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-7100 | Power Function Preserves Ordering in Ordered Group | Let $n \in \N_{>0}$ be a strictly positive integer.
Let $\prec$ be the reflexive reduction of $\preccurlyeq$.
Then the following hold:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
| l = x \preccurlyeq y
| o = \implies
| r = x^n \preccurlyeq y^n
}}
{{eqn | q = \forall x, y \in S
| l = x \prec y
... | An ordered group is an ordered structure which is also a group.
Hence an ordered group is {{afortiori}} an ordered semigroup.
From Power Function Preserves Ordering in Ordered Semigroup:
:$\forall x, y \in S: x \preccurlyeq y \implies x^n \preccurlyeq y^n$
From the Cancellation Laws, every element of a group is cancell... | Let $n \in \N_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Let $\prec$ be the [[Definition:Reflexive Reduction|reflexive reduction]] of $\preccurlyeq$.
Then the following hold:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
| l = x \preccurlyeq y
| o = \implies
| r ... | An [[Definition:Ordered Group|ordered group]] is an [[Definition:Ordered Structure|ordered structure]] which is also a [[Definition:Group|group]].
Hence an [[Definition:Ordered Group|ordered group]] is {{afortiori}} an [[Definition:Ordered Semigroup|ordered semigroup]].
From [[Power Function Preserves Ordering in Or... | Power Function Preserves Ordering in Ordered Group/Proof 2 | https://proofwiki.org/wiki/Power_Function_Preserves_Ordering_in_Ordered_Group | https://proofwiki.org/wiki/Power_Function_Preserves_Ordering_in_Ordered_Group/Proof_2 | [
"Ordered Groups",
"Power Function Preserves Ordering in Ordered Group"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Reflexive Reduction",
"Definition:Power of Element/Semigroup"
] | [
"Definition:Ordered Group",
"Definition:Ordered Structure",
"Definition:Group",
"Definition:Ordered Group",
"Definition:Ordered Semigroup",
"Power Function Preserves Ordering in Ordered Semigroup",
"Cancellation Laws",
"Definition:Element",
"Definition:Group",
"Definition:Cancellable Element",
"... |
proofwiki-7101 | Period of Complex Exponential Function | :$\map \exp {z + 2 k \pi i} = \map \exp z$ | {{begin-eqn}}
{{eqn | l = \map \exp {z + 2 k \pi i}
| r = \map \exp z \, \map \exp {2 k \pi i}
| c = Exponential of Sum: Complex Numbers
}}
{{eqn | r = \map \exp z \times 1
| c = Euler's Formula Example: $e^{2 k i \pi}$
}}
{{eqn | r = \map \exp z
}}
{{end-eqn}}
{{qed}} | :$\map \exp {z + 2 k \pi i} = \map \exp z$ | {{begin-eqn}}
{{eqn | l = \map \exp {z + 2 k \pi i}
| r = \map \exp z \, \map \exp {2 k \pi i}
| c = [[Exponential of Sum/Complex Numbers|Exponential of Sum: Complex Numbers]]
}}
{{eqn | r = \map \exp z \times 1
| c = [[Euler's Formula/Examples/e^2 k i pi|Euler's Formula Example: $e^{2 k i \pi}$]]
}}
... | Period of Complex Exponential Function/Proof 1 | https://proofwiki.org/wiki/Period_of_Complex_Exponential_Function | https://proofwiki.org/wiki/Period_of_Complex_Exponential_Function/Proof_1 | [
"Period of Complex Exponential Function",
"Exponential Function"
] | [] | [
"Exponential of Sum/Complex Numbers",
"Euler's Formula/Examples/e^2 k i pi"
] |
proofwiki-7102 | Period of Complex Exponential Function | :$\map \exp {z + 2 k \pi i} = \map \exp z$ | {{begin-eqn}}
{{eqn | l = e^{i \paren {\theta + 2 k \pi} }
| r = \map \cos {\theta + 2 k \pi} + i \, \map \sin {\theta + 2 k \pi}
| c = Euler's Formula
}}
{{eqn | r = \cos \theta + i \sin \theta
| c = Sine and Cosine are Periodic on Reals
}}
{{eqn | r = e^{i \theta}
| c = Euler's Formula
}}
{{en... | :$\map \exp {z + 2 k \pi i} = \map \exp z$ | {{begin-eqn}}
{{eqn | l = e^{i \paren {\theta + 2 k \pi} }
| r = \map \cos {\theta + 2 k \pi} + i \, \map \sin {\theta + 2 k \pi}
| c = [[Euler's Formula]]
}}
{{eqn | r = \cos \theta + i \sin \theta
| c = [[Sine and Cosine are Periodic on Reals]]
}}
{{eqn | r = e^{i \theta}
| c = [[Euler's Formu... | Period of Complex Exponential Function/Proof 2 | https://proofwiki.org/wiki/Period_of_Complex_Exponential_Function | https://proofwiki.org/wiki/Period_of_Complex_Exponential_Function/Proof_2 | [
"Period of Complex Exponential Function",
"Exponential Function"
] | [] | [
"Euler's Formula",
"Sine and Cosine are Periodic on Reals",
"Euler's Formula"
] |
proofwiki-7103 | Zero Staircase Integral Condition for Primitive | Let $f: D \to \C$ be a continuous complex function, where $D$ is an open and connected domain.
Let $z_0 \in D$.
Suppose that $\ds \oint_C \map f z \rd z = 0$ for all closed staircase contours $C$ in $D$.
Then $f$ has a primitive $F: D \to \C$ defined by:
:$\ds \map F w = \int_{C_w} \map f z \rd z$
where $C_w$ is any st... | From Connected Domain is Connected by Staircase Contours, it follows that there exists a staircase contour $C_w$ in $D$ with start point $z_0$ and end point $w$.
If $C_w'$ is another staircase contour with the same endpoints as $C_w$, then $C_w' \cup \paren {-C_w}$ is a closed staircase contour where $-C_w$ is the reve... | Let $f: D \to \C$ be a [[Definition:Continuous Complex Function|continuous complex function]], where $D$ is an [[Definition:Open Set (Complex Analysis)|open]] and [[Definition:Connected Domain (Complex Analysis)|connected domain]].
Let $z_0 \in D$.
Suppose that $\ds \oint_C \map f z \rd z = 0$ for all [[Definition:C... | From [[Connected Domain is Connected by Staircase Contours]], it follows that there exists a [[Definition:Staircase Contour|staircase contour]] $C_w$ in $D$ with [[Definition:Start Point of Contour (Complex Plane)|start point]] $z_0$ and [[Definition:End Point of Contour (Complex Plane)|end point]] $w$.
If $C_w'$ is a... | Zero Staircase Integral Condition for Primitive | https://proofwiki.org/wiki/Zero_Staircase_Integral_Condition_for_Primitive | https://proofwiki.org/wiki/Zero_Staircase_Integral_Condition_for_Primitive | [
"Complex Analysis"
] | [
"Definition:Continuous Complex Function",
"Definition:Open Set/Complex Analysis",
"Definition:Connected Domain (Complex Analysis)",
"Definition:Contour/Closed/Complex Plane",
"Definition:Staircase Contour",
"Definition:Primitive (Calculus)/Complex",
"Definition:Staircase Contour",
"Definition:Contour/... | [
"Connected Domain is Connected by Staircase Contours",
"Definition:Staircase Contour",
"Definition:Contour/Endpoints/Complex Plane",
"Definition:Contour/Endpoints/Complex Plane",
"Definition:Contour/Endpoints/Complex Plane",
"Definition:Contour/Closed/Complex Plane",
"Definition:Staircase Contour",
"D... |
proofwiki-7104 | Derivative of Complex Polynomial | Let $a_n \in \C$ for $n \in \set {0, 1, \ldots, N}$, where $N \in \N$.
Let $f: \C \to \C$ be a complex polynomial defined by $\ds \map f z = \sum_{n \mathop = 0}^N a_n z^n$.
Then $f$ is complex differentiable and its derivative is:
:$\ds \map {f'} z = \sum_{n \mathop = 1}^N n a_n z^{n - 1}$ | For all $n > N$, put $a_n = 0$.
Then:
:$\ds \map f z = \sum_{n \mathop = 0}^\infty a_n z^n$
The result now follows from Derivative of Complex Power Series.
{{qed}}
Category:Complex Differential Calculus
42ks1h3kb95t4difyockiyrqs87ify7 | Let $a_n \in \C$ for $n \in \set {0, 1, \ldots, N}$, where $N \in \N$.
Let $f: \C \to \C$ be a [[Definition:Complex Number|complex]] [[Definition:Polynomial|polynomial]] defined by $\ds \map f z = \sum_{n \mathop = 0}^N a_n z^n$.
Then $f$ is [[Definition:Complex-Differentiable Function|complex differentiable]] and i... | For all $n > N$, put $a_n = 0$.
Then:
:$\ds \map f z = \sum_{n \mathop = 0}^\infty a_n z^n$
The result now follows from [[Derivative of Complex Power Series]].
{{qed}}
[[Category:Complex Differential Calculus]]
42ks1h3kb95t4difyockiyrqs87ify7 | Derivative of Complex Polynomial | https://proofwiki.org/wiki/Derivative_of_Complex_Polynomial | https://proofwiki.org/wiki/Derivative_of_Complex_Polynomial | [
"Complex Differential Calculus"
] | [
"Definition:Complex Number",
"Definition:Polynomial",
"Definition:Differentiable Mapping/Complex Function",
"Definition:Derivative/Complex Function"
] | [
"Derivative of Complex Power Series",
"Category:Complex Differential Calculus"
] |
proofwiki-7105 | Cantor's Intersection Theorem | Let $M = \struct {A, d}$ be a complete metric space.
Let $\sequence {S_n}$ be a nested sequence of closed balls in $M$ defined by:
:$S_n = \map {B^-_{\rho_n} } {x_n}$
where $\rho_n \to 0$ as $n \to \infty$ and:
:$S_1 \supseteq S_2 \supseteq \cdots \supseteq S_n \supseteq \cdots$
Then there exists a unique $x \in A$ suc... | Let $S_n = \map {B^-_{\rho_n} } {x_n}$ be the closed ball of radius $\rho_n$ about the point $x_n$.
That is, let $S_n = \set {x \in A: \map d {x_n, x} \le \rho_n}$.
Then the sequence $\sequence {x_n}$ forms a Cauchy sequence:
:$\map d {x_n, x_{n + p} } < \rho_n$
for any $p \ge 0$ since $S_{n + p} \subseteq S_n$.
Howeve... | Let $M = \struct {A, d}$ be a [[Definition:Complete Metric Space|complete metric space]].
Let $\sequence {S_n}$ be a [[Definition:Nested Sequence|nested sequence]] of [[Definition:Closed Ball|closed balls]] in $M$ defined by:
:$S_n = \map {B^-_{\rho_n} } {x_n}$
where $\rho_n \to 0$ as $n \to \infty$ and:
:$S_1 \sup... | Let $S_n = \map {B^-_{\rho_n} } {x_n}$ be the [[Definition:Closed Ball|closed ball]] of [[Definition:Radius of Closed Ball|radius]] $\rho_n$ about the point $x_n$.
That is, let $S_n = \set {x \in A: \map d {x_n, x} \le \rho_n}$.
