
Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y... | Let $z = x + i y$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\dfrac {\partial f} {\partial x} } z
| r = \map {\dfrac {\partial u} {\partial x} } {x, y} + i \map {\dfrac {\partial v} {\partial x} } {x, y}
}}
{{eqn | r = \map \Re {\map {f'} z} + i \, \map \Im {\map {f'} z}
| c = from the last part of the [[Cau... | Cauchy-Riemann Equations/Expression of Derivative | https://proofwiki.org/wiki/Cauchy-Riemann_Equations/Expression_of_Derivative | https://proofwiki.org/wiki/Cauchy-Riemann_Equations/Expression_of_Derivative | [
"Cauchy-Riemann Equations"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Subset",
"Definition:Set",
"Definition:Complex Number",
"Definition:Complex Function",
"Definition:Real-Valued Function",
"Definition:Complex Number/Real Part",
"Definition:Complex Number/Imaginary Part",
"Definition:Differentiable Mapping/Complex ... | [
"Cauchy-Riemann Equations/Sufficient Condition",
"Cauchy-Riemann Equations/Sufficient Condition"
] |
proofwiki-6707 | Succeed is Dual to Precede | Let $\left({S, \preceq}\right)$ be an ordered set.
Let $a, b \in S$.
The following are dual statements:
:$a$ succeeds $b$
:$a$ precedes $b$ | By definition, $a$ succeeds $b$ {{iff}}:
:$b \preceq a$
The dual of this statement is:
:$a \preceq b$
by Dual Pairs (Order Theory).
By definition, this means $a$ precedes $b$.
The converse follows from Dual of Dual Statement (Order Theory).
{{qed}} | Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
Let $a, b \in S$.
The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
:$a$ [[Definition:Succeed|succeeds]] $b$
:$a$ [[Definition:Precede|precedes]] $b$ | By definition, $a$ [[Definition:Succeed|succeeds]] $b$ {{iff}}:
:$b \preceq a$
The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
:$a \preceq b$
by [[Dual Pairs (Order Theory)]].
By definition, this means $a$ [[Definition:Precede|precedes]] $b$.
The converse follows from [[Dual of Dual ... | Succeed is Dual to Precede | https://proofwiki.org/wiki/Succeed_is_Dual_to_Precede | https://proofwiki.org/wiki/Succeed_is_Dual_to_Precede | [
"Dual Pairs (Order Theory)"
] | [
"Definition:Ordered Set",
"Definition:Dual Statement (Order Theory)",
"Definition:Succeed",
"Definition:Precede"
] | [
"Definition:Succeed",
"Definition:Dual Statement (Order Theory)",
"Dual Pairs (Order Theory)",
"Definition:Precede",
"Dual of Dual Statement (Order Theory)"
] |
proofwiki-6708 | Strictly Succeed is Dual to Strictly Precede | Let $\struct {S, \preceq}$ be an ordered set.
Let $a, b \in S$.
The following are dual statements:
:$a$ strictly succeeds $b$
:$a$ strictly precedes $b$ | By definition, $a$ strictly succeeds $b$ {{iff}}:
:$b \preceq a$ and $b \ne a$
The dual of this statement is:
:$a \succeq b$ and $b \ne a$
by Dual Pairs (Order Theory).
By definition, this means $a$ strictly precedes $b$.
The converse follows from Dual of Dual Statement (Order Theory).
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $a, b \in S$.
The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
:$a$ [[Definition:Strictly Succeed|strictly succeeds]] $b$
:$a$ [[Definition:Strictly Precede|strictly precedes]] $b$ | By definition, $a$ [[Definition:Strictly Succeed|strictly succeeds]] $b$ {{iff}}:
:$b \preceq a$ and $b \ne a$
The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
:$a \succeq b$ and $b \ne a$
by [[Dual Pairs (Order Theory)]].
By definition, this means $a$ [[Definition:Strictly Precede|stri... | Strictly Succeed is Dual to Strictly Precede | https://proofwiki.org/wiki/Strictly_Succeed_is_Dual_to_Strictly_Precede | https://proofwiki.org/wiki/Strictly_Succeed_is_Dual_to_Strictly_Precede | [
"Dual Pairs (Order Theory)"
] | [
"Definition:Ordered Set",
"Definition:Dual Statement (Order Theory)",
"Definition:Strictly Succeed",
"Definition:Strictly Precede"
] | [
"Definition:Strictly Succeed",
"Definition:Dual Statement (Order Theory)",
"Dual Pairs (Order Theory)",
"Definition:Strictly Precede",
"Dual of Dual Statement (Order Theory)"
] |
proofwiki-6709 | Upper Bound is Dual to Lower Bound | Let $\struct {S, \preceq}$ be an ordered set.
Let $a \in S$ and $T \subseteq S$.
The following are dual statements:
:$a$ is an upper bound for $T$
:$a$ is a lower bound for $T$ | By definition, $a$ is an upper bound for $T$ {{iff}}:
:$\forall t \in T: t \preceq a$
The dual of this statement is:
:$\forall t \in T: a \preceq t$
by Dual Pairs (Order Theory).
By definition, this means $a$ is a lower bound for $T$.
The converse follows from Dual of Dual Statement (Order Theory).
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $a \in S$ and $T \subseteq S$.
The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
:$a$ is an [[Definition:Upper Bound of Set|upper bound]] for $T$
:$a$ is a [[Definition:Lower Bound of Set|lower bound]] for $T$ | By definition, $a$ is an [[Definition:Upper Bound of Set|upper bound]] for $T$ {{iff}}:
:$\forall t \in T: t \preceq a$
The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
:$\forall t \in T: a \preceq t$
by [[Dual Pairs (Order Theory)]].
By definition, this means $a$ is a [[Definition:Lowe... | Upper Bound is Dual to Lower Bound | https://proofwiki.org/wiki/Upper_Bound_is_Dual_to_Lower_Bound | https://proofwiki.org/wiki/Upper_Bound_is_Dual_to_Lower_Bound | [
"Dual Pairs (Order Theory)"
] | [
"Definition:Ordered Set",
"Definition:Dual Statement (Order Theory)",
"Definition:Upper Bound of Set",
"Definition:Lower Bound of Set"
] | [
"Definition:Upper Bound of Set",
"Definition:Dual Statement (Order Theory)",
"Dual Pairs (Order Theory)",
"Definition:Lower Bound of Set",
"Dual of Dual Statement (Order Theory)"
] |
proofwiki-6710 | Supremum is Dual to Infimum | Let $\struct {S, \preceq}$ be an ordered set.
Let $a \in S$ and $T \subseteq S$.
The following are dual statements:
:$a$ is a supremum for $T$
:$a$ is an infimum for $T$ | By definition, $a$ is a supremum for $T$ {{iff}}:
:$a$ is an upper bound for $T$
:$a \preceq b$ for all upper bounds $b$ of $T$
The dual of this statement is:
:$a$ is a lower bound for $T$
:$b \preceq a$ for all lower bounds $b$ of $T$
by Dual Pairs (Order Theory).
By definition, this means $a$ is an infimum for $T$.
T... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $a \in S$ and $T \subseteq S$.
The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
:$a$ is a [[Definition:Supremum of Set|supremum]] for $T$
:$a$ is an [[Definition:Infimum of Set|infimum]] for $T$ | By definition, $a$ is a [[Definition:Supremum of Set|supremum]] for $T$ {{iff}}:
:$a$ is an [[Definition:Upper Bound of Set|upper bound]] for $T$
:$a \preceq b$ for all [[Definition:Upper Bound of Set|upper bounds]] $b$ of $T$
The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
:$a$ is a [[De... | Supremum is Dual to Infimum | https://proofwiki.org/wiki/Supremum_is_Dual_to_Infimum | https://proofwiki.org/wiki/Supremum_is_Dual_to_Infimum | [
"Suprema",
"Infima",
"Dual Pairs (Order Theory)"
] | [
"Definition:Ordered Set",
"Definition:Dual Statement (Order Theory)",
"Definition:Supremum of Set",
"Definition:Infimum of Set"
] | [
"Definition:Supremum of Set",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set",
"Definition:Dual Statement (Order Theory)",
"Definition:Lower Bound of Set",
"Definition:Lower Bound of Set",
"Dual Pairs (Order Theory)",
"Definition:Infimum of Set",
"Dual of Dual Statement (Order Theor... |
proofwiki-6711 | Maximal Element is Dual to Minimal Element | Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$, and $a \in T$.
The following are dual statements:
:$a$ is a maximal element of $T$
:$a$ is a minimal element of $T$ | By definition, $a$ is a maximal element of $T$ {{iff}}:
:$\forall t \in T: a \preceq t$ implies $a = t$
The dual of this statement is:
:$\forall t \in T: t \preceq a$ implies $a = t$
by Dual Pairs (Order Theory).
By definition, this means $a$ is a minimal element of $T$.
The converse follows from Dual of Dual Statement... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$, and $a \in T$.
The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
:$a$ is a [[Definition:Maximal Element|maximal element]] of $T$
:$a$ is a [[Definition:Minimal Element|minimal element]] of $T$ | By definition, $a$ is a [[Definition:Maximal Element|maximal element]] of $T$ {{iff}}:
:$\forall t \in T: a \preceq t$ implies $a = t$
The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
:$\forall t \in T: t \preceq a$ implies $a = t$
by [[Dual Pairs (Order Theory)]].
By definition, this m... | Maximal Element is Dual to Minimal Element | https://proofwiki.org/wiki/Maximal_Element_is_Dual_to_Minimal_Element | https://proofwiki.org/wiki/Maximal_Element_is_Dual_to_Minimal_Element | [
"Maximal Elements",
"Minimal Elements",
"Dual Pairs (Order Theory)"
] | [
"Definition:Ordered Set",
"Definition:Dual Statement (Order Theory)",
"Definition:Maximal/Element",
"Definition:Minimal/Element"
] | [
"Definition:Maximal/Element",
"Definition:Dual Statement (Order Theory)",
"Dual Pairs (Order Theory)",
"Definition:Minimal/Element",
"Dual of Dual Statement (Order Theory)"
] |
proofwiki-6712 | Greatest Element is Dual to Smallest Element | Let $\struct {S, \preceq}$ be an ordered set.
Let $a \in S$.
The following are dual statements:
:$a$ is the greatest element of $S$
:$a$ is the smallest element of $S$ | By definition, $a$ is the greatest element of $S$ {{iff}}:
:$\forall b \in S: b \preceq a$
The dual of this statement is:
:$\forall b \in S: a \preceq b$
by Dual Pairs (Order Theory).
By definition, this means $a$ is the smallest element of $S$.
The converse follows from Dual of Dual Statement (Order Theory).
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $a \in S$.
The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
:$a$ is the [[Definition:Greatest Element|greatest element]] of $S$
:$a$ is the [[Definition:Smallest Element|smallest element]] of $S$ | By definition, $a$ is the [[Definition:Greatest Element|greatest element]] of $S$ {{iff}}:
:$\forall b \in S: b \preceq a$
The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
:$\forall b \in S: a \preceq b$
by [[Dual Pairs (Order Theory)]].
By definition, this means $a$ is the [[Definition... | Greatest Element is Dual to Smallest Element | https://proofwiki.org/wiki/Greatest_Element_is_Dual_to_Smallest_Element | https://proofwiki.org/wiki/Greatest_Element_is_Dual_to_Smallest_Element | [
"Greatest Elements",
"Smallest Elements",
"Dual Pairs (Order Theory)"
] | [
"Definition:Ordered Set",
"Definition:Dual Statement (Order Theory)",
"Definition:Greatest Element",
"Definition:Smallest Element"
] | [
"Definition:Greatest Element",
"Definition:Dual Statement (Order Theory)",
"Dual Pairs (Order Theory)",
"Definition:Smallest Element",
"Dual of Dual Statement (Order Theory)"
] |
proofwiki-6713 | Lower Closure is Dual to Upper Closure | Let $\left({S, \preccurlyeq}\right)$ be an ordered set.
Let $a, b \in S$.
Let $T \subseteq S$
The following are pairs of dual statements:
:$b \in a^\preccurlyeq$, the lower closure of $a$
:$b \in a^\succcurlyeq$, the upper closure of $a$
:$b \in T^\preccurlyeq$, the lower closure of $T$
:$b \in T^\succcurlyeq$, the upp... | === Elements ===
By definition of lower closure, $b \in a^\preccurlyeq$ {{iff}}:
:$b \preccurlyeq a$
The dual of this statement is:
:$a \preccurlyeq b$
by Dual Pairs (Order Theory).
By definition of upper closure, this means $b \in a^\succcurlyeq$.
The converse follows from Dual of Dual Statement (Order Theory).
{{qed|... | Let $\left({S, \preccurlyeq}\right)$ be an [[Definition:Ordered Set|ordered set]].
Let $a, b \in S$.
Let $T \subseteq S$
The following are pairs of [[Definition:Dual Statement (Order Theory)|dual statements]]:
:$b \in a^\preccurlyeq$, the [[Definition:Lower Closure of Element|lower closure]] of $a$
:$b \in a^\succ... | === Elements ===
By definition of [[Definition:Lower Closure of Element|lower closure]], $b \in a^\preccurlyeq$ {{iff}}:
:$b \preccurlyeq a$
The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
:$a \preccurlyeq b$
by [[Dual Pairs (Order Theory)]].
By definition of [[Definition:Upper Closur... | Lower Closure is Dual to Upper Closure | https://proofwiki.org/wiki/Lower_Closure_is_Dual_to_Upper_Closure | https://proofwiki.org/wiki/Lower_Closure_is_Dual_to_Upper_Closure | [
"Upper Closures",
"Lower Closures",
"Dual Pairs (Order Theory)"
] | [
"Definition:Ordered Set",
"Definition:Dual Statement (Order Theory)",
"Definition:Lower Closure/Element",
"Definition:Upper Closure/Element",
"Definition:Lower Closure/Set",
"Definition:Upper Closure/Set"
] | [
"Definition:Lower Closure/Element",
"Definition:Dual Statement (Order Theory)",
"Dual Pairs (Order Theory)",
"Definition:Upper Closure/Element",
"Dual of Dual Statement (Order Theory)",
"Definition:Dual Statement (Order Theory)",
"Dual Pairs (Order Theory)"
] |
proofwiki-6714 | Strict Lower Closure is Dual to Strict Upper Closure | Let $\left({S, \preceq}\right)$ be an ordered set.
Let $a, b \in S$.
The following are dual statements:
:$b \in a^\prec$, the strict lower closure of $a$
:$b \in a^\succ$, the strict upper closure of $a$ | By definition of strict lower closure:
:$b \in a^\prec$
{{iff}}
:$b$ strictly precedes $a$
The dual of this statement is:
:$b$ strictly succeeds $a$
by Dual Pairs (Order Theory).
By definition of strict upper closure, this means:
: $b \in a^\succ$
The converse follows from Dual of Dual Statement (Order Theory).
{{qed}} | Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
Let $a, b \in S$.
The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
:$b \in a^\prec$, the [[Definition:Strict Lower Closure of Element|strict lower closure]] of $a$
:$b \in a^\succ$, the [[Definition:Strict U... | By definition of [[Definition:Strict Lower Closure of Element|strict lower closure]]:
:$b \in a^\prec$
{{iff}}
:$b$ [[Definition:Strictly Precede|strictly precedes]] $a$
The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
:$b$ [[Definition:Strictly Succeed|strictly succeeds]] $a$
by [[Dual Pa... | Strict Lower Closure is Dual to Strict Upper Closure | https://proofwiki.org/wiki/Strict_Lower_Closure_is_Dual_to_Strict_Upper_Closure | https://proofwiki.org/wiki/Strict_Lower_Closure_is_Dual_to_Strict_Upper_Closure | [
"Lower Closures",
"Upper Closures",
"Dual Pairs (Order Theory)"
] | [
"Definition:Ordered Set",
"Definition:Dual Statement (Order Theory)",
"Definition:Strict Lower Closure/Element",
"Definition:Strict Upper Closure/Element"
] | [
"Definition:Strict Lower Closure/Element",
"Definition:Strictly Precede",
"Definition:Dual Statement (Order Theory)",
"Definition:Strictly Succeed",
"Dual Pairs (Order Theory)",
"Definition:Strict Upper Closure/Element",
"Dual of Dual Statement (Order Theory)"
] |
proofwiki-6715 | Join is Dual to Meet | Let $\struct {S, \preceq}$ be an ordered set.
Let $a, b, c \in S$.
The following are dual statements:
:$c = a \vee b$, the join of $a$ and $b$
:$c = a \wedge b$ the meet of $a$ and $b$ | By definition of join, $c = a \vee b$ {{iff}}:
:$c = \sup \set {a, b}$
where $\sup$ denotes supremum.
The dual of this statement is:
:$c = \inf \set {a, b}$
where $\inf$ denotes infimum, by Dual Pairs (Order Theory).
By definition of meet, this means $c = a \wedge b$.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $a, b, c \in S$.