Then the [[Definition:Sequence|sequence]] $\sequence {x_n}$ forms a [[Definition:Cauchy S... | Cantor's Intersection Theorem | https://proofwiki.org/wiki/Cantor's_Intersection_Theorem | https://proofwiki.org/wiki/Cantor's_Intersection_Theorem | [
"Cantor's Intersection Theorem",
"Complete Metric Spaces",
"Named Theorems"
] | [
"Definition:Complete Metric Space",
"Definition:Nested Sequence",
"Definition:Closed Ball",
"Definition:Unique"
] | [
"Definition:Closed Ball",
"Definition:Closed Ball/Metric Space/Radius",
"Definition:Sequence",
"Definition:Cauchy Sequence/Metric Space",
"Definition:Metric Space",
"Definition:Complete Metric Space",
"Definition:Subsequence",
"Definition:Convergent Sequence/Metric Space",
"Definition:Closure (Topol... |
proofwiki-7106 | Principle of Dilemma/Formulation 1 | :$\paren {p \implies q} \land \paren {\neg p \implies q} \dashv \vdash q$ | {{BeginTableau|\paren {p \implies q} \land \paren {\neg p \implies q} \vdash q}}
{{Premise|1|\paren {p \implies q} \land \paren {\neg p \implies q} }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|\neg p \implies q|1|2}}
{{SequentIntro|4|1|p \lor \neg p \implies q \lor q|2, 3|Constructive Dilemma}}
{{Exc... | :$\paren {p \implies q} \land \paren {\neg p \implies q} \dashv \vdash q$ | {{BeginTableau|\paren {p \implies q} \land \paren {\neg p \implies q} \vdash q}}
{{Premise|1|\paren {p \implies q} \land \paren {\neg p \implies q} }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|\neg p \implies q|1|2}}
{{SequentIntro|4|1|p \lor \neg p \implies q \lor q|2, 3|[[Constructive Dilemma]]}}
{... | Principle of Dilemma/Formulation 1/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_1 | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_1/Forward_Implication/Proof_1 | [
"Principle of Dilemma"
] | [] | [
"Constructive Dilemma"
] |
proofwiki-7107 | Principle of Dilemma/Formulation 1 | :$\paren {p \implies q} \land \paren {\neg p \implies q} \dashv \vdash q$ | {{BeginTableau|\paren {p \implies q} \land \paren {\neg p \implies q} \vdash q}}
{{Premise|1|\paren {p \implies q} \land \paren {\neg p \implies q} }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|\neg p \implies q|1|2}}
{{ExcludedMiddle|4|p \lor \neg p}}
{{Assumption|5|p}}
{{ModusPonens|6|1, 5|q|2|5}}
{... | :$\paren {p \implies q} \land \paren {\neg p \implies q} \dashv \vdash q$ | {{BeginTableau|\paren {p \implies q} \land \paren {\neg p \implies q} \vdash q}}
{{Premise|1|\paren {p \implies q} \land \paren {\neg p \implies q} }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|\neg p \implies q|1|2}}
{{ExcludedMiddle|4|p \lor \neg p}}
{{Assumption|5|p}}
{{ModusPonens|6|1, 5|q|2|5}}
{... | Principle of Dilemma/Formulation 1/Forward Implication/Proof 2 | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_1 | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_1/Forward_Implication/Proof_2 | [
"Principle of Dilemma"
] | [] | [] |
proofwiki-7108 | Principle of Dilemma/Formulation 1 | :$\paren {p \implies q} \land \paren {\neg p \implies q} \dashv \vdash q$ | 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:
:$p \implies q, \neg p \implies q \vdash p \lor \neg p \implies q \lor q$
From Law of Excluded Middle, we have:
:$\vdash p \lor \neg p$
From the Rule of Idempot... | :$\paren {p \implies q} \land \paren {\neg p \implies q} \dashv \vdash q$ | 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:
:$p \implies q, \neg p \implies q \vdash p \lor \neg p \implies q \lor q$
From [[Law of Excluded Middle]], we have:
:$\vdash p \lor \neg p$
From the [[Rule... | Principle of Dilemma/Formulation 1/Forward Implication/Proof 3 | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_1 | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_1/Forward_Implication/Proof_3 | [
"Principle of Dilemma"
] | [] | [
"Constructive Dilemma",
"Law of Excluded Middle",
"Rule of Idempotence",
"Hypothetical Syllogism"
] |
proofwiki-7109 | Principle of Dilemma/Formulation 1 | :$\paren {p \implies q} \land \paren {\neg p \implies q} \dashv \vdash q$ | 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>$\begin{array}{|cccccccc||c|} \hline
(p & \implies & q) & \land & (\neg & p & \implies & q) & q \\
\hline
\F & \T & \F & \F & \T & \F & \F & \F & \F \\
\F & \T &... | :$\paren {p \implies q} \land \paren {\neg p \implies q} \dashv \vdash q$ | 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>$\begin{array}{|cccccccc||c|} \hline
... | Principle of Dilemma/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_1 | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_1/Proof_by_Truth_Table | [
"Principle of Dilemma"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7110 | Principle of Dilemma/Formulation 1/Forward Implication | :$\paren {p \implies q} \land \paren {\neg p \implies q} \vdash q$ | {{BeginTableau|\paren {p \implies q} \land \paren {\neg p \implies q} \vdash q}}
{{Premise|1|\paren {p \implies q} \land \paren {\neg p \implies q} }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|\neg p \implies q|1|2}}
{{SequentIntro|4|1|p \lor \neg p \implies q \lor q|2, 3|Constructive Dilemma}}
{{Exc... | :$\paren {p \implies q} \land \paren {\neg p \implies q} \vdash q$ | {{BeginTableau|\paren {p \implies q} \land \paren {\neg p \implies q} \vdash q}}
{{Premise|1|\paren {p \implies q} \land \paren {\neg p \implies q} }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|\neg p \implies q|1|2}}
{{SequentIntro|4|1|p \lor \neg p \implies q \lor q|2, 3|[[Constructive Dilemma]]}}
{... | Principle of Dilemma/Formulation 1/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_1/Forward_Implication/Proof_1 | [
"Principle of Dilemma"
] | [] | [
"Constructive Dilemma"
] |
proofwiki-7111 | Principle of Dilemma/Formulation 1/Forward Implication | :$\paren {p \implies q} \land \paren {\neg p \implies q} \vdash q$ | {{BeginTableau|\paren {p \implies q} \land \paren {\neg p \implies q} \vdash q}}
{{Premise|1|\paren {p \implies q} \land \paren {\neg p \implies q} }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|\neg p \implies q|1|2}}
{{ExcludedMiddle|4|p \lor \neg p}}
{{Assumption|5|p}}
{{ModusPonens|6|1, 5|q|2|5}}
{... | :$\paren {p \implies q} \land \paren {\neg p \implies q} \vdash q$ | {{BeginTableau|\paren {p \implies q} \land \paren {\neg p \implies q} \vdash q}}
{{Premise|1|\paren {p \implies q} \land \paren {\neg p \implies q} }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|\neg p \implies q|1|2}}
{{ExcludedMiddle|4|p \lor \neg p}}
{{Assumption|5|p}}
{{ModusPonens|6|1, 5|q|2|5}}
{... | Principle of Dilemma/Formulation 1/Forward Implication/Proof 2 | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_1/Forward_Implication/Proof_2 | [
"Principle of Dilemma"
] | [] | [] |
proofwiki-7112 | Principle of Dilemma/Formulation 1/Forward Implication | :$\paren {p \implies q} \land \paren {\neg p \implies q} \vdash q$ | 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:
:$p \implies q, \neg p \implies q \vdash p \lor \neg p \implies q \lor q$
From Law of Excluded Middle, we have:
:$\vdash p \lor \neg p$
From the Rule of Idempot... | :$\paren {p \implies q} \land \paren {\neg p \implies q} \vdash q$ | 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:
:$p \implies q, \neg p \implies q \vdash p \lor \neg p \implies q \lor q$
From [[Law of Excluded Middle]], we have:
:$\vdash p \lor \neg p$
From the [[Rule... | Principle of Dilemma/Formulation 1/Forward Implication/Proof 3 | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_1/Forward_Implication/Proof_3 | [
"Principle of Dilemma"
] | [] | [
"Constructive Dilemma",
"Law of Excluded Middle",
"Rule of Idempotence",
"Hypothetical Syllogism"
] |
proofwiki-7113 | Principle of Dilemma/Formulation 1/Reverse Implication | :$q \vdash \paren {p \implies q} \land \paren {\neg p \implies q}$ | {{BeginTableau|q \vdash \paren {p \implies q} \land \paren {\neg p \implies q} }}
{{Premise|1|q}}
{{SequentIntro|2|1|p \implies q|1|True Statement is implied by Every Statement}}
{{SequentIntro|3|1|\neg p \implies q|1|True Statement is implied by Every Statement}}
{{Conjunction|4|1|\paren {p \implies q} \land \paren {\... | :$q \vdash \paren {p \implies q} \land \paren {\neg p \implies q}$ | {{BeginTableau|q \vdash \paren {p \implies q} \land \paren {\neg p \implies q} }}
{{Premise|1|q}}
{{SequentIntro|2|1|p \implies q|1|[[True Statement is implied by Every Statement]]}}
{{SequentIntro|3|1|\neg p \implies q|1|[[True Statement is implied by Every Statement]]}}
{{Conjunction|4|1|\paren {p \implies q} \land \... | Principle of Dilemma/Formulation 1/Reverse Implication | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_1/Reverse_Implication | [
"Principle of Dilemma"
] | [] | [
"True Statement is implied by Every Statement",
"True Statement is implied by Every Statement",
"Category:Principle of Dilemma"
] |
proofwiki-7114 | Principle of Dilemma/Formulation 2 | :$\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \iff q$ | {{BeginTableau|\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q}}
{{Assumption|1|\paren {p \implies q} \land \paren {\neg p \implies q} }}
{{SequentIntro|2|1|q|1|Principle of Dilemma: Formulation 1}}
{{Implication|3||\paren {p \implies q} \land \paren {\neg p \implies q} \implies q|1|2}}
{{EndTa... | :$\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \iff q$ | {{BeginTableau|\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q}}
{{Assumption|1|\paren {p \implies q} \land \paren {\neg p \implies q} }}
{{SequentIntro|2|1|q|1|[[Principle of Dilemma/Formulation 1/Forward Implication|Principle of Dilemma: Formulation 1]]}}
{{Implication|3||\paren {p \implies q... | Principle of Dilemma/Formulation 2/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_2 | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_2/Forward_Implication/Proof_1 | [
"Principle of Dilemma"
] | [] | [
"Principle of Dilemma/Formulation 1/Forward Implication"
] |
proofwiki-7115 | Principle of Dilemma/Formulation 2 | :$\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \iff q$ | {{BeginTableau|\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q}}
{{Assumption|1|\paren {p \implies q} \land \paren {\neg p \implies q} }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|\neg p \implies q|1|2}}
{{Assumption|4|\neg q}}
{{ModusTollens|5|1, 4|\neg p|2|4}}
{{ModusPonens... | :$\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \iff q$ | {{BeginTableau|\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q}}
{{Assumption|1|\paren {p \implies q} \land \paren {\neg p \implies q} }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|\neg p \implies q|1|2}}
{{Assumption|4|\neg q}}
{{ModusTollens|5|1, 4|\neg p|2|4}}
{{ModusPonens... | Principle of Dilemma/Formulation 2/Forward Implication/Proof 2 | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_2 | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_2/Forward_Implication/Proof_2 | [
"Principle of Dilemma"
] | [] | [] |
proofwiki-7116 | Principle of Dilemma/Formulation 2/Forward Implication | :$\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q$ | {{BeginTableau|\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q}}
{{Assumption|1|\paren {p \implies q} \land \paren {\neg p \implies q} }}
{{SequentIntro|2|1|q|1|Principle of Dilemma: Formulation 1}}
{{Implication|3||\paren {p \implies q} \land \paren {\neg p \implies q} \implies q|1|2}}
{{EndTa... | :$\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q$ | {{BeginTableau|\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q}}
{{Assumption|1|\paren {p \implies q} \land \paren {\neg p \implies q} }}
{{SequentIntro|2|1|q|1|[[Principle of Dilemma/Formulation 1/Forward Implication|Principle of Dilemma: Formulation 1]]}}
{{Implication|3||\paren {p \implies q... | Principle of Dilemma/Formulation 2/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_2/Forward_Implication/Proof_1 | [
"Principle of Dilemma"
] | [] | [
"Principle of Dilemma/Formulation 1/Forward Implication"
] |
proofwiki-7117 | Principle of Dilemma/Formulation 2/Forward Implication | :$\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q$ | {{BeginTableau|\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q}}
{{Assumption|1|\paren {p \implies q} \land \paren {\neg p \implies q} }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|\neg p \implies q|1|2}}
{{Assumption|4|\neg q}}
{{ModusTollens|5|1, 4|\neg p|2|4}}
{{ModusPonens... | :$\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q$ | {{BeginTableau|\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q}}
{{Assumption|1|\paren {p \implies q} \land \paren {\neg p \implies q} }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|\neg p \implies q|1|2}}
{{Assumption|4|\neg q}}
{{ModusTollens|5|1, 4|\neg p|2|4}}
{{ModusPonens... | Principle of Dilemma/Formulation 2/Forward Implication/Proof 2 | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_2/Forward_Implication/Proof_2 | [
"Principle of Dilemma"
] | [] | [] |
proofwiki-7118 | Principle of Dilemma/Formulation 2/Reverse Implication | :$\vdash q \implies \paren {\paren {p \implies q} \land \paren {\neg p \implies q} }$ | {{BeginTableau|\vdash q \implies \paren {\paren {p \implies q} \land \paren {\neg p \implies q} } }}
{{Assumption|1|q|}}
{{SequentIntro|2|1|\paren {p \implies q} \land \paren {\neg p \implies q}|1|Principle of Dilemma: Formulation 1: Reverse Implication}}
{{Implication|3||q \implies \paren {\paren {p \implies q} \land ... | :$\vdash q \implies \paren {\paren {p \implies q} \land \paren {\neg p \implies q} }$ | {{BeginTableau|\vdash q \implies \paren {\paren {p \implies q} \land \paren {\neg p \implies q} } }}
{{Assumption|1|q|}}
{{SequentIntro|2|1|\paren {p \implies q} \land \paren {\neg p \implies q}|1|[[Principle of Dilemma/Formulation 1/Reverse Implication|Principle of Dilemma: Formulation 1: Reverse Implication]]}}
{{Imp... | Principle of Dilemma/Formulation 2/Reverse Implication | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_2/Reverse_Implication | https://proofwiki.org/wiki/Principle_of_Dilemma/Formulation_2/Reverse_Implication | [
"Principle of Dilemma"
] | [] | [
"Principle of Dilemma/Formulation 1/Reverse Implication",
"Category:Principle of Dilemma"
] |
proofwiki-7119 | Open Domain is Connected iff it is Path-Connected | Let $D \subseteq \C$ be a open subset of the set of complex numbers.