The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
:$c = a \vee b$, the [[Definition:Join (Order Theory)|join]] of $a$ and $b$
:$c = a \wedge b$ the [[Definition:Meet (Order Theory)|meet]] of $... | By definition of [[Definition:Join (Order Theory)|join]], $c = a \vee b$ {{iff}}:
:$c = \sup \set {a, b}$
where $\sup$ denotes [[Definition:Supremum of Set|supremum]].
The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
:$c = \inf \set {a, b}$
where $\inf$ denotes [[Definition:Infimum of S... | Join is Dual to Meet | https://proofwiki.org/wiki/Join_is_Dual_to_Meet | https://proofwiki.org/wiki/Join_is_Dual_to_Meet | [
"Join and Meet",
"Dual Pairs (Order Theory)"
] | [
"Definition:Ordered Set",
"Definition:Dual Statement (Order Theory)",
"Definition:Join (Order Theory)",
"Definition:Meet (Order Theory)"
] | [
"Definition:Join (Order Theory)",
"Definition:Supremum of Set",
"Definition:Dual Statement (Order Theory)",
"Definition:Infimum of Set",
"Dual Pairs (Order Theory)",
"Definition:Meet (Order Theory)"
] |
proofwiki-6716 | Self-Distributive Law for Conditional/Formulation 1/Forward Implication | :$p \implies \paren {q \implies r} \vdash \paren {p \implies q} \implies \paren {p \implies r}$ | {{BeginTableau|p \implies \paren {q \implies r} \vdash \paren {p \implies q} \implies \paren {p \implies r} }}
{{Premise|1|p \implies \paren {q \implies r} }}
{{Assumption|2|p \implies q}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q \implies r|1|3}}
{{ModusPonens|5|2, 3|q|2|3}}
{{ModusPonens|6|1, 2, 3|r|4|5}}
{{Implicati... | :$p \implies \paren {q \implies r} \vdash \paren {p \implies q} \implies \paren {p \implies r}$ | {{BeginTableau|p \implies \paren {q \implies r} \vdash \paren {p \implies q} \implies \paren {p \implies r} }}
{{Premise|1|p \implies \paren {q \implies r} }}
{{Assumption|2|p \implies q}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q \implies r|1|3}}
{{ModusPonens|5|2, 3|q|2|3}}
{{ModusPonens|6|1, 2, 3|r|4|5}}
{{Implicati... | Self-Distributive Law for Conditional/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_1/Forward_Implication/Proof | [
"Self-Distributive Law for Conditional"
] | [] | [] |
proofwiki-6717 | Self-Distributive Law for Conditional/Formulation 1/Reverse Implication | :$\paren {p \implies q} \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r}$ | {{BeginTableau|\paren {p \implies q} \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r} }}
{{Premise|1|\paren {p \implies q} \implies \paren {p \implies r} }}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{SequentIntro|4|3|p \implies q|3|True Statement is implied by Every Statement}}
{{ModusPonens|5|1, 3|p... | :$\paren {p \implies q} \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r}$ | {{BeginTableau|\paren {p \implies q} \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r} }}
{{Premise|1|\paren {p \implies q} \implies \paren {p \implies r} }}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{SequentIntro|4|3|p \implies q|3|[[True Statement is implied by Every Statement]]}}
{{ModusPonens|5|1,... | Self-Distributive Law for Conditional/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Self-Distributive_Law_for_Conditional/Formulation_1/Reverse_Implication/Proof | [
"Self-Distributive Law for Conditional"
] | [] | [
"True Statement is implied by Every Statement"
] |
proofwiki-6718 | Rule of Exportation/Formulation 1 | :$\paren {p \land q} \implies r \dashv \vdash p \implies \paren {q \implies r}$ | {{BeginTableau|\paren {p \land q} \implies r \vdash p \implies \paren {q \implies r} }}
{{Premise|1|\paren {p \land q} \implies r}}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{Conjunction|4|2, 3|p \land q|2|3}}
{{ModusPonens|5|1, 2, 3|r|1|4}}
{{Implication|6|1, 2|q \implies r|3|5}}
{{Implication|7|1|p \implies \paren {q \i... | :$\paren {p \land q} \implies r \dashv \vdash p \implies \paren {q \implies r}$ | {{BeginTableau|\paren {p \land q} \implies r \vdash p \implies \paren {q \implies r} }}
{{Premise|1|\paren {p \land q} \implies r}}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{Conjunction|4|2, 3|p \land q|2|3}}
{{ModusPonens|5|1, 2, 3|r|1|4}}
{{Implication|6|1, 2|q \implies r|3|5}}
{{Implication|7|1|p \implies \paren {q \i... | Rule of Exportation/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_1/Forward_Implication/Proof | [
"Rule of Exportation"
] | [] | [] |
proofwiki-6719 | Rule of Exportation/Formulation 1 | :$\paren {p \land q} \implies r \dashv \vdash p \implies \paren {q \implies r}$ | === Proof of Forward Implication ===
{{:Rule of Exportation/Formulation 1/Forward Implication/Proof}}
=== Proof of Reverse Implication ===
{{:Rule of Exportation/Formulation 1/Reverse Implication/Proof}} | :$\paren {p \land q} \implies r \dashv \vdash p \implies \paren {q \implies r}$ | === [[Rule of Exportation/Formulation 1/Forward Implication/Proof|Proof of Forward Implication]] ===
{{:Rule of Exportation/Formulation 1/Forward Implication/Proof}}
=== [[Rule of Exportation/Formulation 1/Reverse Implication/Proof|Proof of Reverse Implication]] ===
{{:Rule of Exportation/Formulation 1/Reverse Implica... | Rule of Exportation/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_1/Proof_1 | [
"Rule of Exportation"
] | [] | [
"Rule of Exportation/Formulation 1/Forward Implication/Proof",
"Rule of Exportation/Formulation 1/Reverse Implication/Proof"
] |
proofwiki-6720 | Rule of Exportation/Formulation 1 | :$\paren {p \land q} \implies r \dashv \vdash p \implies \paren {q \implies r}$ | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin{array}{|ccccc||ccccc|} \hline
(p & \land & q) & \implies & r & p & \implies & (q & \implies & r) \\
\hline
\F & \F & \F & \T & \F & \F... | :$\paren {p \land q} \implies r \dashv \vdash p \implies \paren {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 connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|c... | Rule of Exportation/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_1/Proof_by_Truth_Table | [
"Rule of Exportation"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6721 | Rule of Exportation/Formulation 1 | :$\paren {p \land q} \implies r \dashv \vdash p \implies \paren {q \implies r}$ | {{BeginTableau|p \implies \paren {q \implies r} \vdash \paren {p \land q} \implies r}}
{{Premise|1|p \implies \paren {q \implies r} }}
{{Assumption|2|p \land q}}
{{Simplification|3|2|p|2|1}}
{{Simplification|4|2|q|2|2}}
{{ModusPonens|5|1, 2|q \implies r|1|3}}
{{ModusPonens|6|1, 2|r|5|4}}
{{Implication|7|1|\paren {p \la... | :$\paren {p \land q} \implies r \dashv \vdash p \implies \paren {q \implies r}$ | {{BeginTableau|p \implies \paren {q \implies r} \vdash \paren {p \land q} \implies r}}
{{Premise|1|p \implies \paren {q \implies r} }}
{{Assumption|2|p \land q}}
{{Simplification|3|2|p|2|1}}
{{Simplification|4|2|q|2|2}}
{{ModusPonens|5|1, 2|q \implies r|1|3}}
{{ModusPonens|6|1, 2|r|5|4}}
{{Implication|7|1|\paren {p \la... | Rule of Exportation/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_1/Reverse_Implication/Proof | [
"Rule of Exportation"
] | [] | [] |
proofwiki-6722 | Square Matrix with Duplicate Columns has Zero Determinant | If two columns of a square matrix over a commutative ring $\struct {R, +, \circ}$ are identical, then its determinant is zero. | Let $\mathbf A$ be a square matrix over $R$ with two identical columns.
Let $\mathbf A^\intercal$ denote the transpose of $\mathbf A$.
Then $\mathbf A^\intercal$ has two identical rows.
Then:
{{begin-eqn}}
{{eqn | l = \map \det {\mathbf A}
| r = \map \det {\mathbf A^\intercal}
| c = Determinant of Transpose... | If two [[Definition:Column of Matrix|columns]] of a [[Definition:Square Matrix|square matrix]] over a [[Definition:Commutative Ring|commutative ring]] $\struct {R, +, \circ}$ are identical, then its [[Definition:Determinant of Matrix|determinant]] is zero. | Let $\mathbf A$ be a [[Definition:Square Matrix|square matrix]] over $R$ with two identical [[Definition:Column of Matrix|columns]].
Let $\mathbf A^\intercal$ denote the [[Definition:Transpose of Matrix|transpose]] of $\mathbf A$.
Then $\mathbf A^\intercal$ has two identical [[Definition:Row of Matrix|rows]].
Then:
... | Square Matrix with Duplicate Columns has Zero Determinant | https://proofwiki.org/wiki/Square_Matrix_with_Duplicate_Columns_has_Zero_Determinant | https://proofwiki.org/wiki/Square_Matrix_with_Duplicate_Columns_has_Zero_Determinant | [
"Square Matrix with Duplicate Columns has Zero Determinant",
"Determinants",
"Matrix Algebra"
] | [
"Definition:Matrix/Column",
"Definition:Matrix/Square Matrix",
"Definition:Commutative Ring",
"Definition:Determinant/Matrix"
] | [
"Definition:Matrix/Square Matrix",
"Definition:Matrix/Column",
"Definition:Transpose of Matrix",
"Definition:Matrix/Row",
"Determinant of Transpose",
"Square Matrix with Duplicate Rows has Zero Determinant"
] |
proofwiki-6723 | Inverses of Elements Related by Compatible Relation | :$\forall x, y \in G: x \mathrel \RR y \iff y^{-1} \mathrel \RR x^{-1}$ | Let $e$ be the group identity of $G$.
By Relation Compatible with Group Operation is Strongly Compatible: Corollary:
:$(1): \quad x \mathrel \RR y \iff e \mathrel \RR y \circ x^{-1}$
By Relation Compatible with Group Operation is Strongly Compatible: Corollary, also:
:$(2): \quad y^{-1} \mathrel \RR x^{-1} \iff e \math... | :$\forall x, y \in G: x \mathrel \RR y \iff y^{-1} \mathrel \RR x^{-1}$ | Let $e$ be the [[Definition:Identity Element|group identity]] of $G$.
By [[Relation Compatible with Group Operation is Strongly Compatible/Corollary|Relation Compatible with Group Operation is Strongly Compatible: Corollary]]:
:$(1): \quad x \mathrel \RR y \iff e \mathrel \RR y \circ x^{-1}$
By [[Relation Compatible... | Inverses of Elements Related by Compatible Relation | https://proofwiki.org/wiki/Inverses_of_Elements_Related_by_Compatible_Relation | https://proofwiki.org/wiki/Inverses_of_Elements_Related_by_Compatible_Relation | [
"Relations Compatible with Group Operation"
] | [] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Relation Compatible with Group Operation is Strongly Compatible/Corollary",
"Relation Compatible with Group Operation is Strongly Compatible/Corollary",
"Inverse of Group Inverse",
"Category:Relations Compatible with Group Operation"
] |
proofwiki-6724 | Ordering Induced by Join Semilattice | Let $\struct {S, \vee, \preceq}$ be a join semilattice.
By Join Semilattice is Semilattice, $\struct {S, \vee}$ is a semilattice.
By Semilattice Induces Ordering, $\struct {S, \vee}$ induces an ordering $\preceq'$ on $S$, by:
:$a \preceq' b$ {{iff}} $a \vee b = b$
for all $a, b \in S$.
The ordering $\preceq'$ coincides... | It is to be shown that, for all $a, b \in S$:
:$a \preceq b$ {{iff}} $b = \sup \set {a, b}$
by definition of join.
Here $\sup$ denotes supremum.
Since any upper bound $c$ of $\set {a, b}$ must satisfy:
:$b \preceq c$
it suffices to verify that:
:$a \preceq b$ {{iff}} $b$ is an upper bound for $\set {a, b}$
Since $\prec... | Let $\struct {S, \vee, \preceq}$ be a [[Definition:Join Semilattice|join semilattice]].
By [[Join Semilattice is Semilattice]], $\struct {S, \vee}$ is a [[Definition:Semilattice|semilattice]].
By [[Semilattice Induces Ordering]], $\struct {S, \vee}$ induces an [[Definition:Ordering|ordering]] $\preceq'$ on $S$, by:
... | It is to be shown that, for all $a, b \in S$:
:$a \preceq b$ {{iff}} $b = \sup \set {a, b}$
by definition of [[Definition:Join (Order Theory)|join]].
Here $\sup$ denotes [[Definition:Supremum of Set|supremum]].
Since any [[Definition:Upper Bound of Set|upper bound]] $c$ of $\set {a, b}$ must satisfy:
:$b \preceq ... | Ordering Induced by Join Semilattice | https://proofwiki.org/wiki/Ordering_Induced_by_Join_Semilattice | https://proofwiki.org/wiki/Ordering_Induced_by_Join_Semilattice | [
"Order Theory",
"Join Semilattices"
] | [
"Definition:Join Semilattice",
"Join Semilattice is Semilattice",
"Definition:Semilattice",
"Semilattice Induces Ordering",
"Definition:Ordering",
"Definition:Ordering"
] | [
"Definition:Join (Order Theory)",
"Definition:Supremum of Set",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set",
"Definition:Reflexive Relation",
"Definition:Logical Equivalence",
"Ordering Induced by Meet Semilattice",
"Category:Order Theory",
"Category:Join Semilattices"
] |
proofwiki-6725 | Combination Theorem for Complex Derivatives | Let $D$ be an open subset of the set of complex numbers $\C$.
Let $f, g: D \to \C$ be complex-differentiable functions on $D$
Let $z \in D$.
Let $w, c \in \C$ be arbitrary complex numbers.
Then the following results hold:
=== Sum Rule ===
{{:Combination Theorem for Complex Derivatives/Sum Rule}}
=== Multiple Rule ===
{... | Define $k: D \to \C$ as the pointwise product of $f$ and $g$, so $k = fg$.
Let $z_0 \in D$ be a point in $D$.
{{begin-eqn}}
{{eqn | l = \map {k'} {z_0}
| r = \lim_{h \mathop \to 0} \frac {\map k {z_0 + h} - \map k {z_0} } h
| c =
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\map f {z_0 + h} \, \map g {z_0 +... | Let $D$ be an [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$.
Let $f, g: D \to \C$ be [[Definition:Complex-Differentiable Function|complex-differentiable functions]] on $D$
Let $z \in D$.
Let $w, c \in \... | Define $k: D \to \C$ as the [[Definition:Pointwise Multiplication of Complex-Valued Functions|pointwise product]] of $f$ and $g$, so $k = fg$.
Let $z_0 \in D$ be a point in $D$.
{{begin-eqn}}
{{eqn | l = \map {k'} {z_0}
| r = \lim_{h \mathop \to 0} \frac {\map k {z_0 + h} - \map k {z_0} } h
| c =
}}
{{eqn... | Combination Theorem for Complex Derivatives/Product Rule/Proof 1 | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Product_Rule/Proof_1 | [
"Combination Theorem for Complex Derivatives",
"Complex Differential Calculus"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Subset",
"Definition:Set",
"Definition:Complex Number",
"Definition:Differentiable Mapping/Complex Function",
"Definition:Complex Number",
"Combination Theorem for Complex Derivatives/Sum Rule",
"Combination Theorem for Complex Derivatives/Multiple R... | [
"Definition:Pointwise Multiplication of Complex-Valued Functions"
] |
proofwiki-6726 | Combination Theorem for Complex Derivatives | Let $D$ be an open subset of the set of complex numbers $\C$.
Let $f, g: D \to \C$ be complex-differentiable functions on $D$
Let $z \in D$.
Let $w, c \in \C$ be arbitrary complex numbers.
Then the following results hold:
=== Sum Rule ===
{{:Combination Theorem for Complex Derivatives/Sum Rule}}
=== Multiple Rule ===
{... | Denote the open ball of $0$ by radius $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D$.
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
:$\map f {z + h} = \map f z + \map h {\map {f'} z + \map {\epsilon_f}... | Let $D$ be an [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$.
Let $f, g: D \to \C$ be [[Definition:Complex-Differentiable Function|complex-differentiable functions]] on $D$
Let $z \in D$.
Let $w, c \in \... | Denote the [[Definition:Open Ball|open ball]] of $0$ by [[Definition:Radius of Open Ball|radius]] $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D$.
By the [[Epsilon-Function Complex Differentiability Condition]], it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
:$... | Combination Theorem for Complex Derivatives/Product Rule/Proof 2 | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Product_Rule/Proof_2 | [
"Combination Theorem for Complex Derivatives",
"Complex Differential Calculus"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Subset",
"Definition:Set",
"Definition:Complex Number",
"Definition:Differentiable Mapping/Complex Function",
"Definition:Complex Number",
"Combination Theorem for Complex Derivatives/Sum Rule",
"Combination Theorem for Complex Derivatives/Multiple R... | [
"Definition:Open Ball",
"Definition:Open Ball/Radius",
"Epsilon-Function Differentiability Condition/Complex Case",
"Definition:Complex Function",
"Definition:Convergent Mapping/Complex Function",
"Combination Theorem for Limits of Functions/Complex/Product Rule",
"Combination Theorem for Limits of Func... |
proofwiki-6727 | Combination Theorem for Complex Derivatives | Let $D$ be an open subset of the set of complex numbers $\C$.