Then $D$ is connected {{iff}} $D$ is path-connected. | === Necessary Condition ===
Complex Plane is Metric Space shows that $\C$ is topologically equivalent to the Euclidean space $\R^2$.
{{Corollary|Continuous Image of Connected Space is Connected|1}} shows that connectedness is a topological property.
The result follows from Connected Open Subset of Euclidean Space is Pa... | Let $D \subseteq \C$ be a [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Complex Number|set of complex numbers]].
Then $D$ is [[Definition:Connected Set (Topology)|connected]] {{iff}} $D$ is [[Definition:Path-Connected|path-connected]]. | === Necessary Condition ===
[[Complex Plane is Metric Space]] shows that $\C$ is [[Definition:Topologically Equivalent Metric Spaces|topologically equivalent]] to the [[Definition:Euclidean Space|Euclidean space]] $\R^2$.
{{Corollary|Continuous Image of Connected Space is Connected|1}} shows that [[Definition:Connect... | Open Domain is Connected iff it is Path-Connected | https://proofwiki.org/wiki/Open_Domain_is_Connected_iff_it_is_Path-Connected | https://proofwiki.org/wiki/Open_Domain_is_Connected_iff_it_is_Path-Connected | [
"Complex Analysis"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Subset",
"Definition:Complex Number",
"Definition:Connected Set (Topology)",
"Definition:Path-Connected"
] | [
"Complex Plane is Metric Space",
"Definition:Homeomorphism/Metric Spaces",
"Definition:Euclidean Space",
"Definition:Connected Topological Space",
"Definition:Topological Property",
"Connected Open Subset of Euclidean Space is Path-Connected"
] |
proofwiki-7120 | Proof by Contradiction/Variant 2/Formulation 2 | :$\vdash \paren {\paren {p \implies q} \land \paren {p \implies \neg q} } \implies \neg p$ | {{BeginTableau |\vdash \paren {\paren {p \implies q} \land \paren {p \implies \neg q} } \implies \neg p}}
{{Assumption |1|\paren {p \implies q} \land \paren {p \implies \neg q} }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|p \implies \neg q|1|2}}
{{Assumption |4|p}}
{{ModusPonens |5|1, 4|q|2|... | :$\vdash \paren {\paren {p \implies q} \land \paren {p \implies \neg q} } \implies \neg p$ | {{BeginTableau |\vdash \paren {\paren {p \implies q} \land \paren {p \implies \neg q} } \implies \neg p}}
{{Assumption |1|\paren {p \implies q} \land \paren {p \implies \neg q} }}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|p \implies \neg q|1|2}}
{{Assumption |4|p}}
{{ModusPonens |5|1, 4|q|2|... | Proof by Contradiction/Variant 2/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Proof_by_Contradiction/Variant_2/Formulation_2 | https://proofwiki.org/wiki/Proof_by_Contradiction/Variant_2/Formulation_2/Proof_1 | [
"Proof by Contradiction"
] | [] | [] |
proofwiki-7121 | Proof by Contradiction/Variant 3/Formulation 2 | :$\vdash \paren {p \implies \neg p} \implies \neg p$ | {{BeginTableau|\vdash \paren {p \implies \neg p} \implies \neg p}}
{{Assumption|1|p \implies \neg p}}
{{SequentIntro|2|1|\neg p|1|Proof by Contradiction: Variant 3: Formulation 1}}
{{Implication|3||\paren {p \implies \neg p} \implies \neg p|1|2}}
{{EndTableau|qed}} | :$\vdash \paren {p \implies \neg p} \implies \neg p$ | {{BeginTableau|\vdash \paren {p \implies \neg p} \implies \neg p}}
{{Assumption|1|p \implies \neg p}}
{{SequentIntro|2|1|\neg p|1|[[Proof by Contradiction/Variant 3/Formulation 1|Proof by Contradiction: Variant 3: Formulation 1]]}}
{{Implication|3||\paren {p \implies \neg p} \implies \neg p|1|2}}
{{EndTableau|qed}} | Proof by Contradiction/Variant 3/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Proof_by_Contradiction/Variant_3/Formulation_2 | https://proofwiki.org/wiki/Proof_by_Contradiction/Variant_3/Formulation_2/Proof_1 | [
"Proof by Contradiction"
] | [] | [
"Proof by Contradiction/Variant 3/Formulation 1"
] |
proofwiki-7122 | Proof by Contradiction/Variant 3/Formulation 2 | :$\vdash \paren {p \implies \neg p} \implies \neg p$ | {{BeginTableau|\vdash \paren {p \implies \neg p} \implies \neg p}}
{{Premise|1|p \implies \neg p}}
{{Assumption|2|p}}
{{ModusPonens|3|1, 2|\neg p|1|2}}
{{NonContradiction|4|1, 2|2|3}}
{{Contradiction|5|1|\neg p|2|4}}
{{Implication|6||\paren {p \implies \neg p} \implies \neg p|1|5}}
{{EndTableau|qed}} | :$\vdash \paren {p \implies \neg p} \implies \neg p$ | {{BeginTableau|\vdash \paren {p \implies \neg p} \implies \neg p}}
{{Premise|1|p \implies \neg p}}
{{Assumption|2|p}}
{{ModusPonens|3|1, 2|\neg p|1|2}}
{{NonContradiction|4|1, 2|2|3}}
{{Contradiction|5|1|\neg p|2|4}}
{{Implication|6||\paren {p \implies \neg p} \implies \neg p|1|5}}
{{EndTableau|qed}} | Proof by Contradiction/Variant 3/Formulation 2/Proof 2 | https://proofwiki.org/wiki/Proof_by_Contradiction/Variant_3/Formulation_2 | https://proofwiki.org/wiki/Proof_by_Contradiction/Variant_3/Formulation_2/Proof_2 | [
"Proof by Contradiction"
] | [] | [] |
proofwiki-7123 | Proof by Contradiction/Variant 3/Formulation 2 | :$\vdash \paren {p \implies \neg p} \implies \neg p$ | {{BeginTableau|\vdash \paren {p \implies \neg p} \implies \neg p|Instance 2 of the Hilbert-style systems}}
{{TableauLine
| n = 1
| f = \paren {p \lor p} \implies p
| rlnk = Definition:Hilbert Proof System/Instance 2
| rtxt = Axiom $\text A 1$
}}
{{TableauLine
| n = 2
| f = \paren {\neg p \lor \neg p} \implies \ne... | :$\vdash \paren {p \implies \neg p} \implies \neg p$ | {{BeginTableau|\vdash \paren {p \implies \neg p} \implies \neg p|[[Definition:Hilbert Proof System/Instance 2|Instance 2 of the Hilbert-style systems]]}}
{{TableauLine
| n = 1
| f = \paren {p \lor p} \implies p
| rlnk = Definition:Hilbert Proof System/Instance 2
| rtxt = Axiom $\text A 1$
}}
{{TableauLine
| n = 2
... | Proof by Contradiction/Variant 3/Formulation 2/Proof 3 | https://proofwiki.org/wiki/Proof_by_Contradiction/Variant_3/Formulation_2 | https://proofwiki.org/wiki/Proof_by_Contradiction/Variant_3/Formulation_2/Proof_3 | [
"Proof by Contradiction"
] | [] | [
"Definition:Hilbert Proof System/Instance 2"
] |
proofwiki-7124 | Primitive of Function on Connected Domain | Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f$ has a primitive.}}
{{item|(2):|For any two contours $C_1, C_2$ in $D$ with identical start points $z_1 \in D$ and end points $z_2 \in D$, we have:
:$\ds \int_{C_1} \map f z \rd z {{=}} \int_{C... | === $(1)$ implies $(2)$ ===
If $F$ is a primitive of $f$, we have:
{{begin-eqn}}
{{eqn | l = \int_{C_1} \map f z \rd z
| r = \map F {z_2} - \map F {z_1}
| c = Fundamental Theorem of Calculus for Contour Integrals
}}
{{eqn | r = \int_{C_2} \map f z \rd z
}}
{{end-eqn}}
{{qed|lemma}} | Let $f: D \to \C$ be a [[Definition:Continuous Complex Function|continuous complex function]], where $D$ is a [[Definition:Connected Domain (Complex Analysis)|connected domain]].
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f$ has a [[Definition:Complex Primitive|primitive]].}}
{{item|(2):|For any two [[Definition:Contour... | === $(1)$ implies $(2)$ ===
If $F$ is a [[Definition:Complex Primitive|primitive]] of $f$, we have:
{{begin-eqn}}
{{eqn | l = \int_{C_1} \map f z \rd z
| r = \map F {z_2} - \map F {z_1}
| c = [[Fundamental Theorem of Calculus for Contour Integrals]]
}}
{{eqn | r = \int_{C_2} \map f z \rd z
}}
{{end-eqn}... | Primitive of Function on Connected Domain | https://proofwiki.org/wiki/Primitive_of_Function_on_Connected_Domain | https://proofwiki.org/wiki/Primitive_of_Function_on_Connected_Domain | [
"Complex Analysis"
] | [
"Definition:Continuous Complex Function",
"Definition:Connected Domain (Complex Analysis)",
"Definition:Primitive (Calculus)/Complex",
"Definition:Contour/Complex Plane",
"Definition:Contour/Endpoints/Complex Plane",
"Definition:Contour/Endpoints/Complex Plane",
"Definition:Contour/Closed/Complex Plane"... | [
"Definition:Primitive (Calculus)/Complex",
"Fundamental Theorem of Calculus for Contour Integrals",
"Definition:Primitive (Calculus)/Complex"
] |
proofwiki-7125 | Upper Closure is Strict Upper Closure of Immediate Predecessor | Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $b$ be the immediate successor element of $a$:
Then:
:$a^\succ = b^\succcurlyeq$
where:
:$a^\succ$ is the strict upper closure of $a$
:$b^\succcurlyeq$ is the upper closure of $b$. | Let:
:$x \in a^\succ$
By the definition of strict upper closure:
:$a \prec x$
By the definition of total ordering:
:$x \prec b$ or $x \succcurlyeq b$
If $x \prec b$ then $a \prec x \prec b$, contradicting the premise.
Thus:
:$x \succcurlyeq b$
and so:
:$x \in b^\succcurlyeq$
By definition of subset:
:$a^\succ \subseteq... | Let $\struct {S, \preccurlyeq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $b$ be the [[Definition:Immediate Successor Element|immediate successor element]] of $a$:
Then:
:$a^\succ = b^\succcurlyeq$
where:
:$a^\succ$ is the [[Definition:Strict Upper Closure of Element|strict upper closure]] of... | Let:
:$x \in a^\succ$
By the definition of [[Definition:Strict Upper Closure of Element|strict upper closure]]:
:$a \prec x$
By the definition of [[Definition:Total Ordering|total ordering]]:
:$x \prec b$ or $x \succcurlyeq b$
If $x \prec b$ then $a \prec x \prec b$, [[Definition:Contradiction|contradicting]] the [[... | Upper Closure is Strict Upper Closure of Immediate Predecessor | https://proofwiki.org/wiki/Upper_Closure_is_Strict_Upper_Closure_of_Immediate_Predecessor | https://proofwiki.org/wiki/Upper_Closure_is_Strict_Upper_Closure_of_Immediate_Predecessor | [
"Total Orderings",
"Upper Closures"
] | [
"Definition:Totally Ordered Set",
"Definition:Immediate Successor Element",
"Definition:Strict Upper Closure/Element",
"Definition:Upper Closure/Element"
] | [
"Definition:Strict Upper Closure/Element",
"Definition:Total Ordering",
"Definition:Contradiction",
"Definition:Premise",
"Definition:Subset",
"Definition:Upper Closure/Element",
"Extended Transitivity",
"Definition:Subset",
"Definition:Set Equality",
"Category:Total Orderings",
"Category:Upper ... |
proofwiki-7126 | Proof by Contradiction/Sequent Form | :$\paren {p \vdash \bot} \vdash \neg p$ | {{BeginTableau|\paren {p \vdash \bot} \vdash \neg p}}
{{Premise |1|p \vdash \bot}}
{{Assumption |2|p}}
{{Contradiction|3|1|\neg p|2|2}}
{{EndTableau|qed}}
Category:Proof by Contradiction
4ukf0bn4xlan505yq1y2q3x5nvh59xg | :$\paren {p \vdash \bot} \vdash \neg p$ | {{BeginTableau|\paren {p \vdash \bot} \vdash \neg p}}
{{Premise |1|p \vdash \bot}}
{{Assumption |2|p}}
{{Contradiction|3|1|\neg p|2|2}}
{{EndTableau|qed}}
[[Category:Proof by Contradiction]]
4ukf0bn4xlan505yq1y2q3x5nvh59xg | Proof by Contradiction/Sequent Form | https://proofwiki.org/wiki/Proof_by_Contradiction/Sequent_Form | https://proofwiki.org/wiki/Proof_by_Contradiction/Sequent_Form | [
"Proof by Contradiction"
] | [] | [
"Category:Proof by Contradiction"
] |
proofwiki-7127 | Left or Right Inverse of Matrix is Inverse | Let $\mathbf A, \mathbf B$ be square matrices of order $n$ over a commutative ring with unity $\struct {R, +, \circ}$.