Let $f, g: D \to \C$ be complex-differentiable functions on $D$
Let $z \in D$.
Let $w, c \in \C$ be arbitrary complex numbers.
Then the following results hold:
=== Sum Rule ===
{{:Combination Theorem for Complex Derivatives/Sum Rule}}
=== Multiple Rule ===
{... | Let $z_0 \in D$ be a point in $D$.
Define $k : D \to \Z$ by $\map k {z_0} = \map f {z_0} + \map g {z_0}$.
Then:
{{begin-eqn}}
{{eqn | l = \map { k' } {z_0}
| r = \lim_{h \mathop \to 0} \frac {\map k {z_0 + h} - \map k {z_0} } h
| c = {{Defof|Derivative of Complex Function}}
}}
{{eqn | r = \lim_{h \mathop \t... | Let $D$ be an [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$.
Let $f, g: D \to \C$ be [[Definition:Complex-Differentiable Function|complex-differentiable functions]] on $D$
Let $z \in D$.
Let $w, c \in \... | Let $z_0 \in D$ be a point in $D$.
Define $k : D \to \Z$ by $\map k {z_0} = \map f {z_0} + \map g {z_0}$.
Then:
{{begin-eqn}}
{{eqn | l = \map { k' } {z_0}
| r = \lim_{h \mathop \to 0} \frac {\map k {z_0 + h} - \map k {z_0} } h
| c = {{Defof|Derivative of Complex Function}}
}}
{{eqn | r = \lim_{h \mathop... | Combination Theorem for Complex Derivatives/Sum Rule/Proof 1 | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Sum_Rule/Proof_1 | [
"Combination Theorem for Complex Derivatives",
"Complex Differential Calculus"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Subset",
"Definition:Set",
"Definition:Complex Number",
"Definition:Differentiable Mapping/Complex Function",
"Definition:Complex Number",
"Combination Theorem for Complex Derivatives/Sum Rule",
"Combination Theorem for Complex Derivatives/Multiple R... | [
"Complex Multiplication Distributes over Addition",
"Combination Theorem for Limits of Functions/Complex/Sum Rule"
] |
proofwiki-6728 | Combination Theorem for Complex Derivatives | Let $D$ be an open subset of the set of complex numbers $\C$.
Let $f, g: D \to \C$ be complex-differentiable functions on $D$
Let $z \in D$.
Let $w, c \in \C$ be arbitrary complex numbers.
Then the following results hold:
=== Sum Rule ===
{{:Combination Theorem for Complex Derivatives/Sum Rule}}
=== Multiple Rule ===
{... | Denote the open ball of $0$ with radius $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D$.
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
:$\map f {z + h} = \map f z + h \paren {\map {f'} z + \map {\epsilo... | Let $D$ be an [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$.
Let $f, g: D \to \C$ be [[Definition:Complex-Differentiable Function|complex-differentiable functions]] on $D$
Let $z \in D$.
Let $w, c \in \... | Denote the [[Definition:Open Ball|open ball]] of $0$ with [[Definition:Radius of Open Ball|radius]] $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D$.
By the [[Epsilon-Function Complex Differentiability Condition]], it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
... | Combination Theorem for Complex Derivatives/Sum Rule/Proof 2 | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Sum_Rule/Proof_2 | [
"Combination Theorem for Complex Derivatives",
"Complex Differential Calculus"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Subset",
"Definition:Set",
"Definition:Complex Number",
"Definition:Differentiable Mapping/Complex Function",
"Definition:Complex Number",
"Combination Theorem for Complex Derivatives/Sum Rule",
"Combination Theorem for Complex Derivatives/Multiple R... | [
"Definition:Open Ball",
"Definition:Open Ball/Radius",
"Epsilon-Function Differentiability Condition/Complex Case",
"Definition:Complex Function",
"Definition:Convergent Mapping/Complex Function",
"Combination Theorem for Limits of Functions/Complex/Sum Rule",
"Epsilon-Function Differentiability Conditi... |
proofwiki-6729 | Combination Theorem for Complex Derivatives/Sum Rule | :$\map {\paren {f + g}'} z = \map {f'} z + \map {g'} z$ | Let $z_0 \in D$ be a point in $D$.
Define $k : D \to \Z$ by $\map k {z_0} = \map f {z_0} + \map g {z_0}$.
Then:
{{begin-eqn}}
{{eqn | l = \map { k' } {z_0}
| r = \lim_{h \mathop \to 0} \frac {\map k {z_0 + h} - \map k {z_0} } h
| c = {{Defof|Derivative of Complex Function}}
}}
{{eqn | r = \lim_{h \mathop \t... | :$\map {\paren {f + g}'} z = \map {f'} z + \map {g'} z$ | Let $z_0 \in D$ be a point in $D$.
Define $k : D \to \Z$ by $\map k {z_0} = \map f {z_0} + \map g {z_0}$.
Then:
{{begin-eqn}}
{{eqn | l = \map { k' } {z_0}
| r = \lim_{h \mathop \to 0} \frac {\map k {z_0 + h} - \map k {z_0} } h
| c = {{Defof|Derivative of Complex Function}}
}}
{{eqn | r = \lim_{h \mathop... | Combination Theorem for Complex Derivatives/Sum Rule/Proof 1 | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Sum_Rule | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Sum_Rule/Proof_1 | [
"Combination Theorem for Complex Derivatives"
] | [] | [
"Complex Multiplication Distributes over Addition",
"Combination Theorem for Limits of Functions/Complex/Sum Rule"
] |
proofwiki-6730 | Combination Theorem for Complex Derivatives/Sum Rule | :$\map {\paren {f + g}'} z = \map {f'} z + \map {g'} z$ | Denote the open ball of $0$ with radius $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D$.
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
:$\map f {z + h} = \map f z + h \paren {\map {f'} z + \map {\epsilo... | :$\map {\paren {f + g}'} z = \map {f'} z + \map {g'} z$ | Denote the [[Definition:Open Ball|open ball]] of $0$ with [[Definition:Radius of Open Ball|radius]] $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D$.
By the [[Epsilon-Function Complex Differentiability Condition]], it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
... | Combination Theorem for Complex Derivatives/Sum Rule/Proof 2 | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Sum_Rule | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Sum_Rule/Proof_2 | [
"Combination Theorem for Complex Derivatives"
] | [] | [
"Definition:Open Ball",
"Definition:Open Ball/Radius",
"Epsilon-Function Differentiability Condition/Complex Case",
"Definition:Complex Function",
"Definition:Convergent Mapping/Complex Function",
"Combination Theorem for Limits of Functions/Complex/Sum Rule",
"Epsilon-Function Differentiability Conditi... |
proofwiki-6731 | Combination Theorem for Complex Derivatives/Multiple Rule | :$\map {\paren {w f}'} z = w \map {f'} z$ | Denote the open ball of $0$ with radius $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D$.
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
:$\map f {z + h} = \map f z + h \paren {\map {f'} z + \map \epsilon... | :$\map {\paren {w f}'} z = w \map {f'} z$ | Denote the [[Definition:Open Ball|open ball]] of $0$ with [[Definition:Radius of Open Ball|radius]] $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D$.
By the [[Epsilon-Function Complex Differentiability Condition]], it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
... | Combination Theorem for Complex Derivatives/Multiple Rule | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Multiple_Rule | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Multiple_Rule | [
"Combination Theorem for Complex Derivatives"
] | [] | [
"Definition:Open Ball",
"Definition:Open Ball/Radius",
"Epsilon-Function Differentiability Condition/Complex Case",
"Definition:Complex Function",
"Combination Theorem for Limits of Functions/Complex/Multiple Rule",
"Epsilon-Function Differentiability Condition/Complex Case"
] |
proofwiki-6732 | Join Absorbs Meet | Let $\struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $\vee$ denote join.
Then $\vee$ absorbs $\wedge$.
That is, for all $a, b \in S$:
:$a \vee \paren {a \wedge b} = a$ | By Ordering in terms of Join, we have that:
:$a \vee \paren {a \wedge b} = a$ {{iff}} $a \wedge b \preceq a$
The result thus follows from Meet Precedes Operands.
{{qed}} | Let $\struct {S, \wedge, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
Let $\vee$ denote [[Definition:Join (Order Theory)|join]].
Then $\vee$ [[Definition:Absorption Law|absorbs]] $\wedge$.
That is, for all $a, b \in S$:
:$a \vee \paren {a \wedge b} = a$ | By [[Ordering in terms of Join]], we have that:
:$a \vee \paren {a \wedge b} = a$ {{iff}} $a \wedge b \preceq a$
The result thus follows from [[Meet Precedes Operands]].
{{qed}} | Join Absorbs Meet | https://proofwiki.org/wiki/Join_Absorbs_Meet | https://proofwiki.org/wiki/Join_Absorbs_Meet | [
"Join and Meet",
"Absorption Laws (Lattice Theory)"
] | [
"Definition:Meet Semilattice",
"Definition:Join (Order Theory)",
"Definition:Absorption Law"
] | [
"Ordering in terms of Join",
"Meet Precedes Operands"
] |
proofwiki-6733 | Meet Absorbs Join | Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Let $\wedge$ denote meet.
Then $\wedge$ absorbs $\vee$.
That is, for all $a, b \in S$:
:$a \wedge \paren {a \vee b} = a$ | By Dual Pairs (Order Theory), we observe that the theorem statement is dual to that of Join Absorbs Meet.
The result follows by the Global Duality Principle.
{{qed}} | Let $\struct {S, \vee, \preceq}$ be a [[Definition:Join Semilattice|join semilattice]].
Let $\wedge$ denote [[Definition:Meet (Order Theory)|meet]].
Then $\wedge$ [[Definition:Absorption Law|absorbs]] $\vee$.
That is, for all $a, b \in S$:
:$a \wedge \paren {a \vee b} = a$ | By [[Dual Pairs (Order Theory)]], we observe that the theorem statement is [[Definition:Dual Statement (Order Theory)|dual]] to that of [[Join Absorbs Meet]].
The result follows by the [[Duality Principle (Order Theory)/Global Duality|Global Duality Principle]].
{{qed}} | Meet Absorbs Join | https://proofwiki.org/wiki/Meet_Absorbs_Join | https://proofwiki.org/wiki/Meet_Absorbs_Join | [
"Join and Meet",
"Absorption Laws (Lattice Theory)"
] | [
"Definition:Join Semilattice",
"Definition:Meet (Order Theory)",
"Definition:Absorption Law"
] | [
"Dual Pairs (Order Theory)",
"Definition:Dual Statement (Order Theory)",
"Join Absorbs Meet",
"Duality Principle (Order Theory)/Global Duality"
] |
proofwiki-6734 | Rule of Exportation/Formulation 2 | :$\vdash \paren {\paren {p \land q} \implies r} \iff \paren {p \implies \paren {q \implies r} }$ | === Proof of Forward Implication ===
{{:Rule of Exportation/Formulation 2/Forward Implication/Proof 1}} | :$\vdash \paren {\paren {p \land q} \implies r} \iff \paren {p \implies \paren {q \implies r} }$ | === [[Rule of Exportation/Formulation 2/Forward Implication/Proof 1|Proof of Forward Implication]] ===
{{:Rule of Exportation/Formulation 2/Forward Implication/Proof 1}} | Rule of Exportation/Formulation 2 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2 | [
"Rule of Exportation"
] | [] | [
"Rule of Exportation/Formulation 2/Forward Implication/Proof 1",
"Rule of Exportation/Formulation 2/Forward Implication"
] |
proofwiki-6735 | Rule of Exportation/Formulation 2 | :$\vdash \paren {\paren {p \land q} \implies r} \iff \paren {p \implies \paren {q \implies r} }$ | {{BeginTableau|\vdash \paren {\paren {p \land q} \implies r} \implies \paren {p \implies \paren {q \implies r} } }}
{{Assumption|1|\paren {p \land q} \implies r}}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{Conjunction|4|2, 3|p \land q|2|3}}
{{ModusPonens|5|1, 2, 3|r|1|4}}
{{Implication|6|1, 2|q \implies r|3|5}}
{{Implicat... | :$\vdash \paren {\paren {p \land q} \implies r} \iff \paren {p \implies \paren {q \implies r} }$ | {{BeginTableau|\vdash \paren {\paren {p \land q} \implies r} \implies \paren {p \implies \paren {q \implies r} } }}
{{Assumption|1|\paren {p \land q} \implies r}}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{Conjunction|4|2, 3|p \land q|2|3}}
{{ModusPonens|5|1, 2, 3|r|1|4}}
{{Implication|6|1, 2|q \implies r|3|5}}
{{Implicat... | Rule of Exportation/Formulation 2/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Forward_Implication/Proof_1 | [
"Rule of Exportation"
] | [] | [] |
proofwiki-6736 | Rule of Exportation/Formulation 2 | :$\vdash \paren {\paren {p \land q} \implies r} \iff \paren {p \implies \paren {q \implies r} }$ | {{BeginTableau|\vdash \paren {\paren {p \land q} \implies r} \implies \paren {p \implies \paren {q \implies r} } }}
{{Assumption|1|\paren {p \land q} \implies r}}
{{SequentIntro|2|1|p \implies \paren {q \implies r}|1|Rule of Exportation: Formulation 1: Forward Implication}}
{{Implication|3||\paren {\paren {p \land q} \... | :$\vdash \paren {\paren {p \land q} \implies r} \iff \paren {p \implies \paren {q \implies r} }$ | {{BeginTableau|\vdash \paren {\paren {p \land q} \implies r} \implies \paren {p \implies \paren {q \implies r} } }}
{{Assumption|1|\paren {p \land q} \implies r}}
{{SequentIntro|2|1|p \implies \paren {q \implies r}|1|[[Rule of Exportation/Formulation 1/Forward Implication|Rule of Exportation: Formulation 1: Forward Imp... | Rule of Exportation/Formulation 2/Forward Implication/Proof 2 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Forward_Implication/Proof_2 | [
"Rule of Exportation"
] | [] | [
"Rule of Exportation/Formulation 1/Forward Implication"
] |
proofwiki-6737 | Rule of Exportation/Formulation 2 | :$\vdash \paren {\paren {p \land q} \implies r} \iff \paren {p \implies \paren {q \implies r} }$ | 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>$\begin{array}{|ccccc|c|ccccc|} \hline
((p & \land & q) & \implies & r) & \implies & (p & \implies & (q & \implies & r)) \\
\hline
\F & \F &... | :$\vdash \paren {\paren {p \land q} \implies r} \iff \paren {p \implies \paren {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... | Rule of Exportation/Formulation 2/Forward Implication/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Forward_Implication/Proof_by_Truth_Table | [
"Rule of Exportation"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-6738 | Rule of Exportation/Formulation 2 | :$\vdash \paren {\paren {p \land q} \implies r} \iff \paren {p \implies \paren {q \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r} }}
{{Assumption|1|p \implies \paren {q \implies r} }}
{{Assumption|2|p \land q}}
{{Simplification|3|2|p|2|1}}
{{Simplification|4|2|q|2|2}}
{{ModusPonens|5|1, 2|q \implies r|1|3}}
{{ModusPonens|6|1, 2|r|5|4}}... | :$\vdash \paren {\paren {p \land q} \implies r} \iff \paren {p \implies \paren {q \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r} }}
{{Assumption|1|p \implies \paren {q \implies r} }}
{{Assumption|2|p \land q}}
{{Simplification|3|2|p|2|1}}
{{Simplification|4|2|q|2|2}}
{{ModusPonens|5|1, 2|q \implies r|1|3}}
{{ModusPonens|6|1, 2|r|5|4}}... | Rule of Exportation/Formulation 2/Reverse Implication/Proof 1 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Reverse_Implication/Proof_1 | [
"Rule of Exportation"
] | [] | [] |
proofwiki-6739 | Rule of Exportation/Formulation 2 | :$\vdash \paren {\paren {p \land q} \implies r} \iff \paren {p \implies \paren {q \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r} }}
{{Assumption|1|p \implies \paren {q \implies r} }}
{{SequentIntro|2|1|\paren {p \land q} \implies r|1|Rule of Exportation: Formulation 1: Reverse Implication}}
{{Implication|3||\paren {p \implies \paren {... | :$\vdash \paren {\paren {p \land q} \implies r} \iff \paren {p \implies \paren {q \implies r} }$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r} }}
{{Assumption|1|p \implies \paren {q \implies r} }}
{{SequentIntro|2|1|\paren {p \land q} \implies r|1|[[Rule of Exportation/Formulation 1/Reverse Implication|Rule of Exportation: Formulation 1: Reverse Im... | Rule of Exportation/Formulation 2/Reverse Implication/Proof 2 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Reverse_Implication/Proof_2 | [
"Rule of Exportation"
] | [] | [
"Rule of Exportation/Formulation 1/Reverse Implication"
] |