Suppose that:
:$\mathbf A \mathbf B = \mathbf I_n$
where $\mathbf I_n$ is the unit matrix of order $n$.
Then $\mathbf A$ and $\mathbf B$ are nonsingular matrices, and furthermore:
:$\mathbf B = \mathbf... | Let $1_R$ denote the unity of $R$.
We have:
{{begin-eqn}}
{{eqn | l = 1_R
| r = \map \det {\mathbf I_n}
| c = Determinant of Unit Matrix
}}
{{eqn | r = \map \det {\mathbf A \mathbf B}
| c = {{hypothesis}}
}}
{{eqn | r = \map \det {\mathbf A} \map \det {\mathbf B}
| c = Determinant of Matrix Prod... | Let $\mathbf A, \mathbf B$ be [[Definition:Square Matrix|square matrices of order $n$]] over a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] $\struct {R, +, \circ}$.
Suppose that:
:$\mathbf A \mathbf B = \mathbf I_n$
where $\mathbf I_n$ is the [[Definition:Unit Matrix|unit matrix of order $n$... | Let $1_R$ denote the [[Definition:Unity of Ring|unity]] of $R$.
We have:
{{begin-eqn}}
{{eqn | l = 1_R
| r = \map \det {\mathbf I_n}
| c = [[Determinant of Unit Matrix]]
}}
{{eqn | r = \map \det {\mathbf A \mathbf B}
| c = {{hypothesis}}
}}
{{eqn | r = \map \det {\mathbf A} \map \det {\mathbf B}
... | Left or Right Inverse of Matrix is Inverse | https://proofwiki.org/wiki/Left_or_Right_Inverse_of_Matrix_is_Inverse | https://proofwiki.org/wiki/Left_or_Right_Inverse_of_Matrix_is_Inverse | [
"Inverse Matrices"
] | [
"Definition:Matrix/Square Matrix",
"Definition:Commutative and Unitary Ring",
"Definition:Unit Matrix",
"Definition:Nonsingular Matrix",
"Definition:Inverse Matrix"
] | [
"Definition:Unity (Abstract Algebra)/Ring",
"Determinant of Unit Matrix",
"Determinant of Matrix Product",
"Matrix is Nonsingular iff Determinant has Multiplicative Inverse",
"Definition:Nonsingular Matrix",
"Unit Matrix is Unity of Ring of Square Matrices",
"Matrix Multiplication is Associative",
"Un... |
proofwiki-7128 | Banach-Alaoglu Theorem | Let $X$ be a separable normed vector space.
Then the closed unit ball in its normed dual $X^*$ is sequentially compact with respect to the weak-$\ast$ topology. | The aim of this proof is to show the following:
Given a bounded sequence in $X^*$, there exists a weakly convergent subsequence of that bounded sequence.
Let $\sequence {l_n}_{n \mathop \in \N}$ be a bounded sequence in $X^*$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a countable dense subset of $X$.
Choose subsequen... | Let $X$ be a [[Definition:Separable Space|separable]] [[Definition:Normed Vector Space|normed vector space]].
Then the [[Definition:Closed Unit Ball|closed unit ball]] in its [[Definition:Normed Dual Space|normed dual]] $X^*$ is [[Definition:Sequentially Compact In Itself|sequentially compact]] with respect to the [[D... | The aim of this proof is to show the following:
Given a [[Definition:Bounded Sequence in Normed Vector Space|bounded sequence]] in $X^*$, there exists a [[Definition:Weak Convergence|weakly convergent]] [[Definition:Subsequence|subsequence]] of that [[Definition:Bounded Sequence in Normed Vector Space|bounded sequence... | Banach-Alaoglu Theorem/Proof 1 | https://proofwiki.org/wiki/Banach-Alaoglu_Theorem | https://proofwiki.org/wiki/Banach-Alaoglu_Theorem/Proof_1 | [
"Functional Analysis",
"Banach-Alaoglu Theorem",
"Weak-* Topologies"
] | [
"Definition:Separable Space",
"Definition:Normed Vector Space",
"Definition:Closed Unit Ball",
"Definition:Normed Dual Space",
"Definition:Sequentially Compact Space/In Itself",
"Definition:Weak-* Topology"
] | [
"Definition:Bounded Sequence/Normed Vector Space",
"Definition:Weak Convergence",
"Definition:Subsequence",
"Definition:Bounded Sequence/Normed Vector Space",
"Definition:Bounded Sequence/Normed Vector Space",
"Definition:Countable Set",
"Definition:Everywhere Dense",
"Definition:Subsequence",
"Defi... |
proofwiki-7129 | Banach-Alaoglu Theorem | Let $X$ be a separable normed vector space.
Then the closed unit ball in its normed dual $X^*$ is sequentially compact with respect to the weak-$\ast$ topology. | Let $X$ be a normed vector space.
Denote by $B$ the closed unit ball in $X$.
Let $X^*$ be the dual of $X$.
Denote by $B^*$ the closed unit ball in $X^*$.
Let:
:$\map \FF B = \closedint {-1} 1^B$
be the topological space of functions from $B$ to $\closedint {-1} 1$.
By Tychonoff's Theorem:
:$\map \FF B$
is compact wit... | Let $X$ be a [[Definition:Separable Space|separable]] [[Definition:Normed Vector Space|normed vector space]].
Then the [[Definition:Closed Unit Ball|closed unit ball]] in its [[Definition:Normed Dual Space|normed dual]] $X^*$ is [[Definition:Sequentially Compact In Itself|sequentially compact]] with respect to the [[D... | Let $X$ be a [[Definition:Normed Vector Space|normed vector space]].
Denote by $B$ the [[Definition:Closed Unit Ball|closed unit ball]] in $X$.
Let $X^*$ be the [[Definition:Normed Dual Space|dual]] of $X$.
Denote by $B^*$ the [[Definition:Closed Unit Ball|closed unit ball]] in $X^*$.
Let:
:$\map \FF B = \closedint... | Banach-Alaoglu Theorem/Proof 2 | https://proofwiki.org/wiki/Banach-Alaoglu_Theorem | https://proofwiki.org/wiki/Banach-Alaoglu_Theorem/Proof_2 | [
"Functional Analysis",
"Banach-Alaoglu Theorem",
"Weak-* Topologies"
] | [
"Definition:Separable Space",
"Definition:Normed Vector Space",
"Definition:Closed Unit Ball",
"Definition:Normed Dual Space",
"Definition:Sequentially Compact Space/In Itself",
"Definition:Weak-* Topology"
] | [
"Definition:Normed Vector Space",
"Definition:Closed Unit Ball",
"Definition:Normed Dual Space",
"Definition:Closed Unit Ball",
"Tychonoff's Theorem",
"Definition:Product Topology",
"Definition:Restriction/Mapping",
"Banach-Alaoglu Theorem/Lemma 3",
"Banach-Alaoglu Theorem/Lemma 4",
"Banach-Alaogl... |
proofwiki-7130 | Banach-Alaoglu Theorem | Let $X$ be a separable normed vector space.
Then the closed unit ball in its normed dual $X^*$ is sequentially compact with respect to the weak-$\ast$ topology. | Let $B_{X^\ast}$ be the closed unit ball in $X^\ast$.
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
From the definition of the norm of a bounded linear functional, we have:
:$B_{X^\ast} = \set {f : X \to \GF : \cmod {\map f x} \le \norm x \text { and } f \text { is linear} }$
For each $x \in X$, let:
:$K_x ... | Let $X$ be a [[Definition:Separable Space|separable]] [[Definition:Normed Vector Space|normed vector space]].
Then the [[Definition:Closed Unit Ball|closed unit ball]] in its [[Definition:Normed Dual Space|normed dual]] $X^*$ is [[Definition:Sequentially Compact In Itself|sequentially compact]] with respect to the [[D... | Let $B_{X^\ast}$ be the [[Definition:Closed Unit Ball|closed unit ball]] in $X^\ast$.
Let $w^\ast$ be the [[Definition:Weak-* Topology|weak-$\ast$ topology]] on $X^\ast$.
From the definition of the [[Definition:Norm on Bounded Linear Functional|norm of a bounded linear functional]], we have:
:$B_{X^\ast} = \set {... | Banach-Alaoglu Theorem/Proof 3 | https://proofwiki.org/wiki/Banach-Alaoglu_Theorem | https://proofwiki.org/wiki/Banach-Alaoglu_Theorem/Proof_3 | [
"Functional Analysis",
"Banach-Alaoglu Theorem",
"Weak-* Topologies"
] | [
"Definition:Separable Space",
"Definition:Normed Vector Space",
"Definition:Closed Unit Ball",
"Definition:Normed Dual Space",
"Definition:Sequentially Compact Space/In Itself",
"Definition:Weak-* Topology"
] | [
"Definition:Closed Unit Ball",
"Definition:Weak-* Topology",
"Definition:Norm/Bounded Linear Functional",
"Definition:Product Topology",
"Definition:Projection (Mapping Theory)/Family of Sets",
"Definition:Product Topology",
"Definition:Product Topology",
"Definition:Initial Topology",
"Subspace Top... |
proofwiki-7131 | Empty Product is Terminal Object | Let $\mathbf C$ be a metacategory.
Suppose $\mathbf C$ admits a product $\prod \O$ for the empty set.
Then $\prod \O$ is a terminal object of $\mathbf C$. | By definition, $\prod \O$ is the limit of the empty subcategory $\mathbf 0$ of $\mathbf C$.
The result follows from Terminal Object as Limit.
{{qed}}
Category:Limits and Colimits
Category:Products (Category Theory)
2nfoe9f4wxtj0u65bqnab4a3qe0i0aa | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Suppose $\mathbf C$ admits a [[Definition:Product (Category Theory)|product]] $\prod \O$ for the [[Definition:Empty Set|empty set]].
Then $\prod \O$ is a [[Definition:Terminal Object|terminal object]] of $\mathbf C$. | By definition, $\prod \O$ is the [[Definition:Limit (Category Theory)|limit]] of the empty [[Definition:Subcategory|subcategory]] $\mathbf 0$ of $\mathbf C$.
The result follows from [[Terminal Object as Limit]].
{{qed}}
[[Category:Limits and Colimits]]
[[Category:Products (Category Theory)]]
2nfoe9f4wxtj0u65bqnab4a3q... | Empty Product is Terminal Object | https://proofwiki.org/wiki/Empty_Product_is_Terminal_Object | https://proofwiki.org/wiki/Empty_Product_is_Terminal_Object | [
"Limits and Colimits",
"Products (Category Theory)"
] | [
"Definition:Metacategory",
"Definition:Product (Category Theory)",
"Definition:Empty Set",
"Definition:Terminal Object"
] | [
"Definition:Limit (Category Theory)",
"Definition:Subcategory",
"Terminal Object as Limit",
"Category:Limits and Colimits",
"Category:Products (Category Theory)"
] |
proofwiki-7132 | Unary Product for Object is Itself | Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Then $\ds \prod \set C = C$, where $\ds \prod$ denotes product. | Follows directly from Limit of Singleton.
{{qed}}
Category:Objects (Category Theory)
Category:Category Theory
0aw7ydwpy1bhtsooxhhie3203zwv2ok | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $C$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf C$.
Then $\ds \prod \set C = C$, where $\ds \prod$ denotes [[Definition:Product (Category Theory)/General Definition|product]]. | Follows directly from [[Limit of Singleton]].
{{qed}}
[[Category:Objects (Category Theory)]]
[[Category:Category Theory]]
0aw7ydwpy1bhtsooxhhie3203zwv2ok | Unary Product for Object is Itself | https://proofwiki.org/wiki/Unary_Product_for_Object_is_Itself | https://proofwiki.org/wiki/Unary_Product_for_Object_is_Itself | [
"Objects (Category Theory)",
"Category Theory"
] | [
"Definition:Metacategory",
"Definition:Object (Category Theory)",
"Definition:Product (Category Theory)/General Definition"
] | [
"Limit of Singleton",
"Category:Objects (Category Theory)",
"Category:Category Theory"
] |
proofwiki-7133 | Principle of Non-Contradiction/Sequent Form/Formulation 1 | :$p, \neg p \vdash \bot$ | {{BeginTableau|p, \neg p \vdash \bot}}
{{Premise|1|p}}
{{Premise|2|\neg p}}
{{NonContradiction|3|1, 2|1|2}}
{{EndTableau}}
{{Qed}} | :$p, \neg p \vdash \bot$ | {{BeginTableau|p, \neg p \vdash \bot}}
{{Premise|1|p}}
{{Premise|2|\neg p}}
{{NonContradiction|3|1, 2|1|2}}
{{EndTableau}}
{{Qed}} | Principle of Non-Contradiction/Sequent Form/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Principle_of_Non-Contradiction/Sequent_Form/Formulation_1 | https://proofwiki.org/wiki/Principle_of_Non-Contradiction/Sequent_Form/Formulation_1/Proof_1 | [
"Principle of Non-Contradiction"
] | [] | [] |
proofwiki-7134 | Principle of Non-Contradiction/Sequent Form/Formulation 1 | :$p, \neg p \vdash \bot$ | We apply the Method of Truth Tables.
:<nowiki>$\begin {array} {|cccc||c|} \hline
p & \land & \neg & p & \bot \\
\hline
\F & \F & \T & \F & \F \\
\T & \F & \F & \T & \F \\
\hline
\end {array}$</nowiki>
As can be seen by inspection, the truth value of the main connective, that is $\land$, is $F$ for each boolean interpre... | :$p, \neg p \vdash \bot$ | We apply the [[Method of Truth Tables]].
:<nowiki>$\begin {array} {|cccc||c|} \hline
p & \land & \neg & p & \bot \\
\hline
\F & \F & \T & \F & \F \\
\T & \F & \F & \T & \F \\
\hline
\end {array}$</nowiki>
As can be seen by inspection, the [[Definition:Truth Value|truth value]] of the [[Definition:Main Connective (Pro... | Principle of Non-Contradiction/Sequent Form/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Principle_of_Non-Contradiction/Sequent_Form/Formulation_1 | https://proofwiki.org/wiki/Principle_of_Non-Contradiction/Sequent_Form/Formulation_1/Proof_by_Truth_Table | [
"Principle of Non-Contradiction"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7135 | Vector Cross Product is Orthogonal to Factors | Let $\mathbf a$ and $\mathbf b$ be vectors in the real Euclidean space $\R^3$.
Let $\mathbf a \times \mathbf b$ denote the vector cross product.
Then:
{{begin-itemize}}
{{item|(1):|$\mathbf a$ and $\mathbf a \times \mathbf b$ are orthogonal.}}
{{item|(2):|$\mathbf b$ and $\mathbf a \times \mathbf b$ are orthogonal.}}
{... | Let $\mathbf a = \begin {bmatrix} a_1 \\ a_2 \\ a_3 \end {bmatrix}$, and $\mathbf b = \begin {bmatrix} b_1 \\ b_2 \\ b_3 \end {bmatrix}$.
Then the dot product of $\mathbf a$ and $\mathbf a \times \mathbf b$ is:
{{begin-eqn}}
{{eqn | l = \mathbf a \cdot \paren {\mathbf a \times \mathbf b}
| r = a_1 \paren {a_2 b_3... | Let $\mathbf a$ and $\mathbf b$ be [[Definition:Space Vector|vectors]] in the [[Definition:Real Euclidean Space|real Euclidean space]] $\R^3$.
Let $\mathbf a \times \mathbf b$ denote the [[Definition:Vector Cross Product|vector cross product]].
Then:
{{begin-itemize}}
{{item|(1):|$\mathbf a$ and $\mathbf a \times \... | Let $\mathbf a = \begin {bmatrix} a_1 \\ a_2 \\ a_3 \end {bmatrix}$, and $\mathbf b = \begin {bmatrix} b_1 \\ b_2 \\ b_3 \end {bmatrix}$.
Then the [[Definition:Dot Product|dot product]] of $\mathbf a$ and $\mathbf a \times \mathbf b$ is:
{{begin-eqn}}
{{eqn | l = \mathbf a \cdot \paren {\mathbf a \times \mathbf b}
... | Vector Cross Product is Orthogonal to Factors | https://proofwiki.org/wiki/Vector_Cross_Product_is_Orthogonal_to_Factors | https://proofwiki.org/wiki/Vector_Cross_Product_is_Orthogonal_to_Factors | [
"Vector Cross Product"
] | [
"Definition:Vector/Real Euclidean Space/Space Vector",
"Definition:Euclidean Space/Real",
"Definition:Vector Cross Product",
"Definition:Orthogonal (Linear Algebra)",
"Definition:Orthogonal (Linear Algebra)"
] | [
"Definition:Dot Product",
"Definition:Dot Product",
"Definition:Vector/Real Euclidean Space/Space Vector",
"Definition:Orthogonal (Linear Algebra)/Real Vector Space",
"Definition:Orthogonal (Linear Algebra)/Real Vector Space"
] |
proofwiki-7136 | Squeeze Theorem/Sequences/Metric Spaces | Let $M = \struct {S, d}$ be a metric space or pseudometric space.
Let $p \in S$.
Let $\sequence {r_n}$ be a null sequence in $\R$.
Let $\sequence {x_n}$ be a sequence in $S$ such that:
:$\forall n \in \N: \map d {p, x_n} \le r_n$.
Then $\sequence {x_n}$ converges to $p$. | {{begin-eqn}}
{{eqn | q = \forall n \in \N
| l = \map d {p, x_n}
| o = \le
| r = r_n
| c = {{hypothesis}}
}}
{{eqn | q = \forall n \in \N
| l = r_n
| o = \le
| r = \size {r_n}
| c = Negative of Absolute Value
}}
{{eqn | q = \forall \epsilon \in \R_{>0}: \exists N \in \R_{... | Let $M = \struct {S, d}$ be a [[Definition:Metric Space|metric space]] or [[Definition:Pseudometric Space|pseudometric space]].
Let $p \in S$.
Let $\sequence {r_n}$ be a [[Definition:Null Sequence/Real Numbers|null sequence in $\R$]].
Let $\sequence {x_n}$ be a [[Definition:Infinite Sequence|sequence]] in $S$ such t... | {{begin-eqn}}
{{eqn | q = \forall n \in \N
| l = \map d {p, x_n}
| o = \le
| r = r_n
| c = {{hypothesis}}
}}
{{eqn | q = \forall n \in \N
| l = r_n
| o = \le
| r = \size {r_n}
| c = [[Negative of Absolute Value]]
}}
{{eqn | q = \forall \epsilon \in \R_{>0}: \exists N \in ... | Squeeze Theorem/Sequences/Metric Spaces | https://proofwiki.org/wiki/Squeeze_Theorem/Sequences/Metric_Spaces | https://proofwiki.org/wiki/Squeeze_Theorem/Sequences/Metric_Spaces | [
"Metric Spaces",
"Limits of Sequences"
] | [
"Definition:Metric Space",
"Definition:Pseudometric/Pseudometric Space",
"Definition:Null Sequence/Real Numbers",
"Definition:Sequence/Infinite Sequence",
"Definition:Limit of Sequence/Metric Space"
] | [
"Negative of Absolute Value",
"Extended Transitivity",
"Definition:Limit of Sequence/Metric Space",
"Category:Metric Spaces",
"Category:Limits of Sequences"
] |
proofwiki-7137 | Squeeze Theorem/Sequences/Linearly Ordered Space | Let $\struct {S, \le, \tau}$ be a linearly ordered space.
Let $\sequence {x_n}$, $\sequence {y_n}$, and $\sequence {z_n}$ be sequences in $S$.
Let $p \in S$.
Let $\sequence {x_n}$ and $\sequence {z_n}$ both converge to $p$.
Let $\forall n \in \N: x_n \le y_n \le z_n$.
Then $\sequence {y_n}$ converges to $p$. | Let $m \in S$ and $m < p$.
Then $\sequence {x_n}$ eventually succeeds $m$.
Thus by Extended Transitivity, $\sequence {y_n}$ eventually succeeds $m$.
A similar argument using $\sequence {z_n}$ proves the dual statement.
Thus $\sequence {y_n}$ is eventually in each ray containing $p$, so it converges to $p$.
{{qed}}
Cate... | Let $\struct {S, \le, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Let $\sequence {x_n}$, $\sequence {y_n}$, and $\sequence {z_n}$ be [[Definition:Sequence|sequences]] in $S$.
Let $p \in S$.
Let $\sequence {x_n}$ and $\sequence {z_n}$ both converge to $p$.
Let $\forall n \in \N: x_n \le... | Let $m \in S$ and $m < p$.
Then $\sequence {x_n}$ eventually succeeds $m$.
Thus by [[Extended Transitivity]], $\sequence {y_n}$ eventually succeeds $m$.
A similar argument using $\sequence {z_n}$ proves the dual statement.
Thus $\sequence {y_n}$ is eventually in each [[Definition:Ray (Order Theory)|ray]] containing... | Squeeze Theorem/Sequences/Linearly Ordered Space | https://proofwiki.org/wiki/Squeeze_Theorem/Sequences/Linearly_Ordered_Space | https://proofwiki.org/wiki/Squeeze_Theorem/Sequences/Linearly_Ordered_Space | [
"Linearly Ordered Spaces",
"Limits of Sequences"
] | [
"Definition:Linearly Ordered Space",
"Definition:Sequence"
] | [
"Extended Transitivity",
"Definition:Ray (Order Theory)",
"Category:Linearly Ordered Spaces",
"Category:Limits of Sequences"
] |
proofwiki-7138 | Squeeze Theorem for Filter Bases | Let $\struct {S, \le, \tau}$ be a linearly ordered space.
Let $F_1$, $F_2$, and $F_3$ be filter bases in $S$.
Let:
:$\forall T \in F_1: \exists M \in F_2: \forall x \in M: \exists y \in T: y \le x$
That is:
:for each $T \in F_1$, $F_2$ has an element $M$ such that all elements of $M$ succeed some element of $T$.
Simila... | Let $q \in S$ such that $q < p$.
We will show that $F_2$ has an element which is a subset of $q^\ge$.
{{explain|$q^\ge$ is the upper closure in what set?}}
Since $F_1$ converges to $p$, it has an element:
:$A \subseteq q^\ge$.
Thus there is an element $k$ in $A$ and an element $M$ in $F_2$ such that all elements of $M$... | Let $\struct {S, \le, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Let $F_1$, $F_2$, and $F_3$ be [[Definition:Filter Basis|filter bases]] in $S$.
Let:
:$\forall T \in F_1: \exists M \in F_2: \forall x \in M: \exists y \in T: y \le x$
That is:
:for each $T \in F_1$, $F_2$ has an element ... | Let $q \in S$ such that $q < p$.
We will show that $F_2$ has an element which is a subset of $q^\ge$.
{{explain|$q^\ge$ is the [[Definition:Upper Closure of Element|upper closure]] in what set?}}
Since $F_1$ converges to $p$, it has an element:
:$A \subseteq q^\ge$.
Thus there is an element $k$ in $A$ and an elemen... | Squeeze Theorem for Filter Bases | https://proofwiki.org/wiki/Squeeze_Theorem_for_Filter_Bases | https://proofwiki.org/wiki/Squeeze_Theorem_for_Filter_Bases | [
"Filter Theory",
"Linearly Ordered Spaces",
"Convergence"
] | [
"Definition:Linearly Ordered Space",
"Definition:Filter Basis"
] | [
"Definition:Upper Closure/Element",
"Extended Transitivity",
"Category:Filter Theory",
"Category:Linearly Ordered Spaces",
"Category:Convergence"
] |
proofwiki-7139 | Non-Zero Vectors are Orthogonal iff Perpendicular | Let $\mathbf u$, $\mathbf v$ be non-zero vectors in the real Euclidean space $\R^n$.
Then $\mathbf u$ and $\mathbf v$ are orthogonal {{iff}} they are perpendicular. | === Necessary Condition ===
When $\theta$ denotes the angle between $\mathbf u$ and $\mathbf v$ measured in radians, we have:
{{begin-eqn}}
{{eqn | l = 0
| r = \mathbf u \cdot \mathbf v
| c = {{Defof|Orthogonal Vectors}}
}}
{{eqn | r = \norm {\mathbf u} \norm {\mathbf v} \cos \theta
| c = Cosine Formu... | Let $\mathbf u$, $\mathbf v$ be non-[[Definition:Zero Vector|zero]] [[Definition:Vector (Real Euclidean Space)|vectors]] in the [[Definition:Real Euclidean Space|real Euclidean space]] $\R^n$.