proofwiki-6740 | Rule of Exportation/Formulation 2 | :$\vdash \paren {\paren {p \land q} \implies r} \iff \paren {p \implies \paren {q \implies r} }$ | 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>$\begin{array}{|ccccc|c|ccccc|} \hline
(p & \implies & (q & \implies & r)) & \implies & ((p & \land & q) & \implies & r) \\
\hline
\F & \T &... | :$\vdash \paren {\paren {p \land q} \implies r} \iff \paren {p \implies \paren {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... | Rule of Exportation/Formulation 2/Reverse Implication/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Reverse_Implication/Proof_by_Truth_Table | [
"Rule of Exportation"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-6741 | Rule of Exportation/Formulation 2/Forward Implication | :$\vdash \paren {\paren {p \land q} \implies r} \implies \paren {p \implies \paren {q \implies r} }$ | {{BeginTableau|\vdash \paren {\paren {p \land q} \implies r} \implies \paren {p \implies \paren {q \implies r} } }}
{{Assumption|1|\paren {p \land q} \implies r}}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{Conjunction|4|2, 3|p \land q|2|3}}
{{ModusPonens|5|1, 2, 3|r|1|4}}
{{Implication|6|1, 2|q \implies r|3|5}}
{{Implicat... | :$\vdash \paren {\paren {p \land q} \implies r} \implies \paren {p \implies \paren {q \implies r} }$ | {{BeginTableau|\vdash \paren {\paren {p \land q} \implies r} \implies \paren {p \implies \paren {q \implies r} } }}
{{Assumption|1|\paren {p \land q} \implies r}}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{Conjunction|4|2, 3|p \land q|2|3}}
{{ModusPonens|5|1, 2, 3|r|1|4}}
{{Implication|6|1, 2|q \implies r|3|5}}
{{Implicat... | Rule of Exportation/Formulation 2/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Forward_Implication/Proof_1 | [
"Rule of Exportation"
] | [] | [] |
proofwiki-6742 | Rule of Exportation/Formulation 2/Forward Implication | :$\vdash \paren {\paren {p \land q} \implies r} \implies \paren {p \implies \paren {q \implies r} }$ | {{BeginTableau|\vdash \paren {\paren {p \land q} \implies r} \implies \paren {p \implies \paren {q \implies r} } }}
{{Assumption|1|\paren {p \land q} \implies r}}
{{SequentIntro|2|1|p \implies \paren {q \implies r}|1|Rule of Exportation: Formulation 1: Forward Implication}}
{{Implication|3||\paren {\paren {p \land q} \... | :$\vdash \paren {\paren {p \land q} \implies r} \implies \paren {p \implies \paren {q \implies r} }$ | {{BeginTableau|\vdash \paren {\paren {p \land q} \implies r} \implies \paren {p \implies \paren {q \implies r} } }}
{{Assumption|1|\paren {p \land q} \implies r}}
{{SequentIntro|2|1|p \implies \paren {q \implies r}|1|[[Rule of Exportation/Formulation 1/Forward Implication|Rule of Exportation: Formulation 1: Forward Imp... | Rule of Exportation/Formulation 2/Forward Implication/Proof 2 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Forward_Implication/Proof_2 | [
"Rule of Exportation"
] | [] | [
"Rule of Exportation/Formulation 1/Forward Implication"
] |
proofwiki-6743 | Rule of Exportation/Formulation 2/Forward Implication | :$\vdash \paren {\paren {p \land q} \implies r} \implies \paren {p \implies \paren {q \implies r} }$ | 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>$\begin{array}{|ccccc|c|ccccc|} \hline
((p & \land & q) & \implies & r) & \implies & (p & \implies & (q & \implies & r)) \\
\hline
\F & \F &... | :$\vdash \paren {\paren {p \land q} \implies r} \implies \paren {p \implies \paren {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... | Rule of Exportation/Formulation 2/Forward Implication/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Forward_Implication/Proof_by_Truth_Table | [
"Rule of Exportation"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-6744 | Rule of Exportation/Formulation 2/Reverse Implication | :$\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r}$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r} }}
{{Assumption|1|p \implies \paren {q \implies r} }}
{{Assumption|2|p \land q}}
{{Simplification|3|2|p|2|1}}
{{Simplification|4|2|q|2|2}}
{{ModusPonens|5|1, 2|q \implies r|1|3}}
{{ModusPonens|6|1, 2|r|5|4}}... | :$\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r}$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r} }}
{{Assumption|1|p \implies \paren {q \implies r} }}
{{Assumption|2|p \land q}}
{{Simplification|3|2|p|2|1}}
{{Simplification|4|2|q|2|2}}
{{ModusPonens|5|1, 2|q \implies r|1|3}}
{{ModusPonens|6|1, 2|r|5|4}}... | Rule of Exportation/Formulation 2/Reverse Implication/Proof 1 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Reverse_Implication | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Reverse_Implication/Proof_1 | [
"Rule of Exportation"
] | [] | [] |
proofwiki-6745 | Rule of Exportation/Formulation 2/Reverse Implication | :$\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r}$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r} }}
{{Assumption|1|p \implies \paren {q \implies r} }}
{{SequentIntro|2|1|\paren {p \land q} \implies r|1|Rule of Exportation: Formulation 1: Reverse Implication}}
{{Implication|3||\paren {p \implies \paren {... | :$\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r}$ | {{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r} }}
{{Assumption|1|p \implies \paren {q \implies r} }}
{{SequentIntro|2|1|\paren {p \land q} \implies r|1|[[Rule of Exportation/Formulation 1/Reverse Implication|Rule of Exportation: Formulation 1: Reverse Im... | Rule of Exportation/Formulation 2/Reverse Implication/Proof 2 | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Reverse_Implication | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Reverse_Implication/Proof_2 | [
"Rule of Exportation"
] | [] | [
"Rule of Exportation/Formulation 1/Reverse Implication"
] |
proofwiki-6746 | Rule of Exportation/Formulation 2/Reverse Implication | :$\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r}$ | 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>$\begin{array}{|ccccc|c|ccccc|} \hline
(p & \implies & (q & \implies & r)) & \implies & ((p & \land & q) & \implies & r) \\
\hline
\F & \T &... | :$\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land 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... | Rule of Exportation/Formulation 2/Reverse Implication/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Reverse_Implication | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_2/Reverse_Implication/Proof_by_Truth_Table | [
"Rule of Exportation"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-6747 | Rule of Exportation/Formulation 1/Forward Implication | :$\paren {p \land q} \implies r \vdash p \implies \paren {q \implies r}$ | {{BeginTableau|\paren {p \land q} \implies r \vdash p \implies \paren {q \implies r} }}
{{Premise|1|\paren {p \land q} \implies r}}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{Conjunction|4|2, 3|p \land q|2|3}}
{{ModusPonens|5|1, 2, 3|r|1|4}}
{{Implication|6|1, 2|q \implies r|3|5}}
{{Implication|7|1|p \implies \paren {q \i... | :$\paren {p \land q} \implies r \vdash p \implies \paren {q \implies r}$ | {{BeginTableau|\paren {p \land q} \implies r \vdash p \implies \paren {q \implies r} }}
{{Premise|1|\paren {p \land q} \implies r}}
{{Assumption|2|p}}
{{Assumption|3|q}}
{{Conjunction|4|2, 3|p \land q|2|3}}
{{ModusPonens|5|1, 2, 3|r|1|4}}
{{Implication|6|1, 2|q \implies r|3|5}}
{{Implication|7|1|p \implies \paren {q \i... | Rule of Exportation/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_1/Forward_Implication/Proof | [
"Rule of Exportation"
] | [] | [] |
proofwiki-6748 | Rule of Exportation/Formulation 1/Reverse Implication | :$p \implies \paren {q \implies r} \vdash \paren {p \land q} \implies r$ | {{BeginTableau|p \implies \paren {q \implies r} \vdash \paren {p \land q} \implies r}}
{{Premise|1|p \implies \paren {q \implies r} }}
{{Assumption|2|p \land q}}
{{Simplification|3|2|p|2|1}}
{{Simplification|4|2|q|2|2}}
{{ModusPonens|5|1, 2|q \implies r|1|3}}
{{ModusPonens|6|1, 2|r|5|4}}
{{Implication|7|1|\paren {p \la... | :$p \implies \paren {q \implies r} \vdash \paren {p \land q} \implies r$ | {{BeginTableau|p \implies \paren {q \implies r} \vdash \paren {p \land q} \implies r}}
{{Premise|1|p \implies \paren {q \implies r} }}
{{Assumption|2|p \land q}}
{{Simplification|3|2|p|2|1}}
{{Simplification|4|2|q|2|2}}
{{ModusPonens|5|1, 2|q \implies r|1|3}}
{{ModusPonens|6|1, 2|r|5|4}}
{{Implication|7|1|\paren {p \la... | Rule of Exportation/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Rule_of_Exportation/Formulation_1/Reverse_Implication/Proof | [
"Rule of Exportation"
] | [] | [] |
proofwiki-6749 | Rule of Commutation/Conjunction/Formulation 1 | :$p \land q \dashv \vdash q \land p$ | {{BeginTableau|p \land q \vdash q \land p}}
{{Premise|1|p \land q}}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|q|1|2}}
{{Conjunction|4|1|q \land p|3|2}}
{{EndTableau}}
{{qed|lemma}}
{{BeginTableau|q \land p \vdash p \land q}}
{{Premise|1|q \land p}}
{{Simplification|2|1|q|1|1}}
{{Simplification|3|1|p|1|2}}
{{Con... | :$p \land q \dashv \vdash q \land p$ | {{BeginTableau|p \land q \vdash q \land p}}
{{Premise|1|p \land q}}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|q|1|2}}
{{Conjunction|4|1|q \land p|3|2}}
{{EndTableau}}
{{qed|lemma}}
{{BeginTableau|q \land p \vdash p \land q}}
{{Premise|1|q \land p}}
{{Simplification|2|1|q|1|1}}
{{Simplification|3|1|p|1|2}}
{{C... | Rule of Commutation/Conjunction/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Commutation/Conjunction/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Commutation/Conjunction/Formulation_1/Proof_1 | [
"Rule of Commutation"
] | [] | [] |
proofwiki-6750 | Rule of Commutation/Conjunction/Formulation 1 | :$p \land q \dashv \vdash q \land p$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccc||ccc|} \hline
p & \land & q & q & \land & p \\
\hline
\F & \F & \F & \F & \F & \F \\
\F & \F & \T & \T & \F & \F \\
\T & \F & \F & \F & \F & \T \\
\T ... | :$p \land q \dashv \vdash q \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 all [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|ccc||ccc|} \hline
p & \land & ... | Rule of Commutation/Conjunction/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Commutation/Conjunction/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Commutation/Conjunction/Formulation_1/Proof_by_Truth_Table | [
"Rule of Commutation"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6751 | Rule of Commutation/Disjunction/Formulation 1 | :$p \lor q \dashv \vdash q \lor p$ | {{BeginTableau|p \lor q \vdash q \lor p}}
{{Premise|1|p \lor q}}
{{Assumption|2|p}}
{{Addition|3|2|q \lor p|2|2}}
{{Assumption|4|p}}
{{Addition|5|4|q \lor p|4|1}}
{{ProofByCases|6|1|q \lor p|1|2|3|4|5}}
{{EndTableau}}
{{qed}}
{{BeginTableau|q \lor p \vdash p \lor q}}
{{Premise|1|q \lor p}}
{{Assumption|2|q}}
{{Addition... | :$p \lor q \dashv \vdash q \lor p$ | {{BeginTableau|p \lor q \vdash q \lor p}}
{{Premise|1|p \lor q}}
{{Assumption|2|p}}
{{Addition|3|2|q \lor p|2|2}}
{{Assumption|4|p}}
{{Addition|5|4|q \lor p|4|1}}
{{ProofByCases|6|1|q \lor p|1|2|3|4|5}}
{{EndTableau}}
{{qed}}
{{BeginTableau|q \lor p \vdash p \lor q}}
{{Premise|1|q \lor p}}
{{Assumption|2|q}}
{{Additi... | Rule of Commutation/Disjunction/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Commutation/Disjunction/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Commutation/Disjunction/Formulation_1/Proof_1 | [
"Rule of Commutation"
] | [] | [] |
proofwiki-6752 | Rule of Commutation/Disjunction/Formulation 1 | :$p \lor q \dashv \vdash q \lor p$ | We apply the Method of Truth Tables.
As can be seen by inspection, in both cases, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin{array}{|ccc||ccc|} \hline
p & \lor & q & q & \lor & p \\
\hline
\F & \F & \F & \F & \F & \F \\
\F & \T & \T & \T & \T & \F \\
\T & \T & \F... | :$p \lor q \dashv \vdash q \lor p$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, in both 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>$\begin{array}{|ccc||c... | Rule of Commutation/Disjunction/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Commutation/Disjunction/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Commutation/Disjunction/Formulation_1/Proof_by_Truth_Table | [
"Rule of Commutation"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6753 | Rule of Commutation/Conjunction/Formulation 2 | :$\vdash \paren {p \land q} \iff \paren {q \land p}$ | {{BeginTableau|\vdash \paren {p \land q} \iff \paren {q \land p} }}
{{Assumption |1|p \land q}}
{{Commutation|2|1|q \land p|1|Conjunction}}
{{Implication|3||\paren {p \land q} \implies \paren {q \land p}|1|2}}
{{Assumption |4|q \land p}}
{{Commutation|5|4|p \land q|4|Conjunction}}
{{Implication|6||\paren {q \land p} \i... | :$\vdash \paren {p \land q} \iff \paren {q \land p}$ | {{BeginTableau|\vdash \paren {p \land q} \iff \paren {q \land p} }}
{{Assumption |1|p \land q}}
{{Commutation|2|1|q \land p|1|Conjunction}}
{{Implication|3||\paren {p \land q} \implies \paren {q \land p}|1|2}}
{{Assumption |4|q \land p}}
{{Commutation|5|4|p \land q|4|Conjunction}}
{{Implication|6||\paren {q \land p} \i... | Rule of Commutation/Conjunction/Formulation 2/Proof 2 | https://proofwiki.org/wiki/Rule_of_Commutation/Conjunction/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Commutation/Conjunction/Formulation_2/Proof_2 | [
"Rule of Commutation"
] | [] | [] |
proofwiki-6754 | Rule of Commutation/Disjunction/Formulation 2 | :$\vdash \paren {p \lor q} \iff \paren {q \lor p}$ | {{BeginTableau|\vdash \paren {p \lor q} \iff \paren {q \lor p} }}
{{Assumption|1|p \lor q}}
{{Commutation|2|1|q \lor p|1|Disjunction}}
{{Implication|3||\paren {p \lor q} \implies \paren {q \lor p}|1|2}}
{{Assumption|4|q \lor p}}
{{Commutation|5|4|p \lor q|4|Disjunction}}
{{Implication|6||\paren {q \lor p} \implies \par... | :$\vdash \paren {p \lor q} \iff \paren {q \lor p}$ | {{BeginTableau|\vdash \paren {p \lor q} \iff \paren {q \lor p} }}
{{Assumption|1|p \lor q}}
{{Commutation|2|1|q \lor p|1|Disjunction}}
{{Implication|3||\paren {p \lor q} \implies \paren {q \lor p}|1|2}}
{{Assumption|4|q \lor p}}
{{Commutation|5|4|p \lor q|4|Disjunction}}
{{Implication|6||\paren {q \lor p} \implies \par... | Rule of Commutation/Disjunction/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Rule_of_Commutation/Disjunction/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Commutation/Disjunction/Formulation_2/Proof_1 | [
"Rule of Commutation"
] | [] | [] |
proofwiki-6755 | Rule of Commutation/Disjunction/Formulation 2 | :$\vdash \paren {p \lor q} \iff \paren {q \lor p}$ | 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.
:<nowiki>$\begin {array} {|ccc|c|ccc|} \hline
(p & \lor & q) & \iff & (q & \lor & p) \\
\hline
\F & \F & \F & \T & \F & \F & \F \\
\F & \T & \T & \T & \T & \T & \F \\
\T &... | :$\vdash \paren {p \lor q} \iff \paren {q \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 connective]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin {array} {|ccc|c|ccc|} \hline
(... | Rule of Commutation/Disjunction/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Commutation/Disjunction/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Commutation/Disjunction/Formulation_2/Proof_by_Truth_Table | [
"Rule of Commutation"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6756 | Combination Theorem for Complex Derivatives/Product Rule | :$\map {\paren {f g}'} z = \map {f'} z \map g z + \map f z \map {g'} z$ | Define $k: D \to \C$ as the pointwise product of $f$ and $g$, so $k = fg$.
Let $z_0 \in D$ be a point in $D$.