Then $\mathbf u$ and $\mathbf v$ are [[Definition:Orthogonal (Linear Algebra)|orthogonal]] {{iff}} they are [[Definition:Perp... | === Necessary Condition ===
When $\theta$ denotes the [[Definition:Angle Between Vectors|angle]] between $\mathbf u$ and $\mathbf v$ measured in [[Definition:Radian|radians]], we have:
{{begin-eqn}}
{{eqn | l = 0
| r = \mathbf u \cdot \mathbf v
| c = {{Defof|Orthogonal Vectors}}
}}
{{eqn | r = \norm {\mat... | Non-Zero Vectors are Orthogonal iff Perpendicular | https://proofwiki.org/wiki/Non-Zero_Vectors_are_Orthogonal_iff_Perpendicular | https://proofwiki.org/wiki/Non-Zero_Vectors_are_Orthogonal_iff_Perpendicular | [
"Linear Algebra",
"Analytic Geometry"
] | [
"Definition:Zero Vector",
"Definition:Vector/Real Euclidean Space",
"Definition:Euclidean Space/Real",
"Definition:Orthogonal (Linear Algebra)",
"Definition:Perpendicular (Linear Algebra)"
] | [
"Definition:Angle between Vectors",
"Definition:Angular Measure/Radian",
"Cosine Formula for Dot Product",
"Definition:Euclidean Norm",
"Definition:Zero Vector",
"Definition:Norm/Vector Space",
"Zeroes of Sine and Cosine",
"Definition:Angle between Vectors",
"Definition:Perpendicular (Linear Algebra... |
proofwiki-7140 | Exponential Function is Continuous/Complex | :$\forall z_0 \in \C: \ds \lim_{z \mathop \to z_0} \exp z = \exp z_0$ | This proof depends on the differential equation definition of the exponential function.
The result follows from Complex-Differentiable Function is Continuous.
{{qed}}
Category:Exponential Function is Continuous
nxshz8om78e0hbqwlg95tfzbpc71tr7 | :$\forall z_0 \in \C: \ds \lim_{z \mathop \to z_0} \exp z = \exp z_0$ | This proof depends on the [[Definition:Exponential Function/Complex/Differential Equation|differential equation definition of the exponential function]].
The result follows from [[Complex-Differentiable Function is Continuous]].
{{qed}}
[[Category:Exponential Function is Continuous]]
nxshz8om78e0hbqwlg95tfzbpc71tr7 | Exponential Function is Continuous/Complex | https://proofwiki.org/wiki/Exponential_Function_is_Continuous/Complex | https://proofwiki.org/wiki/Exponential_Function_is_Continuous/Complex | [
"Exponential Function is Continuous"
] | [] | [
"Definition:Exponential Function/Complex/Differential Equation",
"Complex-Differentiable Function is Continuous",
"Category:Exponential Function is Continuous"
] |
proofwiki-7141 | Relation Compatible with Group Operation is Reflexive or Antireflexive | Let $\struct {G, \circ}$ be a group.
Let $\RR$ be a relation on $G$ that is compatible with $\circ$.
Then $\RR$ is reflexive or antireflexive. | Suppose that $\RR$ is not antireflexive.
Then there is some $x \in G$ such that $x \mathrel \RR x$.
Let $y \in G$.
Then by the definition of compatibility:
:$\paren {x \circ \paren {x^{-1} \circ y} } \mathrel \RR \paren {x \circ \paren {x^{-1} \circ y} }$
By {{Group-axiom|1}} and {{Group-axiom|3}}:
:$y \mathrel \RR y$
... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\RR$ be a [[Definition:Endorelation|relation]] on $G$ that is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Then $\RR$ is [[Definition:Reflexive Relation|reflexive]] or [[Definition:Antireflexive Relation|antireflexive]]. | Suppose that $\RR$ is not [[Definition:Antireflexive Relation|antireflexive]].
Then there is some $x \in G$ such that $x \mathrel \RR x$.
Let $y \in G$.
Then by the definition of [[Definition:Relation Compatible with Operation|compatibility]]:
:$\paren {x \circ \paren {x^{-1} \circ y} } \mathrel \RR \paren {x \circ ... | Relation Compatible with Group Operation is Reflexive or Antireflexive | https://proofwiki.org/wiki/Relation_Compatible_with_Group_Operation_is_Reflexive_or_Antireflexive | https://proofwiki.org/wiki/Relation_Compatible_with_Group_Operation_is_Reflexive_or_Antireflexive | [
"Relations Compatible with Group Operation",
"Reflexive Relations",
"Antireflexive Relations"
] | [
"Definition:Group",
"Definition:Endorelation",
"Definition:Relation Compatible with Operation",
"Definition:Reflexive Relation",
"Definition:Antireflexive Relation"
] | [
"Definition:Antireflexive Relation",
"Definition:Relation Compatible with Operation",
"Definition:Reflexive Relation",
"Definition:Reflexive Relation",
"Definition:Antireflexive Relation",
"Category:Relations Compatible with Group Operation",
"Category:Reflexive Relations",
"Category:Antireflexive Rel... |
proofwiki-7142 | Linearly Ordered Space on Order-Convex Subset is Subspace Topology | Let $\struct {S, \preceq,\tau}$ be a linearly ordered space.
Let $A \subseteq S$ be a order-convex set in $S$.
Let $\upsilon$ be the order topology on $A$.
Let $\tau'$ be the $\tau$-relative subspace topology on $A$.
Then $\upsilon = \tau'$. | {{improve|Reword so as to be less wordy. Add pictures so as to be easier to visualize.}}
By the definition of the order topology, the sets of open rays in $\struct {S, \preceq}$ and $\struct {A, \preceq}$ form sub-bases for $\tau$ and $\upsilon$, respectively.
By Sub-Basis for Topological Subspace, we need only show th... | Let $\struct {S, \preceq,\tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Let $A \subseteq S$ be a [[Definition:Order-Convex Set|order-convex set]] in $S$.
Let $\upsilon$ be the [[Definition:Order Topology|order topology]] on $A$.
Let $\tau'$ be the $\tau$-relative [[Definition:Subspace Topo... | {{improve|Reword so as to be less wordy. Add pictures so as to be easier to visualize.}}
By the definition of the [[Definition:Order Topology|order topology]], the sets of [[Definition:Open Ray|open rays]] in $\struct {S, \preceq}$ and $\struct {A, \preceq}$ form [[Definition:Sub-Basis|sub-bases]] for $\tau$ and $\ups... | Linearly Ordered Space on Order-Convex Subset is Subspace Topology | https://proofwiki.org/wiki/Linearly_Ordered_Space_on_Order-Convex_Subset_is_Subspace_Topology | https://proofwiki.org/wiki/Linearly_Ordered_Space_on_Order-Convex_Subset_is_Subspace_Topology | [
"Linearly Ordered Spaces",
"Topological Subspaces"
] | [
"Definition:Linearly Ordered Space",
"Definition:Order-Convex Set",
"Definition:Order Topology",
"Definition:Topological Subspace"
] | [
"Definition:Order Topology",
"Definition:Ray (Order Theory)/Open",
"Definition:Sub-Basis",
"Sub-Basis for Topological Subspace",
"Definition:Ray (Order Theory)/Open",
"Definition:Ray (Order Theory)/Open",
"Definition:Ray (Order Theory)/Open",
"Definition:Ray (Order Theory)/Open",
"Definition:Ray (Or... |
proofwiki-7143 | Exponential Function is Continuous/Real Numbers | The real exponential function is continuous.
That is:
:$\forall x_0 \in \R: \ds \lim_{x \mathop \to x_0} \exp x = \exp x_0$ | This proof depends on the limit definition of the exponential function.
Let:
:$\ds \exp x = \lim_{n \mathop \to \infty} \paren {1 + \dfrac x n}^n$
Fix $x_0 \in \R$.
Consider $I := \closedint {x_0 - 1} {x_0 + 1}$.
From Closed Bounded Subset of Real Numbers is Compact, $I$ is compact.
From Exponential Sequence is Uniform... | The [[Definition:Real Exponential Function|real exponential function]] is [[Definition:Continuous Function|continuous]].
That is:
:$\forall x_0 \in \R: \ds \lim_{x \mathop \to x_0} \exp x = \exp x_0$ | This proof depends on the [[Definition:Exponential Function/Real/Limit of Sequence|limit definition of the exponential function]].
Let:
:$\ds \exp x = \lim_{n \mathop \to \infty} \paren {1 + \dfrac x n}^n$
Fix $x_0 \in \R$.
Consider $I := \closedint {x_0 - 1} {x_0 + 1}$.
From [[Closed Bounded Subset of Real Numbers... | Exponential Function is Continuous/Real Numbers/Proof 1 | https://proofwiki.org/wiki/Exponential_Function_is_Continuous/Real_Numbers | https://proofwiki.org/wiki/Exponential_Function_is_Continuous/Real_Numbers/Proof_1 | [
"Exponential Function is Continuous",
"Continuous Real Functions"
] | [
"Definition:Exponential Function/Real",
"Definition:Continuous Function"
] | [
"Definition:Exponential Function/Real/Limit of Sequence",
"Closed Bounded Subset of Real Numbers is Compact",
"Definition:Compact Space/Real Analysis",
"Exponential Sequence is Uniformly Convergent on Compact Sets",
"Definition:Uniform Convergence/Metric Space",
"Uniform Limit Theorem"
] |
proofwiki-7144 | Exponential Function is Continuous/Real Numbers | The real exponential function is continuous.
That is:
:$\forall x_0 \in \R: \ds \lim_{x \mathop \to x_0} \exp x = \exp x_0$ | This proof depends on the definition of the exponential function as the function inverse of the natural logarithm.
From Logarithm is Strictly Increasing, $\ln$ is strictly monotone on $\R_{>0}$.
From Real Natural Logarithm Function is Continuous, $\ln$ is continuous on $\R_{>0}$
Thus, from the Continuous Inverse Theore... | The [[Definition:Real Exponential Function|real exponential function]] is [[Definition:Continuous Function|continuous]].
That is:
:$\forall x_0 \in \R: \ds \lim_{x \mathop \to x_0} \exp x = \exp x_0$ | This proof depends on the [[Definition:Exponential Function/Real/Inverse of Natural Logarithm|definition of the exponential function as the function inverse]] of the [[Definition:Natural Logarithm|natural logarithm]].
From [[Logarithm is Strictly Increasing]], $\ln$ is [[Definition:Strictly Monotone Real Function|str... | Exponential Function is Continuous/Real Numbers/Proof 2 | https://proofwiki.org/wiki/Exponential_Function_is_Continuous/Real_Numbers | https://proofwiki.org/wiki/Exponential_Function_is_Continuous/Real_Numbers/Proof_2 | [
"Exponential Function is Continuous",
"Continuous Real Functions"
] | [
"Definition:Exponential Function/Real",
"Definition:Continuous Function"
] | [
"Definition:Exponential Function/Real/Inverse of Natural Logarithm",
"Definition:Natural Logarithm",
"Logarithm is Strictly Increasing",
"Definition:Strictly Monotone/Real Function",
"Real Natural Logarithm Function is Continuous",
"Definition:Continuous Real Function/Interval",
"Continuous Inverse Theo... |
proofwiki-7145 | Exponential Function is Continuous/Real Numbers | The real exponential function is continuous.
That is:
:$\forall x_0 \in \R: \ds \lim_{x \mathop \to x_0} \exp x = \exp x_0$ | This proof depends on the differential equation definition of the exponential function.
The result follows from Differentiable Function is Continuous.
{{qed}} | The [[Definition:Real Exponential Function|real exponential function]] is [[Definition:Continuous Function|continuous]].
That is:
:$\forall x_0 \in \R: \ds \lim_{x \mathop \to x_0} \exp x = \exp x_0$ | This proof depends on the [[Definition:Exponential Function/Real/Differential Equation|differential equation definition of the exponential function]].
The result follows from [[Differentiable Function is Continuous]].
{{qed}} | Exponential Function is Continuous/Real Numbers/Proof 3 | https://proofwiki.org/wiki/Exponential_Function_is_Continuous/Real_Numbers | https://proofwiki.org/wiki/Exponential_Function_is_Continuous/Real_Numbers/Proof_3 | [
"Exponential Function is Continuous",
"Continuous Real Functions"
] | [
"Definition:Exponential Function/Real",
"Definition:Continuous Function"
] | [
"Definition:Exponential Function/Real/Differential Equation",
"Differentiable Function is Continuous"
] |
proofwiki-7146 | Exponential Function is Continuous/Real Numbers | The real exponential function is continuous.