{{begin-eqn}}
{{eqn | l = \map {k'} {z_0}
| r = \lim_{h \mathop \to 0} \frac {\map k {z_0 + h} - \map k {z_0} } h
| c =
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\map f {z_0 + h} \, \map g {z_0 +... | :$\map {\paren {f g}'} z = \map {f'} z \map g z + \map f z \map {g'} z$ | Define $k: D \to \C$ as the [[Definition:Pointwise Multiplication of Complex-Valued Functions|pointwise product]] of $f$ and $g$, so $k = fg$.
Let $z_0 \in D$ be a point in $D$.
{{begin-eqn}}
{{eqn | l = \map {k'} {z_0}
| r = \lim_{h \mathop \to 0} \frac {\map k {z_0 + h} - \map k {z_0} } h
| c =
}}
{{eqn... | Combination Theorem for Complex Derivatives/Product Rule/Proof 1 | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Product_Rule | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Product_Rule/Proof_1 | [
"Combination Theorem for Complex Derivatives"
] | [] | [
"Definition:Pointwise Multiplication of Complex-Valued Functions"
] |
proofwiki-6757 | Combination Theorem for Complex Derivatives/Product Rule | :$\map {\paren {f g}'} z = \map {f'} z \map g z + \map f z \map {g'} z$ | Denote the open ball of $0$ by radius $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D$.
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
:$\map f {z + h} = \map f z + \map h {\map {f'} z + \map {\epsilon_f}... | :$\map {\paren {f g}'} z = \map {f'} z \map g z + \map f z \map {g'} z$ | Denote the [[Definition:Open Ball|open ball]] of $0$ by [[Definition:Radius of Open Ball|radius]] $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D$.
By the [[Epsilon-Function Complex Differentiability Condition]], it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
:$... | Combination Theorem for Complex Derivatives/Product Rule/Proof 2 | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Product_Rule | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Product_Rule/Proof_2 | [
"Combination Theorem for Complex Derivatives"
] | [] | [
"Definition:Open Ball",
"Definition:Open Ball/Radius",
"Epsilon-Function Differentiability Condition/Complex Case",
"Definition:Complex Function",
"Definition:Convergent Mapping/Complex Function",
"Combination Theorem for Limits of Functions/Complex/Product Rule",
"Combination Theorem for Limits of Func... |
proofwiki-6758 | Dedekind Completion is Unique up to Isomorphism | Let $S$ be an ordered set.
Let $\struct {X, f}$ and $\struct {Y, g}$ be Dedekind completions of $S$.
Then there exists a unique order isomorphism $\phi: X \to Y$ such that $\phi \circ f = g$. | By assumption, $\struct {X, f}$ is a Dedekind completion of $S$.
Also, $g: S \to Y$ is an order embedding and $Y$ is Dedekind complete.
Hence by definition of Dedekind completion, there exists a unique $\phi: X \to Y$ such that:
:$\phi \circ f = g$
It only remains to show that $\psi$ is an order isomorphism.
By reversi... | Let $S$ be an [[Definition:Ordered Set|ordered set]].
Let $\struct {X, f}$ and $\struct {Y, g}$ be [[Definition:Dedekind Completion|Dedekind completions]] of $S$.
Then there exists a [[Definition:Unique|unique]] [[Definition:Order Isomorphism|order isomorphism]] $\phi: X \to Y$ such that $\phi \circ f = g$. | By assumption, $\struct {X, f}$ is a [[Definition:Dedekind Completion|Dedekind completion]] of $S$.
Also, $g: S \to Y$ is an [[Definition:Order Embedding|order embedding]] and $Y$ is [[Definition:Dedekind Complete|Dedekind complete]].
Hence by definition of [[Definition:Dedekind Completion|Dedekind completion]], the... | Dedekind Completion is Unique up to Isomorphism | https://proofwiki.org/wiki/Dedekind_Completion_is_Unique_up_to_Isomorphism | https://proofwiki.org/wiki/Dedekind_Completion_is_Unique_up_to_Isomorphism | [
"Order Theory"
] | [
"Definition:Ordered Set",
"Definition:Dedekind Completion",
"Definition:Unique",
"Definition:Order Isomorphism"
] | [
"Definition:Dedekind Completion",
"Definition:Order Embedding",
"Definition:Dedekind Completeness Property",
"Definition:Dedekind Completion",
"Definition:Unique",
"Definition:Order Isomorphism",
"Definition:Unique",
"Definition:Dedekind Completion",
"Definition:Identity Mapping",
"Definition:Orde... |
proofwiki-6759 | Combination Theorem for Complex Derivatives/Combined Sum Rule | Let $\map {\dfrac \d {\d z} } {c f + w g}$ denote the derivative of $c f + w g$.
Then:
:$\map {\map {\dfrac \d {\d z} } {c f + w g} } z = c \dfrac \d {\d z} \map f z + w \dfrac \d {\d z} \map g z$ | {{begin-eqn}}
{{eqn | l = c \dfrac \d {\d z} \map f z + w \dfrac \d {\d z} \map g z
| r = \map {\map {\dfrac \d {\d z} } {c f} } z + \map {\map {\dfrac \d {\d z} } {w g} } z
| c = Combination Theorem for Complex-Differentiable Functions: Multiple Rule
}}
{{eqn | r = \map {\map {\dfrac \d {\d z} } {c f + w g... | Let $\map {\dfrac \d {\d z} } {c f + w g}$ denote the [[Definition:Derivative of Complex Function|derivative]] of $c f + w g$.
Then:
:$\map {\map {\dfrac \d {\d z} } {c f + w g} } z = c \dfrac \d {\d z} \map f z + w \dfrac \d {\d z} \map g z$ | {{begin-eqn}}
{{eqn | l = c \dfrac \d {\d z} \map f z + w \dfrac \d {\d z} \map g z
| r = \map {\map {\dfrac \d {\d z} } {c f} } z + \map {\map {\dfrac \d {\d z} } {w g} } z
| c = [[Combination Theorem for Complex-Differentiable Functions/Multiple Rule|Combination Theorem for Complex-Differentiable Function... | Combination Theorem for Complex Derivatives/Combined Sum Rule | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Combined_Sum_Rule | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Combined_Sum_Rule | [
"Combination Theorem for Complex Derivatives"
] | [
"Definition:Derivative/Complex Function"
] | [
"Combination Theorem for Complex Derivatives/Multiple Rule",
"Combination Theorem for Complex Derivatives/Sum Rule"
] |
proofwiki-6760 | Existence of Dedekind Completion | Let $\struct {S, \preceq}$ be an ordered set.
Then there exists a Dedekind completion of $S$.
That is, there exists a Dedekind complete ordered set $\tilde S$ and an order embedding $\phi: S \to \tilde S$ such that:
:For all Dedekind complete ordered sets $X$, and for all order embeddings $f: S \to X$, there exists an ... | For all subsets $I \subseteq S$, define:
:$\map {\operatorname U} I = \left\{{x \in S: x}\right.$ is an upper bound for $\left.{I}\right\}$
:$\map {\operatorname L} I = \left\{{x \in S: x}\right.$ is a lower bound for $\left.{I}\right\}$
Note that, for all $I, J \subseteq S$:
:$I \subseteq \map {\operatorname {LU} } I$... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Then there exists a [[Definition:Dedekind Completion|Dedekind completion]] of $S$.
That is, there exists a [[Definition:Dedekind Complete|Dedekind complete]] [[Definition:Ordered Set|ordered set]] $\tilde S$ and an [[Definition:Order Embedding... | For all [[Definition:Subset|subsets]] $I \subseteq S$, define:
:$\map {\operatorname U} I = \left\{{x \in S: x}\right.$ is an [[Definition:Upper Bound of Set|upper bound]] for $\left.{I}\right\}$
:$\map {\operatorname L} I = \left\{{x \in S: x}\right.$ is a [[Definition:Lower Bound of Set|lower bound]] for $\left.{I}... | Existence of Dedekind Completion | https://proofwiki.org/wiki/Existence_of_Dedekind_Completion | https://proofwiki.org/wiki/Existence_of_Dedekind_Completion | [
"Order Theory"
] | [
"Definition:Ordered Set",
"Definition:Dedekind Completion",
"Definition:Dedekind Completeness Property",
"Definition:Ordered Set",
"Definition:Order Embedding",
"Definition:Dedekind Completeness Property",
"Definition:Ordered Set",
"Definition:Order Embedding",
"Definition:Order Embedding"
] | [
"Definition:Subset",
"Definition:Upper Bound of Set",
"Definition:Lower Bound of Set",
"Definition:Non-Empty Set",
"Definition:Bounded Above Set",
"Subset Relation is Ordering",
"Definition:Non-Empty Set",
"Definition:Bounded Above Set",
"Union is Smallest Superset/General Result",
"Definition:Non... |
proofwiki-6761 | Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering | Let $S$ be a set.
Let $\RR$ be a transitive, antisymmetric relation on $S$.
Let $\RR^\ne$ denote the reflexive reduction of $\RR$.
Then $\RR^\ne$ is a strict ordering. | To show that $\RR^\ne$ is a strict ordering, it is sufficient to show that $\RR^\ne$ is antireflexive and transitive. | Let $S$ be a [[Definition:set|set]].
Let $\RR$ be a [[Definition:Transitive Relation|transitive]], [[Definition:Antisymmetric Relation|antisymmetric]] [[Definition:Endorelation|relation]] on $S$.
Let $\RR^\ne$ denote the [[Definition:Reflexive Reduction|reflexive reduction]] of $\RR$.
Then $\RR^\ne$ is a [[Definiti... | To show that $\RR^\ne$ is a [[Definition:Strict Ordering|strict ordering]], it is sufficient to show that $\RR^\ne$ is [[Definition:Antireflexive Relation|antireflexive]] and [[Definition:Transitive Relation|transitive]]. | Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering | https://proofwiki.org/wiki/Reflexive_Reduction_of_Transitive_Antisymmetric_Relation_is_Strict_Ordering | https://proofwiki.org/wiki/Reflexive_Reduction_of_Transitive_Antisymmetric_Relation_is_Strict_Ordering | [
"Reflexive Reductions",
"Strict Orderings"
] | [
"Definition:set",
"Definition:Transitive Relation",
"Definition:Antisymmetric Relation",
"Definition:Endorelation",
"Definition:Reflexive Reduction",
"Definition:Strict Ordering"
] | [
"Definition:Strict Ordering",
"Definition:Antireflexive Relation",
"Definition:Transitive Relation",
"Definition:Transitive Relation"
] |
proofwiki-6762 | Reflexive Reduction of Antisymmetric Relation is Asymmetric | Let $S$ be a set.
Let $\RR$ be an antisymmetric relation on $S$.
Let $\RR^\ne$ be the reflexive reduction of $\RR$.
Then $\RR^\ne$ is asymmetric. | {{AimForCont}} $\RR^\ne$ is not asymmetric.
That is:
:$\exists a, b \in S: a \mathrel {\RR^\ne} b$ and $b \mathrel {\RR^\ne} a$
Then by the definition of reflexive reduction:
:$a \mathrel \RR b$, $b \mathrel \RR a$
and $a \ne b$.
But this contradicts the antisymmetry of $\RR$.
Thus, by definition, $\RR^\ne$ is an asymm... | Let $S$ be a [[Definition:Set|set]].
Let $\RR$ be an [[Definition:Antisymmetric Relation|antisymmetric relation]] on $S$.
Let $\RR^\ne$ be the [[Definition:Reflexive Reduction|reflexive reduction]] of $\RR$.
Then $\RR^\ne$ is [[Definition:Asymmetric Relation|asymmetric]]. | {{AimForCont}} $\RR^\ne$ is not [[Definition:Asymmetric Relation|asymmetric]].
That is:
:$\exists a, b \in S: a \mathrel {\RR^\ne} b$ and $b \mathrel {\RR^\ne} a$
Then by the definition of [[Definition:Reflexive Reduction|reflexive reduction]]:
:$a \mathrel \RR b$, $b \mathrel \RR a$
and $a \ne b$.
But this [[Proof ... | Reflexive Reduction of Antisymmetric Relation is Asymmetric | https://proofwiki.org/wiki/Reflexive_Reduction_of_Antisymmetric_Relation_is_Asymmetric | https://proofwiki.org/wiki/Reflexive_Reduction_of_Antisymmetric_Relation_is_Asymmetric | [
"Reflexive Reductions"
] | [
"Definition:Set",
"Definition:Antisymmetric Relation",
"Definition:Reflexive Reduction",
"Definition:Asymmetric Relation"
] | [
"Definition:Asymmetric Relation",
"Definition:Reflexive Reduction",
"Proof by Contradiction",
"Definition:Antisymmetric Relation",
"Definition:Asymmetric Relation"
] |
proofwiki-6763 | Reductio ad Absurdum/Variant 1 | :$\neg p \implies \bot \vdash p$ | {{BeginTableau|\neg p \implies \bot \vdash p}}
{{Premise|1|\neg p \implies \bot}}
{{Assumption|2|\neg p}}
{{ModusPonens|3|1, 2|\bot|1|2}}
{{Contradiction|4|1|\neg \neg p|2|3}}
{{DoubleNegElimination|5|1|p|4}}
{{EndTableau|qed}} | :$\neg p \implies \bot \vdash p$ | {{BeginTableau|\neg p \implies \bot \vdash p}}
{{Premise|1|\neg p \implies \bot}}
{{Assumption|2|\neg p}}
{{ModusPonens|3|1, 2|\bot|1|2}}
{{Contradiction|4|1|\neg \neg p|2|3}}
{{DoubleNegElimination|5|1|p|4}}
{{EndTableau|qed}} | Reductio ad Absurdum/Variant 1/Proof 1 | https://proofwiki.org/wiki/Reductio_ad_Absurdum/Variant_1 | https://proofwiki.org/wiki/Reductio_ad_Absurdum/Variant_1/Proof_1 | [
"Reductio ad Absurdum"
] | [] | [] |
proofwiki-6764 | Reductio ad Absurdum/Variant 1 | :$\neg p \implies \bot \vdash p$ | As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|cccc||c|} \hline
\neg & p & \implies & \bot & p \\
\hline
\T & \F & \F & \F & \F \\
\F & \T & \T & \F & \T \\
\hline
\end{array}$
{{qed}} | :$\neg p \implies \bot \vdash p$ | 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]].
$\begin{array}{|cccc||c|} \hline
\neg & p & \implies & \bot & p \\
\hline
\T & \F & \F &... | Reductio ad Absurdum/Variant 1/Proof by Truth Table | https://proofwiki.org/wiki/Reductio_ad_Absurdum/Variant_1 | https://proofwiki.org/wiki/Reductio_ad_Absurdum/Variant_1/Proof_by_Truth_Table | [
"Reductio ad Absurdum"
] | [] | [
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6765 | Reductio ad Absurdum/Variant 2 | :$\neg p \implies \paren {q \land \neg q} \vdash p$ | {{BeginTableau|\neg p \implies \left({q \land \neg q}\right) \vdash p}}
{{Premise|1|\neg p \implies \left({q \land \neg q}\right)}}
{{Assumption|2|\neg p}}
{{ModusPonens|3|1, 2|q \land \neg q|1|2}}
{{Simplification|4|1, 2|q|3|1}}
{{Simplification|5|1, 2|\neg q|3|2}}
{{NonContradiction|6|1, 2|4|5}}
{{Contradiction|7|1|\... | :$\neg p \implies \paren {q \land \neg q} \vdash p$ | {{BeginTableau|\neg p \implies \left({q \land \neg q}\right) \vdash p}}
{{Premise|1|\neg p \implies \left({q \land \neg q}\right)}}
{{Assumption|2|\neg p}}
{{ModusPonens|3|1, 2|q \land \neg q|1|2}}
{{Simplification|4|1, 2|q|3|1}}
{{Simplification|5|1, 2|\neg q|3|2}}
{{NonContradiction|6|1, 2|4|5}}
{{Contradiction|7|1|\... | Reductio ad Absurdum/Variant 2/Proof 1 | https://proofwiki.org/wiki/Reductio_ad_Absurdum/Variant_2 | https://proofwiki.org/wiki/Reductio_ad_Absurdum/Variant_2/Proof_1 | [
"Reductio ad Absurdum"
] | [] | [] |
proofwiki-6766 | Reductio ad Absurdum/Variant 2 | :$\neg p \implies \paren {q \land \neg q} \vdash p$ | As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccccccc||c|} \hline
\neg & p & \implies & (q & \land & \neg & q) & p \\
\hline
\T & \F & \F & \F & \F & \T & \F & \F \\
\T & \F & \F & \T & \F & \F & \T & \F \\
\F & \T & \T & \F & \F & \T & ... | :$\neg p \implies \paren {q \land \neg q} \vdash p$ | 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]].