That is:
:$\forall x_0 \in \R: \ds \lim_{x \mathop \to x_0} \exp x = \exp x_0$ | This proof depends on the continuous extension definition of the exponential function.
Let $\exp$ be the unique continuous extension of $e^x$ from $\Q$ to $\R$.
By definition, $\exp$ is continuous.
Hence the result.
{{qed}} | The [[Definition:Real Exponential Function|real exponential function]] is [[Definition:Continuous Function|continuous]].
That is:
:$\forall x_0 \in \R: \ds \lim_{x \mathop \to x_0} \exp x = \exp x_0$ | This proof depends on the [[Definition:Exponential Function/Real/Extension of Rational Exponential|continuous extension definition of the exponential function]].
Let $\exp$ be the unique [[Definition:Continuous Extension|continuous extension]] of $e^x$ from $\Q$ to $\R$.
By definition, $\exp$ is continuous.
Hence th... | Exponential Function is Continuous/Real Numbers/Proof 4 | https://proofwiki.org/wiki/Exponential_Function_is_Continuous/Real_Numbers | https://proofwiki.org/wiki/Exponential_Function_is_Continuous/Real_Numbers/Proof_4 | [
"Exponential Function is Continuous",
"Continuous Real Functions"
] | [
"Definition:Exponential Function/Real",
"Definition:Continuous Function"
] | [
"Definition:Exponential Function/Real/Extension of Rational Exponential",
"Definition:Continuous Extension"
] |
proofwiki-7147 | Exponential Function is Continuous/Real Numbers | The real exponential function is continuous.
That is:
:$\forall x_0 \in \R: \ds \lim_{x \mathop \to x_0} \exp x = \exp x_0$ | This proof depends on the series definition of $\exp$.
That is, let:
:$\ds \exp x = \sum_{k \mathop = 0}^ \infty \frac {x^k} {k!}$
From Series of Power over Factorial Converges, the radius of convergence of $\exp$ is $\infty$.
Thus, from Power Series Converges to Continuous Function, $\exp$ is continuous on $\R$.
{{qed... | The [[Definition:Real Exponential Function|real exponential function]] is [[Definition:Continuous Function|continuous]].
That is:
:$\forall x_0 \in \R: \ds \lim_{x \mathop \to x_0} \exp x = \exp x_0$ | This proof depends on the [[Definition:Exponential Function/Real/Power Series Expansion|series definition of $\exp$]].
That is, let:
:$\ds \exp x = \sum_{k \mathop = 0}^ \infty \frac {x^k} {k!}$
From [[Series of Power over Factorial Converges]], the [[Definition:Radius of Convergence|radius of convergence]] of $\exp... | Exponential Function is Continuous/Real Numbers/Proof 5 | https://proofwiki.org/wiki/Exponential_Function_is_Continuous/Real_Numbers | https://proofwiki.org/wiki/Exponential_Function_is_Continuous/Real_Numbers/Proof_5 | [
"Exponential Function is Continuous",
"Continuous Real Functions"
] | [
"Definition:Exponential Function/Real",
"Definition:Continuous Function"
] | [
"Definition:Exponential Function/Real/Power Series Expansion",
"Series of Power over Factorial Converges",
"Definition:Radius of Convergence",
"Power Series Converges to Continuous Function"
] |
proofwiki-7148 | Edge of Tree is Bridge | Let $T$ be a tree.
Let $e$ be an edge of $T$.
Then $e$ is a bridge of $T$. | From Condition for Edge to be Bridge, $e$ is a bridge {{iff}} $e$ does not lie on any circuit.
Since $T$ is a tree, there are no circuits in $T$.
The result follows.
{{qed}}
Category:Tree Theory
2gv9ezkowkeum6iuxybx22zdv45vg9z | Let $T$ be a [[Definition:Tree (Graph Theory)|tree]].
Let $e$ be an [[Definition:Edge of Graph|edge]] of $T$.
Then $e$ is a [[Definition:Bridge (Graph Theory)|bridge]] of $T$. | From [[Condition for Edge to be Bridge]], $e$ is a [[Definition:Bridge (Graph Theory)|bridge]] {{iff}} $e$ does not lie on any [[Definition:Circuit (Graph Theory)|circuit]].
Since $T$ is a [[Definition:Tree (Graph Theory)|tree]], there are no [[Definition:Circuit (Graph Theory)|circuits]] in $T$.
The result follows.... | Edge of Tree is Bridge | https://proofwiki.org/wiki/Edge_of_Tree_is_Bridge | https://proofwiki.org/wiki/Edge_of_Tree_is_Bridge | [
"Tree Theory"
] | [
"Definition:Tree (Graph Theory)",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Bridge (Graph Theory)"
] | [
"Condition for Edge to be Bridge",
"Definition:Bridge (Graph Theory)",
"Definition:Circuit (Graph Theory)",
"Definition:Tree (Graph Theory)",
"Definition:Circuit (Graph Theory)",
"Category:Tree Theory"
] |
proofwiki-7149 | Sub-Basis for Topological Subspace | Let $\struct {X, \tau}$ be a topological space.
Let $K$ be a sub-basis for $\tau$.
Let $\struct {S, \tau'}$ be a subspace of $\struct {X, \tau}$.
Let $K' = \set {U \cap S: U \in K}$.
That is, $K'$ consists of the $\tau'$-open sets in $S$ corresponding to elements of $K$.
Then $K'$ is a sub-basis for $\tau'$. | Let $B$ be the basis for $\tau$ generated by $K$.
By Basis for Topological Subspace, $B$ generates a basis, $B'$, for $\tau'$.
We will show that $K'$ generates $B'$.
Let $V' \in B'$.
Then for some $V \in B$:
:$V' = S \cap V$
By the definition of a sub-basis, there is a finite subset $K_V$ of $K$ such that:
:$\ds V = \b... | Let $\struct {X, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $K$ be a [[Definition:Sub-Basis|sub-basis]] for $\tau$.
Let $\struct {S, \tau'}$ be a [[Definition:Topological Subspace|subspace]] of $\struct {X, \tau}$.
Let $K' = \set {U \cap S: U \in K}$.
That is, $K'$ consists of the [[Definit... | Let $B$ be the [[Definition:Basis (Topology)|basis]] for $\tau$ generated by $K$.
By [[Basis for Topological Subspace]], $B$ generates a basis, $B'$, for $\tau'$.
We will show that $K'$ generates $B'$.
Let $V' \in B'$.
Then for some $V \in B$:
:$V' = S \cap V$
By the definition of a [[Definition:Sub-Basis|sub-basi... | Sub-Basis for Topological Subspace | https://proofwiki.org/wiki/Sub-Basis_for_Topological_Subspace | https://proofwiki.org/wiki/Sub-Basis_for_Topological_Subspace | [
"Topological Bases",
"Topological Subspaces"
] | [
"Definition:Topological Space",
"Definition:Sub-Basis",
"Definition:Topological Subspace",
"Definition:Open Set/Topology",
"Definition:Element",
"Definition:Sub-Basis"
] | [
"Definition:Basis (Topology)",
"Basis for Topological Subspace",
"Definition:Sub-Basis",
"Definition:Finite Set",
"Definition:Subset",
"Category:Topological Bases",
"Category:Topological Subspaces"
] |
proofwiki-7150 | Modus Ponendo Ponens/Variant 2 | :$\vdash p \implies \paren {\paren {p \implies q} \implies q}$ | {{BeginTableau|\vdash p \implies \paren {\paren {p \implies q} \implies q} }}
{{Assumption|1|p}}
{{Assumption|2|p \implies q}}
{{ModusPonens|3|1, 2|q|2|1}}
{{Implication|4|1|\paren {p \implies q} \implies q|2|3}}
{{Implication|5||p \implies \paren {\paren {p \implies q} \implies q}|1|4}}
{{EndTableau|qed}} | :$\vdash p \implies \paren {\paren {p \implies q} \implies q}$ | {{BeginTableau|\vdash p \implies \paren {\paren {p \implies q} \implies q} }}
{{Assumption|1|p}}
{{Assumption|2|p \implies q}}
{{ModusPonens|3|1, 2|q|2|1}}
{{Implication|4|1|\paren {p \implies q} \implies q|2|3}}
{{Implication|5||p \implies \paren {\paren {p \implies q} \implies q}|1|4}}
{{EndTableau|qed}} | Modus Ponendo Ponens/Variant 2/Proof 1 | https://proofwiki.org/wiki/Modus_Ponendo_Ponens/Variant_2 | https://proofwiki.org/wiki/Modus_Ponendo_Ponens/Variant_2/Proof_1 | [
"Modus Ponendo Ponens"
] | [] | [] |
proofwiki-7151 | Modus Ponendo Ponens/Variant 2 | :$\vdash p \implies \paren {\paren {p \implies q} \implies q}$ | We apply the Method of Truth Tables.
:<nowiki>$\begin{array}{|c|c|ccccc|} \hline
p & \implies & ((p & \implies & q) & \implies & q)\\
\hline
\F & \T & \F & \T & \F & \F & \F \\
\F & \T & \F & \T & \T & \T & \T \\
\T & \T & \T & \F & \F & \T & \F \\
\T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$</nowiki>
As can... | :$\vdash p \implies \paren {\paren {p \implies q} \implies q}$ | We apply the [[Method of Truth Tables]].