$\begin{array}{|ccccccc||c|} \hline
\neg & p & \implies & (q & \land & \neg & q) & p \\
... | Reductio ad Absurdum/Variant 2/Proof by Truth Table | https://proofwiki.org/wiki/Reductio_ad_Absurdum/Variant_2 | https://proofwiki.org/wiki/Reductio_ad_Absurdum/Variant_2/Proof_by_Truth_Table | [
"Reductio ad Absurdum"
] | [] | [
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6767 | Proof by Contradiction/Variant 1 | :$\paren {p \vdash \paren {q \land \neg q} } \vdash \neg p$ | {{BeginTableau|\paren {p \vdash \paren {q \land \neg q} } \vdash \neg p}}
{{Premise|1|p \vdash \paren {q \land \neg q} }}
{{Assumption|2|p}}
{{SequentIntro|3|1, 2|q \land \neg q|1, 2|{{hypothesis}} }}
{{Simplification|4|1, 2|q|3|1}}
{{Simplification|5|1, 2|\neg q|3|2}}
{{NonContradiction|6|1, 2|4|5}}
{{Contradiction|7|... | :$\paren {p \vdash \paren {q \land \neg q} } \vdash \neg p$ | {{BeginTableau|\paren {p \vdash \paren {q \land \neg q} } \vdash \neg p}}
{{Premise|1|p \vdash \paren {q \land \neg q} }}
{{Assumption|2|p}}
{{SequentIntro|3|1, 2|q \land \neg q|1, 2|{{hypothesis}} }}
{{Simplification|4|1, 2|q|3|1}}
{{Simplification|5|1, 2|\neg q|3|2}}
{{NonContradiction|6|1, 2|4|5}}
{{Contradiction|7|... | Proof by Contradiction/Variant 1 | https://proofwiki.org/wiki/Proof_by_Contradiction/Variant_1 | https://proofwiki.org/wiki/Proof_by_Contradiction/Variant_1 | [
"Proof by Contradiction"
] | [] | [] |
proofwiki-6768 | Equivalence of Definitions of Topological Group | Let $\struct {G, \odot}$ be a group.
On its underlying set $G$, let $\struct {G, \tau}$ be a topological space.
{{TFAE|def = Topological Group}} | === Definition 1 implies Definition 2 ===
Let $\struct {G, \odot, \tau}$ be a topological group by Definition 1.
Let $\phi: \struct {G, \tau} \to \struct {G, \tau}$ be the mapping defined as:
:$\forall x \in G: \map \phi x = x^{-1}$
By definition:
:$\odot: \struct {G, \tau} \times \struct {G, \tau} \to \struct {G, \tau... | Let $\struct {G, \odot}$ be a [[Definition:Group|group]].
On its [[Definition:Underlying Set of Topological Space|underlying set]] $G$, let $\struct {G, \tau}$ be a [[Definition:Topological Space|topological space]].
{{TFAE|def = Topological Group}} | === Definition 1 implies Definition 2 ===
Let $\struct {G, \odot, \tau}$ be a [[Definition:Topological Group/Definition 1|topological group by Definition 1]].
Let $\phi: \struct {G, \tau} \to \struct {G, \tau}$ be the [[Definition:Mapping|mapping]] defined as:
:$\forall x \in G: \map \phi x = x^{-1}$
By definition:... | Equivalence of Definitions of Topological Group | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Topological_Group | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Topological_Group | [
"Topological Groups"
] | [
"Definition:Group",
"Definition:Underlying Set/Topological Space",
"Definition:Topological Space"
] | [
"Definition:Topological Group/Definition 1",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)",
"Continuous Mapping to Product Space",
"Definition:Continuous Mapping (Topology)",
"Definition:Mapping"... |
proofwiki-6769 | Conditional is Left Distributive over Conjunction/Formulation 1/Forward Implication | :$p \implies \paren {q \land r} \vdash \paren {p \implies q} \land \paren {p \implies r}$ | {{BeginTableau|p \implies \paren {q \land r} \vdash \paren {p \implies q} \land \paren {p \implies r} }}
{{Premise|1|p \implies \paren {q \land r} }}
{{Assumption|2|p}}
{{ModusPonens|3|1, 2|q \land r|1|2}}
{{Simplification|4|1, 2|q|3|1}}
{{Simplification|5|1, 2|r|3|2}}
{{Implication|6|1|p \implies q|2|4}}
{{Implication... | :$p \implies \paren {q \land r} \vdash \paren {p \implies q} \land \paren {p \implies r}$ | {{BeginTableau|p \implies \paren {q \land r} \vdash \paren {p \implies q} \land \paren {p \implies r} }}
{{Premise|1|p \implies \paren {q \land r} }}
{{Assumption|2|p}}
{{ModusPonens|3|1, 2|q \land r|1|2}}
{{Simplification|4|1, 2|q|3|1}}
{{Simplification|5|1, 2|r|3|2}}
{{Implication|6|1|p \implies q|2|4}}
{{Implication... | Conditional is Left Distributive over Conjunction/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_1/Forward_Implication/Proof | [
"Conditional is Left Distributive over Conjunction"
] | [] | [] |
proofwiki-6770 | Conditional is Left Distributive over Conjunction/Formulation 1/Reverse Implication | :$\paren {p \implies q} \land \paren {p \implies r} \vdash p \implies \paren {q \land r}$ | {{BeginTableau|\paren {p \implies q} \land \paren {p \implies r} \vdash p \implies \paren {q \land r} }}
{{Premise|1|\paren {p \implies q} \land \paren {p \implies r} }}
{{SequentIntro|2|1|\paren {p \land p} \implies \paren {q \land r}|1|Praeclarum Theorema}}
{{Assumption|3|p}}
{{Idempotence|4|3|p \land p|3|Conjunction... | :$\paren {p \implies q} \land \paren {p \implies r} \vdash p \implies \paren {q \land r}$ | {{BeginTableau|\paren {p \implies q} \land \paren {p \implies r} \vdash p \implies \paren {q \land r} }}
{{Premise|1|\paren {p \implies q} \land \paren {p \implies r} }}
{{SequentIntro|2|1|\paren {p \land p} \implies \paren {q \land r}|1|[[Praeclarum Theorema]]}}
{{Assumption|3|p}}
{{Idempotence|4|3|p \land p|3|Conjunc... | Conditional is Left Distributive over Conjunction/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_1/Reverse_Implication/Proof | [
"Conditional is Left Distributive over Conjunction"
] | [] | [
"Praeclarum Theorema"
] |
proofwiki-6771 | Conditional is Left Distributive over Conjunction/Formulation 1 | :$p \implies \paren {q \land r} \dashv \vdash \paren {p \implies q} \land \paren {p \implies r}$ | {{BeginTableau|p \implies \paren {q \land r} \vdash \paren {p \implies q} \land \paren {p \implies r} }}
{{Premise|1|p \implies \paren {q \land r} }}
{{Assumption|2|p}}
{{ModusPonens|3|1, 2|q \land r|1|2}}
{{Simplification|4|1, 2|q|3|1}}
{{Simplification|5|1, 2|r|3|2}}
{{Implication|6|1|p \implies q|2|4}}
{{Implication... | :$p \implies \paren {q \land r} \dashv \vdash \paren {p \implies q} \land \paren {p \implies r}$ | {{BeginTableau|p \implies \paren {q \land r} \vdash \paren {p \implies q} \land \paren {p \implies r} }}
{{Premise|1|p \implies \paren {q \land r} }}
{{Assumption|2|p}}
{{ModusPonens|3|1, 2|q \land r|1|2}}
{{Simplification|4|1, 2|q|3|1}}
{{Simplification|5|1, 2|r|3|2}}
{{Implication|6|1|p \implies q|2|4}}
{{Implication... | Conditional is Left Distributive over Conjunction/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_1 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_1/Forward_Implication/Proof | [
"Conditional is Left Distributive over Conjunction"
] | [] | [] |
proofwiki-6772 | Conditional is Left Distributive over Conjunction/Formulation 1 | :$p \implies \paren {q \land r} \dashv \vdash \paren {p \implies q} \land \paren {p \implies r}$ | === Proof of Forward Implication ===
{{:Conditional is Left Distributive over Conjunction/Formulation 1/Forward Implication/Proof}}
=== Proof of Reverse Implication ===
{{:Conditional is Left Distributive over Conjunction/Formulation 1/Reverse Implication/Proof}} | :$p \implies \paren {q \land r} \dashv \vdash \paren {p \implies q} \land \paren {p \implies r}$ | === [[Conditional is Left Distributive over Conjunction/Formulation 1/Forward Implication/Proof|Proof of Forward Implication]] ===
{{:Conditional is Left Distributive over Conjunction/Formulation 1/Forward Implication/Proof}}
=== [[Conditional is Left Distributive over Conjunction/Formulation 1/Reverse Implication/Pro... | Conditional is Left Distributive over Conjunction/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_1 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_1/Proof_1 | [
"Conditional is Left Distributive over Conjunction"
] | [] | [
"Conditional is Left Distributive over Conjunction/Formulation 1/Forward Implication/Proof",
"Conditional is Left Distributive over Conjunction/Formulation 1/Reverse Implication/Proof"
] |
proofwiki-6773 | Conditional is Left Distributive over Conjunction/Formulation 1 | :$p \implies \paren {q \land r} \dashv \vdash \paren {p \implies q} \land \paren {p \implies r}$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccccc||ccccccc|} \hline
p & \implies & (q & \land & r) & (p & \implies & q) & \land & (p & \implies & r) \\
\hline
\F & \T & \F & \F & \F & \F & \T & \F &... | :$p \implies \paren {q \land r} \dashv \vdash \paren {p \implies q} \land \paren {p \implies r}$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|ccccc||ccccccc|} \hline
p & \i... | Conditional is Left Distributive over Conjunction/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_1 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_1/Proof_by_Truth_Table | [
"Conditional is Left Distributive over Conjunction"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6774 | Conditional is Left Distributive over Conjunction/Formulation 1 | :$p \implies \paren {q \land r} \dashv \vdash \paren {p \implies q} \land \paren {p \implies r}$ | {{BeginTableau|\paren {p \implies q} \land \paren {p \implies r} \vdash p \implies \paren {q \land r} }}
{{Premise|1|\paren {p \implies q} \land \paren {p \implies r} }}
{{SequentIntro|2|1|\paren {p \land p} \implies \paren {q \land r}|1|Praeclarum Theorema}}
{{Assumption|3|p}}
{{Idempotence|4|3|p \land p|3|Conjunction... | :$p \implies \paren {q \land r} \dashv \vdash \paren {p \implies q} \land \paren {p \implies r}$ | {{BeginTableau|\paren {p \implies q} \land \paren {p \implies r} \vdash p \implies \paren {q \land r} }}
{{Premise|1|\paren {p \implies q} \land \paren {p \implies r} }}
{{SequentIntro|2|1|\paren {p \land p} \implies \paren {q \land r}|1|[[Praeclarum Theorema]]}}
{{Assumption|3|p}}
{{Idempotence|4|3|p \land p|3|Conjunc... | Conditional is Left Distributive over Conjunction/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_1 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_1/Reverse_Implication/Proof | [
"Conditional is Left Distributive over Conjunction"
] | [] | [
"Praeclarum Theorema"
] |
proofwiki-6775 | Conditional is Left Distributive over Conjunction/Formulation 2/Forward Implication | :$\vdash \paren {p \implies \paren {q \land r} } \implies \paren {\paren {p \implies q} \land \paren {p \implies r} }$ | Let us use the following abbreviations
{{begin-eqn}}
{{eqn | l = \phi
| o = \text {for}
| r = p \implies \paren {q \land r}
| c =
}}
{{eqn | l = \psi
| o = \text {for}
| r = \paren {p \implies q} \land \paren {p \implies r}
| c =
}}
{{end-eqn}}
{{BeginTableau|\vdash \phi \implies \... | :$\vdash \paren {p \implies \paren {q \land r} } \implies \paren {\paren {p \implies q} \land \paren {p \implies r} }$ | Let us use the following abbreviations
{{begin-eqn}}
{{eqn | l = \phi
| o = \text {for}
| r = p \implies \paren {q \land r}
| c =
}}
{{eqn | l = \psi
| o = \text {for}
| r = \paren {p \implies q} \land \paren {p \implies r}
| c =
}}
{{end-eqn}}
{{BeginTableau|\vdash \phi \implie... | Conditional is Left Distributive over Conjunction/Formulation 2/Forward Implication/Proof 2 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_2/Forward_Implication/Proof_2 | [
"Conditional is Left Distributive over Conjunction"
] | [] | [
"Conditional is Left Distributive over Conjunction/Formulation 1/Forward Implication"
] |
proofwiki-6776 | Conditional is Left Distributive over Conjunction/Formulation 2/Reverse Implication | :$\vdash \paren {\paren {p \implies q} \land \paren {p \implies r} } \implies \paren {p \implies \paren {q \land r} }$ | Let us use the following abbreviations
{{begin-eqn}}
{{eqn | l = \phi
| o = \text {for}
| r = \paren {p \implies q} \land \paren {p \implies r}
| c =
}}
{{eqn | l = \psi
| o = \text {for}
| r = p \implies \paren {q \land r}
| c =
}}
{{end-eqn}}
{{BeginTableau|\vdash \phi \implies \... | :$\vdash \paren {\paren {p \implies q} \land \paren {p \implies r} } \implies \paren {p \implies \paren {q \land r} }$ | Let us use the following abbreviations
{{begin-eqn}}
{{eqn | l = \phi
| o = \text {for}
| r = \paren {p \implies q} \land \paren {p \implies r}
| c =
}}
{{eqn | l = \psi
| o = \text {for}
| r = p \implies \paren {q \land r}
| c =
}}
{{end-eqn}}
{{BeginTableau|\vdash \phi \implie... | Conditional is Left Distributive over Conjunction/Formulation 2/Reverse Implication/Proof 2 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_2/Reverse_Implication | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_2/Reverse_Implication/Proof_2 | [
"Conditional is Left Distributive over Conjunction"
] | [] | [
"Conditional is Left Distributive over Conjunction/Formulation 1/Reverse Implication"
] |
proofwiki-6777 | Conditional is Left Distributive over Conjunction/Formulation 2 | :$\vdash \paren {p \implies \paren {q \land r} } \iff \paren {\paren {p \implies q} \land \paren {p \implies r} }$ | Let us use the following abbreviations
{{begin-eqn}}
{{eqn | l = \phi
| o = \text {for}
| r = p \implies \paren {q \land r}
| c =
}}
{{eqn | l = \psi
| o = \text {for}
| r = \paren {p \implies q} \land \paren {p \implies r}
| c =
}}
{{end-eqn}}
{{BeginTableau|\vdash \phi \implies \... | :$\vdash \paren {p \implies \paren {q \land r} } \iff \paren {\paren {p \implies q} \land \paren {p \implies r} }$ | Let us use the following abbreviations
{{begin-eqn}}
{{eqn | l = \phi
| o = \text {for}
| r = p \implies \paren {q \land r}
| c =
}}
{{eqn | l = \psi
| o = \text {for}
| r = \paren {p \implies q} \land \paren {p \implies r}
| c =
}}
{{end-eqn}}
{{BeginTableau|\vdash \phi \implie... | Conditional is Left Distributive over Conjunction/Formulation 2/Forward Implication/Proof 2 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_2 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_2/Forward_Implication/Proof_2 | [
"Conditional is Left Distributive over Conjunction"
] | [] | [
"Conditional is Left Distributive over Conjunction/Formulation 1/Forward Implication"
] |
proofwiki-6778 | Conditional is Left Distributive over Conjunction/Formulation 2 | :$\vdash \paren {p \implies \paren {q \land r} } \iff \paren {\paren {p \implies q} \land \paren {p \implies r} }$ | === Proof of Forward Implication ===
{{:Conditional is Left Distributive over Conjunction/Formulation 2/Forward Implication/Proof 1}}
=== Proof of Reverse Implication ===
{{:Conditional is Left Distributive over Conjunction/Formulation 2/Reverse Implication/Proof 1}}
{{BeginTableau|\vdash \paren {p \implies \paren {q \... | :$\vdash \paren {p \implies \paren {q \land r} } \iff \paren {\paren {p \implies q} \land \paren {p \implies r} }$ | === [[Conditional is Left Distributive over Conjunction/Formulation 2/Forward Implication/Proof 1|Proof of Forward Implication]] ===
{{:Conditional is Left Distributive over Conjunction/Formulation 2/Forward Implication/Proof 1}}
=== [[Conditional is Left Distributive over Conjunction/Formulation 2/Reverse Implication... | Conditional is Left Distributive over Conjunction/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_2 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_2/Proof_1 | [
"Conditional is Left Distributive over Conjunction"
] | [] | [
"Conditional is Left Distributive over Conjunction/Formulation 2/Forward Implication/Proof 1",
"Conditional is Left Distributive over Conjunction/Formulation 2/Reverse Implication/Proof 1",
"Conditional is Left Distributive over Conjunction/Formulation 2/Forward Implication",
"Conditional is Left Distributive... |
proofwiki-6779 | Conditional is Left Distributive over Conjunction/Formulation 2 | :$\vdash \paren {p \implies \paren {q \land r} } \iff \paren {\paren {p \implies q} \land \paren {p \implies r} }$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations.
:<nowiki>$\begin{array}{|ccccc|c|ccccccc|} \hline
(p & \implies & (q & \land & r)) & \iff & ((p & \implies & q) & \land & (p & \implies & r)) \\
\hline
\F & \T & \F & \F... | :$\vdash \paren {p \implies \paren {q \land r} } \iff \paren {\paren {p \implies q} \land \paren {p \implies r} }$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|... | Conditional is Left Distributive over Conjunction/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_2 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_2/Proof_by_Truth_Table | [
"Conditional is Left Distributive over Conjunction"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-6780 | Conditional is Left Distributive over Conjunction/Formulation 2 | :$\vdash \paren {p \implies \paren {q \land r} } \iff \paren {\paren {p \implies q} \land \paren {p \implies r} }$ | Let us use the following abbreviations
{{begin-eqn}}
{{eqn | l = \phi
| o = \text {for}
| r = \paren {p \implies q} \land \paren {p \implies r}
| c =
}}
{{eqn | l = \psi
| o = \text {for}
| r = p \implies \paren {q \land r}
| c =
}}
{{end-eqn}}
{{BeginTableau|\vdash \phi \implies \... | :$\vdash \paren {p \implies \paren {q \land r} } \iff \paren {\paren {p \implies q} \land \paren {p \implies r} }$ | Let us use the following abbreviations
{{begin-eqn}}
{{eqn | l = \phi
| o = \text {for}
| r = \paren {p \implies q} \land \paren {p \implies r}
| c =
}}
{{eqn | l = \psi
| o = \text {for}
| r = p \implies \paren {q \land r}
| c =
}}
{{end-eqn}}
{{BeginTableau|\vdash \phi \implie... | Conditional is Left Distributive over Conjunction/Formulation 2/Reverse Implication/Proof 2 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_2 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_2/Reverse_Implication/Proof_2 | [
"Conditional is Left Distributive over Conjunction"
] | [] | [
"Conditional is Left Distributive over Conjunction/Formulation 1/Reverse Implication"
] |
proofwiki-6781 | Criterion for Ring with Unity to be Topological Ring | Let $\struct {R, +, \circ}$ be a ring with unity.
Let $\tau$ be a topology over $R$.
Suppose that $+$ and $\circ$ are $\tau$-continuous mappings.
Then $\struct {R, +, \circ, \tau}$ is a topological ring. | As we presume $\circ$ to be continuous, we need only prove that $\struct {R, +, \tau}$ is a topological group.
As we presume $+$ to be continuous, we need only show that negation is continuous.
As $\struct {R, \circ}$ is a semigroup and $\circ$ is continuous:
:$\struct{R, \circ, \tau}$ is a topological semigroup.
From ... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]].
Let $\tau$ be a [[Definition:Topology|topology]] over $R$.
Suppose that $+$ and $\circ$ are $\tau$-[[Definition:Continuous Mapping|continuous mappings]].
Then $\struct {R, +, \circ, \tau}$ is a [[Definition:Topological Ring|topological... | As we presume $\circ$ to be [[Definition:Continuous Mapping|continuous]], we need only prove that $\struct {R, +, \tau}$ is a [[Definition:Topological Group|topological group]].
As we presume $+$ to be [[Definition:Continuous Mapping|continuous]], we need only show that negation is [[Definition:Continuous Mapping|cont... | Criterion for Ring with Unity to be Topological Ring | https://proofwiki.org/wiki/Criterion_for_Ring_with_Unity_to_be_Topological_Ring | https://proofwiki.org/wiki/Criterion_for_Ring_with_Unity_to_be_Topological_Ring | [] | [
"Definition:Ring with Unity",
"Definition:Topology",
"Definition:Continuous Mapping",
"Definition:Topological Ring"
] | [
"Definition:Continuous Mapping",
"Definition:Topological Group",
"Definition:Continuous Mapping",
"Definition:Continuous Mapping",
"Definition:Semigroup",
"Definition:Continuous Mapping",
"Definition:Topological Group",
"Identity Mapping is Homeomorphism",
"Definition:Identity Mapping",
"Definitio... |
proofwiki-6782 | Union of Relations is Relation | Let $S$ and $T$ be sets.
Let $\FF$ be a family of relations from $S$ to $T$.
Let $\ds \RR = \bigcup \FF$, the union of all the elements of $\FF$.
Then $\RR$ is a relation from $S$ to $T$.
{{expand|Binary case}} | By the definition of a relation from $S$ to $T$, each element of $\FF$ is a subset of $S \times T$.
By Union of Subsets is Subset: Set of Sets:
:$\RR \subseteq S \times T$
Therefore, by the definition of a relation from $S$ to $T$, $\RR$ is a relation from $S$ to $T$.
{{qed}}
Category:Relation Theory
703ie71rc52tci47k0... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\FF$ be a family of [[Definition:Relation|relations]] from $S$ to $T$.
Let $\ds \RR = \bigcup \FF$, the [[Definition:Union of Set of Sets|union]] of all the [[Definition:Element|elements]] of $\FF$.
Then $\RR$ is a [[Definition:Relation|relation]] from $S$ to $T$.
{... | By the definition of a [[Definition:Relation|relation]] from $S$ to $T$, each [[Definition:Element|element]] of $\FF$ is a [[Definition:Subset|subset]] of $S \times T$.
By [[Union of Subsets is Subset/Set of Sets|Union of Subsets is Subset: Set of Sets]]:
:$\RR \subseteq S \times T$
Therefore, by the definition of a... | Union of Relations is Relation | https://proofwiki.org/wiki/Union_of_Relations_is_Relation | https://proofwiki.org/wiki/Union_of_Relations_is_Relation | [
"Relation Theory"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Set Union/Set of Sets",
"Definition:Element",
"Definition:Relation"
] | [
"Definition:Relation",
"Definition:Element",
"Definition:Subset",
"Union of Subsets is Subset/Set of Sets",
"Definition:Relation",
"Definition:Relation",
"Category:Relation Theory"
] |
proofwiki-6783 | Combination Theorem for Complex Derivatives/Quotient Rule | For all $z \in D$ with $\map g z \ne 0$:
:$\map {\paren {\dfrac f g}'} z = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2}$ | Denote the open ball of $0$ with radius $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D \setminus \set {x \in D: \map g z = 0}$.
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r_0 \in \R_{>0}$ such that for all $h \in \map {B_{r_0} } 0 \setminus \set 0$:
:$\map f {z + h} = \m... | For all $z \in D$ with $\map g z \ne 0$:
:$\map {\paren {\dfrac f g}'} z = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2}$ | Denote the [[Definition:Open Ball|open ball]] of $0$ with [[Definition:Radius of Open Ball|radius]] $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D \setminus \set {x \in D: \map g z = 0}$.
By the [[Epsilon-Function Complex Differentiability Condition]], it follows that there exists $r_0 \in \R_{>0}$ such that for all... | Combination Theorem for Complex Derivatives/Quotient Rule | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Quotient_Rule | https://proofwiki.org/wiki/Combination_Theorem_for_Complex_Derivatives/Quotient_Rule | [
"Combination Theorem for Complex Derivatives"
] | [] | [
"Definition:Open Ball",
"Definition:Open Ball/Radius",
"Epsilon-Function Differentiability Condition/Complex Case",
"Definition:Complex Function",
"Definition:Convergent Complex Function ",
"Complex-Differentiable Function is Continuous",
"Definition:Continuous Complex Function",
"Reverse Triangle Ine... |
proofwiki-6784 | Complex-Differentiable Function is Continuous | Let $f: D \to \C$ be a complex function, where $D \subseteq \C$ is an open set.
Let $f$ be complex-differentiable at $a \in D$.
Then $f$ is continuous at $a$. | Let $\map {N_r} 0$ denote the $r$-neighborhood of $0$ in $\C$.
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {N_r} 0 \setminus \set 0$:
:$(1): \quad \map f {a + h} = \map f a + h \paren {\map {f'} a + \map \epsilon h}$
where $\eps... | Let $f: D \to \C$ be a [[Definition:Complex Function|complex function]], where $D \subseteq \C$ is an [[Definition:Open Set (Complex Analysis)|open set]].
Let $f$ be [[Definition:Complex-Differentiable Function|complex-differentiable]] at $a \in D$.
Then $f$ is [[Definition:Continuous Complex Function|continuous]] a... | Let $\map {N_r} 0$ denote the [[Definition:Neighborhood (Complex Analysis)|$r$-neighborhood]] of $0$ in $\C$.
By the [[Epsilon-Function Complex Differentiability Condition]], it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {N_r} 0 \setminus \set 0$:
:$(1): \quad \map f {a + h} = \map f a + ... | Complex-Differentiable Function is Continuous/Proof 1 | https://proofwiki.org/wiki/Complex-Differentiable_Function_is_Continuous | https://proofwiki.org/wiki/Complex-Differentiable_Function_is_Continuous/Proof_1 | [
"Complex-Differentiable Function is Continuous",
"Complex Differential Calculus",
"Continuous Mappings"
] | [
"Definition:Complex Function",
"Definition:Open Set/Complex Analysis",
"Definition:Differentiable Mapping/Complex Function",
"Definition:Continuous Complex Function"
] | [
"Definition:Neighborhood (Complex Analysis)",
"Epsilon-Function Differentiability Condition/Complex Case",
"Definition:Complex Function",
"Combination Theorem for Limits of Functions/Complex",
"Definition:Continuous Complex Function/Using Limit",
"Definition:Continuous Complex Function"
] |
proofwiki-6785 | Complex-Differentiable Function is Continuous | Let $f: D \to \C$ be a complex function, where $D \subseteq \C$ is an open set.
Let $f$ be complex-differentiable at $a \in D$.
Then $f$ is continuous at $a$. | For each $z \in D$:
{{begin-eqn}}
{{eqn | l = \lim_{w \mathop \to z} \map f w
| r = \map f z + \lim_{w \mathop \to z} \paren {\map f w - \map f z}
| c = Sum Rule for Limits of Complex Functions
}}
{{eqn | r = \map f z + \lim_{w \mathop \to z} \paren {\frac {\map f w - \map f z} {w - z} \paren {w - z} }
}}
{... | Let $f: D \to \C$ be a [[Definition:Complex Function|complex function]], where $D \subseteq \C$ is an [[Definition:Open Set (Complex Analysis)|open set]].
Let $f$ be [[Definition:Complex-Differentiable Function|complex-differentiable]] at $a \in D$.
Then $f$ is [[Definition:Continuous Complex Function|continuous]] a... | For each $z \in D$:
{{begin-eqn}}
{{eqn | l = \lim_{w \mathop \to z} \map f w
| r = \map f z + \lim_{w \mathop \to z} \paren {\map f w - \map f z}
| c = [[Sum Rule for Limits of Complex Functions]]
}}
{{eqn | r = \map f z + \lim_{w \mathop \to z} \paren {\frac {\map f w - \map f z} {w - z} \paren {w - z} }
... | Complex-Differentiable Function is Continuous/Proof 2 | https://proofwiki.org/wiki/Complex-Differentiable_Function_is_Continuous | https://proofwiki.org/wiki/Complex-Differentiable_Function_is_Continuous/Proof_2 | [
"Complex-Differentiable Function is Continuous",
"Complex Differential Calculus",
"Continuous Mappings"
] | [
"Definition:Complex Function",
"Definition:Open Set/Complex Analysis",
"Definition:Differentiable Mapping/Complex Function",
"Definition:Continuous Complex Function"
] | [
"Combination Theorem for Limits of Functions/Complex/Sum Rule",
"Combination Theorem for Limits of Functions/Complex/Product Rule"
] |
proofwiki-6786 | Proof by Cases/Formulation 1/Forward Implication/Proof 1 | :$\paren {p \implies r} \land \paren {q \implies r} \vdash \paren {p \lor q} \implies r$ | {{BeginTableau|\paren {p \implies r} \land \paren {q \implies r} \vdash \paren {p \lor q} \implies r}}
{{Premise|1|\paren {p \implies r} \land \paren {q \implies r} }}
{{Simplification|2|1|p \implies r|1|1}}
{{Simplification|3|1|q \implies r|1|2}}
{{SequentIntro|4|1|p \lor q \implies r \lor r|2, 3|Constructive Dilemma}... | :$\paren {p \implies r} \land \paren {q \implies r} \vdash \paren {p \lor q} \implies r$ | {{BeginTableau|\paren {p \implies r} \land \paren {q \implies r} \vdash \paren {p \lor q} \implies r}}
{{Premise|1|\paren {p \implies r} \land \paren {q \implies r} }}
{{Simplification|2|1|p \implies r|1|1}}
{{Simplification|3|1|q \implies r|1|2}}
{{SequentIntro|4|1|p \lor q \implies r \lor r|2, 3|[[Constructive Dilemm... | Proof by Cases/Formulation 1/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_1/Forward_Implication/Proof_1 | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_1/Forward_Implication/Proof_1 | [
"Proof by Cases"
] | [] | [
"Constructive Dilemma"
] |
proofwiki-6787 | Closed Ball in Euclidean Space is Compact | Let $x \in \R_n$ be a point in the Euclidean space $\R^n$.
Let $\epsilon \in \R_{>0}$.
Then the closed $\epsilon$-ball $\map {B_\epsilon^-} x$ is compact. | From Closed Ball is Closed in Metric Space, it follows that $\map {B_\epsilon^-} x$ is closed in $\R^n$.
For all $a \in \map {B_\epsilon^-} x$ we have $\map d {x, a} \le \epsilon$, where $d$ denotes the Euclidean metric.
Then $\map {B_\epsilon^-} x$ is bounded in $\R^n$.
From the Heine-Borel Theorem, it follows that $\... | Let $x \in \R_n$ be a point in the [[Definition:Euclidean Space|Euclidean space]] $\R^n$.
Let $\epsilon \in \R_{>0}$.
Then the [[Definition:Closed Ball|closed $\epsilon$-ball]] $\map {B_\epsilon^-} x$ is [[Definition:Compact Subspace|compact]]. | From [[Closed Ball is Closed in Metric Space]], it follows that $\map {B_\epsilon^-} x$ is [[Definition:Closed Set (Metric Space)|closed]] in $\R^n$.
For all $a \in \map {B_\epsilon^-} x$ we have $\map d {x, a} \le \epsilon$, where $d$ denotes the [[Definition:Euclidean Metric on Real Vector Space|Euclidean metric]].
... | Closed Ball in Euclidean Space is Compact | https://proofwiki.org/wiki/Closed_Ball_in_Euclidean_Space_is_Compact | https://proofwiki.org/wiki/Closed_Ball_in_Euclidean_Space_is_Compact | [
"Closed Balls",
"Compact Topological Spaces",
"Euclidean Spaces"
] | [
"Definition:Euclidean Space",
"Definition:Closed Ball",
"Definition:Compact Topological Space/Subspace"
] | [
"Closed Ball is Closed/Metric Space",
"Definition:Closed Set/Metric Space",
"Definition:Euclidean Metric/Real Vector Space",
"Definition:Bounded Metric Space",
"Heine-Borel Theorem/Euclidean Space",
"Definition:Compact Topological Space/Subspace",
"Category:Closed Balls",
"Category:Compact Topological... |
proofwiki-6788 | Isometric Image of Cauchy Sequence is Cauchy Sequence | Let $\struct {S_1, d_1}$ and $\struct {S_2, d_2}$ be metric spaces.
Let $f: S_1 \to S_2$ be an isometry.
Let $\sequence {x_n}$ be a Cauchy sequence in $S_1$.
Let $\sequence {y_n} = \sequence {\map f {x_n} }$ be the image of $\sequence {x_n}$ under $f$.
Then $\sequence {y_n}$ is a Cauchy sequence. | Let $\epsilon \in \R_{>0}$.
By the definition of Cauchy sequence, there is an $N \in \R$ such that:
:$\paren{m > N} \land \paren {n > N} \implies \map {d_1} {x_m, x_n} < \epsilon$
Since $f$ is an isometry, $\map {d_2} {y_m , y_n} = \map {d_1} {x_m, x_n}$ for all $m$ and $n$.
Thus:
:$\paren {m > N} \land \paren {n > N} ... | Let $\struct {S_1, d_1}$ and $\struct {S_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
Let $f: S_1 \to S_2$ be an [[Definition:Isometry (Metric Spaces)|isometry]].
Let $\sequence {x_n}$ be a [[Definition:Cauchy Sequence|Cauchy sequence]] in $S_1$.
Let $\sequence {y_n} = \sequence {\map f {x_n} }$ be the ima... | Let $\epsilon \in \R_{>0}$.
By the definition of [[Definition:Cauchy Sequence|Cauchy sequence]], there is an $N \in \R$ such that:
:$\paren{m > N} \land \paren {n > N} \implies \map {d_1} {x_m, x_n} < \epsilon$
Since $f$ is an [[Definition:Isometry (Metric Spaces)|isometry]], $\map {d_2} {y_m , y_n} = \map {d_1} {x_... | Isometric Image of Cauchy Sequence is Cauchy Sequence | https://proofwiki.org/wiki/Isometric_Image_of_Cauchy_Sequence_is_Cauchy_Sequence | https://proofwiki.org/wiki/Isometric_Image_of_Cauchy_Sequence_is_Cauchy_Sequence | [
"Isometries (Metric Spaces)",
"Cauchy Sequences"
] | [
"Definition:Metric Space",
"Definition:Isometry (Metric Spaces)",
"Definition:Cauchy Sequence",
"Definition:Cauchy Sequence"
] | [
"Definition:Cauchy Sequence",
"Definition:Isometry (Metric Spaces)",
"Definition:Cauchy Sequence",
"Category:Isometries (Metric Spaces)",
"Category:Cauchy Sequences"
] |
proofwiki-6789 | Union of Subsets is Subset | Let $S_1$, $S_2$, and $T$ be sets.
Let $S_1$ and $S_2$ both be subsets of $T$.
Then:
:$S_1 \cup S_2 \subseteq T$
That is:
:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \implies \paren {S_1 \cup S_2} \subseteq T$ | Let:
:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T}$
Then:
{{begin-eqn}}
{{eqn | l = S_1 \cup S_2
| o = \subseteq
| r = T \cup T
| c = Set Union Preserves Subsets
}}
{{eqn | ll= \leadsto
| l = S_1 \cup S_2
| o = \subseteq
| r = T
| c = Set Union is Idempotent
}}
{{en... | Let $S_1$, $S_2$, and $T$ be [[Definition:Set|sets]].
Let $S_1$ and $S_2$ both be [[Definition:Subset|subsets]] of $T$.
Then:
:$S_1 \cup S_2 \subseteq T$
That is:
:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \implies \paren {S_1 \cup S_2} \subseteq T$ | Let:
:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T}$
Then:
{{begin-eqn}}
{{eqn | l = S_1 \cup S_2
| o = \subseteq
| r = T \cup T
| c = [[Set Union Preserves Subsets]]
}}
{{eqn | ll= \leadsto
| l = S_1 \cup S_2
| o = \subseteq
| r = T
| c = [[Set Union is Idempotent... | Union of Subsets is Subset/Proof 1 | https://proofwiki.org/wiki/Union_of_Subsets_is_Subset | https://proofwiki.org/wiki/Union_of_Subsets_is_Subset/Proof_1 | [
"Set Union",
"Subsets",
"Union of Subsets is Subset"
] | [
"Definition:Set",
"Definition:Subset"
] | [
"Set Union Preserves Subsets",
"Set Union is Idempotent"
] |
proofwiki-6790 | Union of Subsets is Subset | Let $S_1$, $S_2$, and $T$ be sets.
Let $S_1$ and $S_2$ both be subsets of $T$.
Then:
:$S_1 \cup S_2 \subseteq T$
That is:
:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \implies \paren {S_1 \cup S_2} \subseteq T$ | Let $x \in S_1 \cup S_2$.
By the definition of union, either $x \in S_1$ or $x \in S_2$.
By hypothesis, $S_1 \subseteq T$ and $S_2 \subseteq T$.
By definition of subset:
:$x \in S_1 \implies x \in T$
:$x \in S_2 \implies x \in T$
By Proof by Cases it follows that $x \in T$.
Hence the result by definition of subset.
{{q... | Let $S_1$, $S_2$, and $T$ be [[Definition:Set|sets]].
Let $S_1$ and $S_2$ both be [[Definition:Subset|subsets]] of $T$.
Then:
:$S_1 \cup S_2 \subseteq T$
That is:
:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \implies \paren {S_1 \cup S_2} \subseteq T$ | Let $x \in S_1 \cup S_2$.
By the definition of [[Definition:Set Union|union]], either $x \in S_1$ or $x \in S_2$.
[[Definition:By Hypothesis|By hypothesis]], $S_1 \subseteq T$ and $S_2 \subseteq T$.
By definition of [[Definition:Subset|subset]]:
:$x \in S_1 \implies x \in T$
:$x \in S_2 \implies x \in T$
By [[Proof... | Union of Subsets is Subset/Proof 2 | https://proofwiki.org/wiki/Union_of_Subsets_is_Subset | https://proofwiki.org/wiki/Union_of_Subsets_is_Subset/Proof_2 | [
"Set Union",
"Subsets",
"Union of Subsets is Subset"
] | [
"Definition:Set",
"Definition:Subset"
] | [
"Definition:Set Union",
"Definition:By Hypothesis",
"Definition:Subset",
"Proof by Cases",
"Definition:Subset"
] |
proofwiki-6791 | Proof by Cases/Formulation 1/Forward Implication | :$\paren {p \implies r} \land \paren {q \implies r} \vdash \paren {p \lor q} \implies r$ | {{BeginTableau|\paren {p \implies r} \land \paren {q \implies r} \vdash \paren {p \lor q} \implies r}}
{{Premise|1|\paren {p \implies r} \land \paren {q \implies r} }}
{{Simplification|2|1|p \implies r|1|1}}
{{Simplification|3|1|q \implies r|1|2}}
{{SequentIntro|4|1|p \lor q \implies r \lor r|2, 3|Constructive Dilemma}... | :$\paren {p \implies r} \land \paren {q \implies r} \vdash \paren {p \lor q} \implies r$ | {{BeginTableau|\paren {p \implies r} \land \paren {q \implies r} \vdash \paren {p \lor q} \implies r}}
{{Premise|1|\paren {p \implies r} \land \paren {q \implies r} }}
{{Simplification|2|1|p \implies r|1|1}}
{{Simplification|3|1|q \implies r|1|2}}
{{SequentIntro|4|1|p \lor q \implies r \lor r|2, 3|[[Constructive Dilemm... | Proof by Cases/Formulation 1/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_1/Forward_Implication/Proof_1 | [
"Proof by Cases"
] | [] | [
"Constructive Dilemma"
] |
proofwiki-6792 | Proof by Cases/Formulation 1/Forward Implication | :$\paren {p \implies r} \land \paren {q \implies r} \vdash \paren {p \lor q} \implies r$ | From the Constructive Dilemma we have:
:$p \implies q, r \implies s \vdash p \lor r \implies q \lor s$
from which, changing the names of letters strategically:
:$p \implies r, q \implies r \vdash p \lor q \implies r \lor r$
From the Rule of Idempotence we have:
:$r \lor r \vdash r$
and the result follows by Hypothetica... | :$\paren {p \implies r} \land \paren {q \implies r} \vdash \paren {p \lor q} \implies r$ | From the [[Constructive Dilemma]] we have:
:$p \implies q, r \implies s \vdash p \lor r \implies q \lor s$
from which, changing the names of letters strategically:
:$p \implies r, q \implies r \vdash p \lor q \implies r \lor r$
From the [[Rule of Idempotence]] we have:
:$r \lor r \vdash r$
and the result follows by [[... | Proof by Cases/Formulation 1/Forward Implication/Proof 2 | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_1/Forward_Implication/Proof_2 | [
"Proof by Cases"
] | [] | [
"Constructive Dilemma",
"Rule of Idempotence",
"Hypothetical Syllogism"
] |
proofwiki-6793 | Proof by Cases/Formulation 1 | :$\paren {p \implies r} \land \paren {q \implies r} \dashv \vdash \paren {p \lor q} \implies r$ | {{BeginTableau|\paren {p \implies r} \land \paren {q \implies r} \vdash \paren {p \lor q} \implies r}}
{{Premise|1|\paren {p \implies r} \land \paren {q \implies r} }}
{{Simplification|2|1|p \implies r|1|1}}
{{Simplification|3|1|q \implies r|1|2}}
{{SequentIntro|4|1|p \lor q \implies r \lor r|2, 3|Constructive Dilemma}... | :$\paren {p \implies r} \land \paren {q \implies r} \dashv \vdash \paren {p \lor q} \implies r$ | {{BeginTableau|\paren {p \implies r} \land \paren {q \implies r} \vdash \paren {p \lor q} \implies r}}
{{Premise|1|\paren {p \implies r} \land \paren {q \implies r} }}
{{Simplification|2|1|p \implies r|1|1}}
{{Simplification|3|1|q \implies r|1|2}}
{{SequentIntro|4|1|p \lor q \implies r \lor r|2, 3|[[Constructive Dilemm... | Proof by Cases/Formulation 1/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_1 | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_1/Forward_Implication/Proof_1 | [
"Proof by Cases"
] | [] | [
"Constructive Dilemma"
] |
proofwiki-6794 | Proof by Cases/Formulation 1 | :$\paren {p \implies r} \land \paren {q \implies r} \dashv \vdash \paren {p \lor q} \implies r$ | From the Constructive Dilemma we have:
:$p \implies q, r \implies s \vdash p \lor r \implies q \lor s$
from which, changing the names of letters strategically:
:$p \implies r, q \implies r \vdash p \lor q \implies r \lor r$
From the Rule of Idempotence we have:
:$r \lor r \vdash r$
and the result follows by Hypothetica... | :$\paren {p \implies r} \land \paren {q \implies r} \dashv \vdash \paren {p \lor q} \implies r$ | From the [[Constructive Dilemma]] we have:
:$p \implies q, r \implies s \vdash p \lor r \implies q \lor s$
from which, changing the names of letters strategically:
:$p \implies r, q \implies r \vdash p \lor q \implies r \lor r$
From the [[Rule of Idempotence]] we have:
:$r \lor r \vdash r$
and the result follows by [[... | Proof by Cases/Formulation 1/Forward Implication/Proof 2 | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_1 | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_1/Forward_Implication/Proof_2 | [
"Proof by Cases"
] | [] | [
"Constructive Dilemma",
"Rule of Idempotence",
"Hypothetical Syllogism"
] |
proofwiki-6795 | Proof by Cases/Formulation 1 | :$\paren {p \implies r} \land \paren {q \implies r} \dashv \vdash \paren {p \lor q} \implies r$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin{array}{|ccccccc||ccccc|} \hline
(p & \implies & r) & \land & (q & \implies & r) & (p & \lor & q) & \implies & r \\
\hline
\F & \T & \F & \T & \F & \T & \... | :$\paren {p \implies r} \land \paren {q \implies r} \dashv \vdash \paren {p \lor q} \implies r$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|ccccccc||ccccc|} \hli... | Proof by Cases/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_1 | https://proofwiki.org/wiki/Proof_by_Cases/Formulation_1/Proof_by_Truth_Table | [
"Proof by Cases"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-6796 | Distance between Closed Sets in Euclidean Space | Let $S, T \subseteq \R^n$ be closed, non-empty subsets of the real Euclidean space $R^n$.
Suppose that $S$ is bounded, and $S$ and $T$ are disjoint.
Then there exists $x \in S$ and $y \in T$ such that:
:$\map d {x, y} = \map d {S, T} > 0$
where:
:$d$ denotes the Euclidean metric
:$\map d {S, T}$ is the distance between... | By definition of distance from subset, we can for all $n \in \N$ find $x_n \in S, y_n \in T$ such that:
:$\map d {S, T} \le \map d {x_n, y_n} < \map d {S, T} + \dfrac 1 n$
so:
:$\ds \lim_{n \mathop \to \infty} \map d {x_n, y_n} = \map d {S, T}$
By definition of bounded space, there exists $a \in S$ and $K \in \R$ such ... | Let $S, T \subseteq \R^n$ be [[Definition:Closed Set of Metric Space|closed]], [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subsets]] of the [[Definition:Real Euclidean Space|real Euclidean space]] $R^n$.
Suppose that $S$ is [[Definition:Bounded Metric Space|bounded]], and $S$ and $T$ are [[Definition:D... | By definition of [[Definition:Distance from Subset|distance from subset]], we can for all $n \in \N$ find $x_n \in S, y_n \in T$ such that:
:$\map d {S, T} \le \map d {x_n, y_n} < \map d {S, T} + \dfrac 1 n$
so:
:$\ds \lim_{n \mathop \to \infty} \map d {x_n, y_n} = \map d {S, T}$
By definition of [[Definition:Bounde... | Distance between Closed Sets in Euclidean Space | https://proofwiki.org/wiki/Distance_between_Closed_Sets_in_Euclidean_Space | https://proofwiki.org/wiki/Distance_between_Closed_Sets_in_Euclidean_Space | [
"Real Euclidean Spaces"
] | [
"Definition:Closed Set/Metric Space",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Euclidean Space/Real",
"Definition:Bounded Metric Space",
"Definition:Disjoint Sets",
"Definition:Euclidean Metric/Real Vector Space",
"Definition:Distance/Sets"
] | [
"Definition:Distance/Sets",
"Definition:Bounded Metric Space",
"Definition:Bounded Sequence",
"Definition:Bounded Sequence",
"Triangle Inequality/Vectors in Euclidean Space",
"Definition:Sequence",
"Minkowski's Inequality for Sums",
"Bounded Sequence in Euclidean Space has Convergent Subsequence",
"... |
proofwiki-6797 | Complement in Distributive Lattice is Unique | Let $\struct {S, \vee, \wedge, \preceq}$ be a bounded distributive lattice.
Then every $a \in S$ admits at most one complement. | Let $a \in S$, and suppose that $b, c \in S$ are complements for $a$.
Then:
{{begin-eqn}}
{{eqn | l = b
| r = \top \wedge b
| c = $\top$ is the identity for $\wedge$
}}
{{eqn | r = \paren {c \vee a} \wedge b
| c = $c$ is a complement for $a$
}}
{{eqn | r = \paren {c \wedge b} \vee \paren {a \wedge b}
... | Let $\struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Bounded Lattice|bounded]] [[Definition:Distributive Lattice|distributive lattice]].
Then every $a \in S$ admits at most one [[Definition:Complement (Lattice Theory)|complement]]. | Let $a \in S$, and suppose that $b, c \in S$ are [[Definition:Complement (Lattice Theory)|complements]] for $a$.
Then:
{{begin-eqn}}
{{eqn | l = b
| r = \top \wedge b
| c = $\top$ is the [[Definition:Identity Element|identity]] for $\wedge$
}}
{{eqn | r = \paren {c \vee a} \wedge b
| c = $c$ is a [[... | Complement in Distributive Lattice is Unique | https://proofwiki.org/wiki/Complement_in_Distributive_Lattice_is_Unique | https://proofwiki.org/wiki/Complement_in_Distributive_Lattice_is_Unique | [
"Distributive Lattices",
"Bounded Lattices"
] | [
"Definition:Bounded Lattice",
"Definition:Distributive Lattice",
"Definition:Complement (Lattice Theory)"
] | [
"Definition:Complement (Lattice Theory)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Complement (Lattice Theory)",
"Definition:Distributive Lattice",
"Definition:Complement (Lattice Theory)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Category:Distributiv... |
proofwiki-6798 | Equivalence of Definitions of Top of Lattice | {{TFAE|def = Top of Lattice}}
Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice. | By definition, $\top$ is the greatest element of $S$ {{iff}} for all $a \in S$:
:$a \preceq \top$
By Ordering in terms of Meet, this is equivalent to:
:$a \wedge \top = a$
If this equality holds for all $a \in S$, then by definition $\top$ is an identity for $\wedge$.
The result follows.
{{qed}}
Category:Top of Lattice... | {{TFAE|def = Top of Lattice}}
Let $\struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Lattice (Order Theory)|lattice]]. | By definition, $\top$ is the [[Definition:Greatest Element|greatest element]] of $S$ {{iff}} for all $a \in S$:
:$a \preceq \top$
By [[Ordering in terms of Meet]], this is equivalent to:
:$a \wedge \top = a$
If this equality holds for all $a \in S$, then by definition $\top$ is an [[Definition:Identity Element|iden... | Equivalence of Definitions of Top of Lattice | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Top_of_Lattice | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Top_of_Lattice | [
"Top of Lattice"
] | [
"Definition:Lattice (Order Theory)"
] | [
"Definition:Greatest Element",
"Ordering in terms of Meet",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Category:Top of Lattice"
] |
proofwiki-6799 | Equivalence of Definitions of Bottom of Lattice | Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $\bot$ be a bottom of $\struct {S, \vee, \wedge, \preceq}$.
{{TFAE|def = Bottom of Lattice|view = Bottom|context = Lattice Theory}} | By definition, $\bot$ is the smallest element of $S$ {{iff}} for all $a \in S$:
:$\bot \preceq a$
By Ordering in terms of Join, this is equivalent to:
:$a \vee \bot = a$
If this equality holds for all $a \in S$, then by definition $\bot$ is an identity for $\vee$.
The result follows.
{{qed}}
Category:Bottom of Lattice
... | Let $\struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Lattice (Order Theory)|lattice]].
Let $\bot$ be a [[Definition:Bottom of Lattice|bottom]] of $\struct {S, \vee, \wedge, \preceq}$.
{{TFAE|def = Bottom of Lattice|view = Bottom|context = Lattice Theory}} | By definition, $\bot$ is the [[Definition:Smallest Element|smallest element]] of $S$ {{iff}} for all $a \in S$:
:$\bot \preceq a$
By [[Ordering in terms of Join]], this is equivalent to:
:$a \vee \bot = a$
If this equality holds for all $a \in S$, then by definition $\bot$ is an [[Definition:Identity Element|identi... | Equivalence of Definitions of Bottom of Lattice | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Bottom_of_Lattice | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Bottom_of_Lattice | [
"Bottom of Lattice"
] | [
"Definition:Lattice (Order Theory)",
"Definition:Bottom of Lattice"
] | [
"Definition:Smallest Element",
"Ordering in terms of Join",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Category:Bottom of Lattice"
] |
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