:<nowiki>$\begin{array}{|c|c|ccccc|} \hline
p & \implies & ((p & \implies & q) & \implies & q)\\
\hline
\F & \T & \F & \T & \F & \F & \F \\
\F & \T & \F & \T & \T & \T & \T \\
\T & \T & \T & \F & \F & \T & \F \\
\T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$</nowiki>
... | Modus Ponendo Ponens/Variant 2/Proof by Truth Table | https://proofwiki.org/wiki/Modus_Ponendo_Ponens/Variant_2 | https://proofwiki.org/wiki/Modus_Ponendo_Ponens/Variant_2/Proof_by_Truth_Table | [
"Modus Ponendo Ponens"
] | [] | [
"Method of Truth Tables",
"Definition:Main Connective",
"Definition:True"
] |
proofwiki-7152 | Rule of Commutation/Conjunction | Conjunction is commutative:
=== Formulation 1 ===
{{:Rule of Commutation/Conjunction/Formulation 1}}
=== Formulation 2 ===
{{:Rule of Commutation/Conjunction/Formulation 2}} | {{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... | [[Definition:Conjunction|Conjunction]] is [[Definition:Commutative Operation|commutative]]:
=== [[Rule of Commutation/Conjunction/Formulation 1|Formulation 1]] ===
{{:Rule of Commutation/Conjunction/Formulation 1}}
=== [[Rule of Commutation/Conjunction/Formulation 2|Formulation 2]] ===
{{:Rule of Commutation/Conjuncti... | {{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 | https://proofwiki.org/wiki/Rule_of_Commutation/Conjunction/Formulation_1/Proof_1 | [
"Conjunction",
"Rule of Commutation"
] | [
"Definition:Conjunction",
"Definition:Commutative/Operation",
"Rule of Commutation/Conjunction/Formulation 1",
"Rule of Commutation/Conjunction/Formulation 2"
] | [] |
proofwiki-7153 | Rule of Commutation/Conjunction | Conjunction is commutative:
=== Formulation 1 ===
{{:Rule of Commutation/Conjunction/Formulation 1}}
=== Formulation 2 ===
{{:Rule of Commutation/Conjunction/Formulation 2}} | 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 ... | [[Definition:Conjunction|Conjunction]] is [[Definition:Commutative Operation|commutative]]:
=== [[Rule of Commutation/Conjunction/Formulation 1|Formulation 1]] ===
{{:Rule of Commutation/Conjunction/Formulation 1}}
=== [[Rule of Commutation/Conjunction/Formulation 2|Formulation 2]] ===
{{:Rule of Commutation/Conjuncti... | 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 | https://proofwiki.org/wiki/Rule_of_Commutation/Conjunction/Formulation_1/Proof_by_Truth_Table | [
"Conjunction",
"Rule of Commutation"
] | [
"Definition:Conjunction",
"Definition:Commutative/Operation",
"Rule of Commutation/Conjunction/Formulation 1",
"Rule of Commutation/Conjunction/Formulation 2"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7154 | Rule of Commutation/Conjunction | Conjunction is commutative:
=== Formulation 1 ===
{{:Rule of Commutation/Conjunction/Formulation 1}}
=== Formulation 2 ===
{{:Rule of Commutation/Conjunction/Formulation 2}} | {{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... | [[Definition:Conjunction|Conjunction]] is [[Definition:Commutative Operation|commutative]]:
=== [[Rule of Commutation/Conjunction/Formulation 1|Formulation 1]] ===
{{:Rule of Commutation/Conjunction/Formulation 1}}
=== [[Rule of Commutation/Conjunction/Formulation 2|Formulation 2]] ===
{{:Rule of Commutation/Conjuncti... | {{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 | https://proofwiki.org/wiki/Rule_of_Commutation/Conjunction/Formulation_2/Proof_2 | [
"Conjunction",
"Rule of Commutation"
] | [
"Definition:Conjunction",
"Definition:Commutative/Operation",
"Rule of Commutation/Conjunction/Formulation 1",
"Rule of Commutation/Conjunction/Formulation 2"
] | [] |
proofwiki-7155 | Rule of Commutation/Disjunction | Disjunction is commutative:
=== Formulation 1 ===
{{:Rule of Commutation/Disjunction/Formulation 1}}
=== Formulation 2 ===
{{:Rule of Commutation/Disjunction/Formulation 2}} | {{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... | [[Definition:Disjunction|Disjunction]] is [[Definition:Commutative Operation|commutative]]:
=== [[Rule of Commutation/Disjunction/Formulation 1|Formulation 1]] ===
{{:Rule of Commutation/Disjunction/Formulation 1}}
=== [[Rule of Commutation/Disjunction/Formulation 2|Formulation 2]] ===
{{:Rule of Commutation/Disjuncti... | {{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 | https://proofwiki.org/wiki/Rule_of_Commutation/Disjunction/Formulation_1/Proof_1 | [
"Disjunction",
"Rule of Commutation"
] | [
"Definition:Disjunction",
"Definition:Commutative/Operation",
"Rule of Commutation/Disjunction/Formulation 1",
"Rule of Commutation/Disjunction/Formulation 2"
] | [] |
proofwiki-7156 | Rule of Commutation/Disjunction | Disjunction is commutative:
=== Formulation 1 ===
{{:Rule of Commutation/Disjunction/Formulation 1}}
=== Formulation 2 ===
{{:Rule of Commutation/Disjunction/Formulation 2}} | 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... | [[Definition:Disjunction|Disjunction]] is [[Definition:Commutative Operation|commutative]]:
=== [[Rule of Commutation/Disjunction/Formulation 1|Formulation 1]] ===
{{:Rule of Commutation/Disjunction/Formulation 1}}
=== [[Rule of Commutation/Disjunction/Formulation 2|Formulation 2]] ===
{{:Rule of Commutation/Disjuncti... | 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 | https://proofwiki.org/wiki/Rule_of_Commutation/Disjunction/Formulation_1/Proof_by_Truth_Table | [
"Disjunction",
"Rule of Commutation"
] | [
"Definition:Disjunction",
"Definition:Commutative/Operation",
"Rule of Commutation/Disjunction/Formulation 1",
"Rule of Commutation/Disjunction/Formulation 2"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7157 | Rule of Commutation/Disjunction | Disjunction is commutative:
=== Formulation 1 ===
{{:Rule of Commutation/Disjunction/Formulation 1}}
=== Formulation 2 ===
{{:Rule of Commutation/Disjunction/Formulation 2}} | {{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... | [[Definition:Disjunction|Disjunction]] is [[Definition:Commutative Operation|commutative]]:
=== [[Rule of Commutation/Disjunction/Formulation 1|Formulation 1]] ===
{{:Rule of Commutation/Disjunction/Formulation 1}}
=== [[Rule of Commutation/Disjunction/Formulation 2|Formulation 2]] ===
{{:Rule of Commutation/Disjuncti... | {{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 | https://proofwiki.org/wiki/Rule_of_Commutation/Disjunction/Formulation_2/Proof_1 | [
"Disjunction",
"Rule of Commutation"
] | [
"Definition:Disjunction",
"Definition:Commutative/Operation",
"Rule of Commutation/Disjunction/Formulation 1",
"Rule of Commutation/Disjunction/Formulation 2"
] | [] |
proofwiki-7158 | Rule of Commutation/Disjunction | Disjunction is commutative:
=== Formulation 1 ===
{{:Rule of Commutation/Disjunction/Formulation 1}}
=== Formulation 2 ===
{{:Rule of Commutation/Disjunction/Formulation 2}} | 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 &... | [[Definition:Disjunction|Disjunction]] is [[Definition:Commutative Operation|commutative]]:
=== [[Rule of Commutation/Disjunction/Formulation 1|Formulation 1]] ===
{{:Rule of Commutation/Disjunction/Formulation 1}}
=== [[Rule of Commutation/Disjunction/Formulation 2|Formulation 2]] ===
{{:Rule of Commutation/Disjuncti... | 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>$\begin {array} {|ccc|c|ccc|} \hline
(... | Rule of Commutation/Disjunction/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Commutation/Disjunction | https://proofwiki.org/wiki/Rule_of_Commutation/Disjunction/Formulation_2/Proof_by_Truth_Table | [
"Disjunction",
"Rule of Commutation"
] | [
"Definition:Disjunction",
"Definition:Commutative/Operation",
"Rule of Commutation/Disjunction/Formulation 1",
"Rule of Commutation/Disjunction/Formulation 2"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7159 | Factor Principles/Conjunction on Right/Formulation 1 | :$p \implies q \vdash \paren {p \land r} \implies \paren {q \land r}$ | {{BeginTableau|p \implies q \vdash \paren {p \land r} \implies \paren {q \land r} }}
{{Premise|1|p \implies q}}
{{IdentityLaw|2||r \implies r|(None)|This is a theorem so depends on nothing}}
{{Conjunction|3|1|\paren {p \implies q} \land \paren {r \implies r}|1|2}}
{{SequentIntro|4|1|\paren {p \land r} \implies \paren {... | :$p \implies q \vdash \paren {p \land r} \implies \paren {q \land r}$ | {{BeginTableau|p \implies q \vdash \paren {p \land r} \implies \paren {q \land r} }}
{{Premise|1|p \implies q}}
{{IdentityLaw|2||r \implies r|(None)|This is a theorem so depends on nothing}}
{{Conjunction|3|1|\paren {p \implies q} \land \paren {r \implies r}|1|2}}
{{SequentIntro|4|1|\paren {p \land r} \implies \paren {... | Factor Principles/Conjunction on Right/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Factor_Principles/Conjunction_on_Right/Formulation_1 | https://proofwiki.org/wiki/Factor_Principles/Conjunction_on_Right/Formulation_1/Proof_1 | [
"Factor Principles"
] | [] | [
"Praeclarum Theorema"
] |
proofwiki-7160 | Factor Principles/Conjunction on Right/Formulation 1 | :$p \implies q \vdash \paren {p \land r} \implies \paren {q \land r}$ | {{BeginTableau|p \implies q \vdash \paren {p \land r} \implies \paren {q \land r} }}
{{Premise |1|p \implies q}}
{{Assumption |2|p \land r}}
{{Simplification|3|2|p|2|1}}
{{ModusPonens |4|1, 2|q|1|3}}
{{Simplification|5|2|r|2|2}}
{{Conjunction |6|1, 2|q \land r|4|5}}
{{Implication |7|1|\paren {p \land r} ... | :$p \implies q \vdash \paren {p \land r} \implies \paren {q \land r}$ | {{BeginTableau|p \implies q \vdash \paren {p \land r} \implies \paren {q \land r} }}
{{Premise |1|p \implies q}}
{{Assumption |2|p \land r}}
{{Simplification|3|2|p|2|1}}
{{ModusPonens |4|1, 2|q|1|3}}
{{Simplification|5|2|r|2|2}}
{{Conjunction |6|1, 2|q \land r|4|5}}
{{Implication |7|1|\paren {p \land r} ... | Factor Principles/Conjunction on Right/Formulation 1/Proof 2 | https://proofwiki.org/wiki/Factor_Principles/Conjunction_on_Right/Formulation_1 | https://proofwiki.org/wiki/Factor_Principles/Conjunction_on_Right/Formulation_1/Proof_2 | [
"Factor Principles"
] | [] | [] |
proofwiki-7161 | Factor Principles/Conjunction on Right/Formulation 1 | :$p \implies q \vdash \paren {p \land r} \implies \paren {q \land r}$ | Proof by Truth Table:
$\begin{array}{|ccc||ccccccccccc|}
\hline
p & q & r & (p & \implies & q) & \implies & ((p & \land & r) & \implies & (q & \land & r)) \\
\hline
\T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\
\T & \T & \F & \T & \T & \T & \T & \T & \F & \F & \T & \T & \F & \F \\
\T & \F & \T ... | :$p \implies q \vdash \paren {p \land r} \implies \paren {q \land r}$ | Proof by [[Definition:Truth Table|Truth Table]]:
$\begin{array}{|ccc||ccccccccccc|}
\hline
p & q & r & (p & \implies & q) & \implies & ((p & \land & r) & \implies & (q & \land & r)) \\
\hline
\T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\
\T & \T & \F & \T & \T & \T & \T & \T & \F & \F & \T & \... | Factor Principles/Conjunction on Right/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Factor_Principles/Conjunction_on_Right/Formulation_1 | https://proofwiki.org/wiki/Factor_Principles/Conjunction_on_Right/Formulation_1/Proof_by_Truth_Table | [
"Factor Principles"
] | [] | [
"Definition:Truth Table"
] |
proofwiki-7162 | Interior of Convex Angle is Convex Set | Let $\mathbf v, \mathbf w$ be two non-zero vectors in $\R^2$, and let $p$ be a point in $\R^2$.
Suppose that the angle between $\mathbf v$ and $\mathbf w$ is a convex angle.
Then the set
:$U = \set {p + s t \mathbf v + \paren {1 - s} t \mathbf w : s \in \openint 0 1, t \in \R_{>0} }$
is a convex set. | Let $p_1, p_2 \in U$.
Then for $i \in \set {1, 2}$, $p_i = p + s_i t_i \mathbf v + \paren {1 - s_i} t_i \mathbf w$ for some $s_i \in \openint 0 1, t_i \in \R_{>0}$.
:File:InteriorOfConvexAngle.png
{{WLOG}}, assume that $t_1 \le t_2$.
Suppose that $q \in \R^2$ lies on the line segment joining $p_1$ and $p_2$, so:
{{begi... | Let $\mathbf v, \mathbf w$ be two non-[[Definition:Zero Vector|zero]] [[Definition:Vector (Linear Algebra)|vectors]] in $\R^2$, and let $p$ be a [[Definition:Point|point]] in $\R^2$.
Suppose that the [[Definition:Angle Between Vectors|angle]] between $\mathbf v$ and $\mathbf w$ is a [[Definition:Convex Angle|convex a... | Let $p_1, p_2 \in U$.
Then for $i \in \set {1, 2}$, $p_i = p + s_i t_i \mathbf v + \paren {1 - s_i} t_i \mathbf w$ for some $s_i \in \openint 0 1, t_i \in \R_{>0}$.
:[[File:InteriorOfConvexAngle.png]]
{{WLOG}}, assume that $t_1 \le t_2$.
Suppose that $q \in \R^2$ lies on the [[Definition:Straight Line Segment (Vec... | Interior of Convex Angle is Convex Set | https://proofwiki.org/wiki/Interior_of_Convex_Angle_is_Convex_Set | https://proofwiki.org/wiki/Interior_of_Convex_Angle_is_Convex_Set | [
"Vector Spaces",
"Convex Sets (Vector Spaces)"
] | [
"Definition:Zero Vector",
"Definition:Vector/Linear Algebra",
"Definition:Point",
"Definition:Angle between Vectors",
"Definition:Convex Angle",
"Definition:Set",
"Definition:Convex Set (Vector Space)"
] | [
"File:InteriorOfConvexAngle.png",
"Definition:Convex Set (Vector Space)/Line Segment",
"Definition:Convex Set (Vector Space)",
"Category:Vector Spaces",
"Category:Convex Sets (Vector Spaces)"
] |
proofwiki-7163 | Factor Principles/Conjunction on Left/Formulation 1 | :$p \implies q \vdash \paren {r \land p} \implies \paren {r \land q}$ | {{BeginTableau|p \implies q \vdash \paren {r \land p} \implies \paren {r \land q} }}
{{Premise|1|p \implies q}}
{{IdentityLaw|2||r \implies r|(None)|This is a theorem so depends on nothing}}
{{Conjunction|3|1|\paren {r \implies r} \land \paren {p \implies q}|2|1}}
{{SequentIntro|4|1|\paren {r \land p} \implies \paren {... | :$p \implies q \vdash \paren {r \land p} \implies \paren {r \land q}$ | {{BeginTableau|p \implies q \vdash \paren {r \land p} \implies \paren {r \land q} }}
{{Premise|1|p \implies q}}
{{IdentityLaw|2||r \implies r|(None)|This is a theorem so depends on nothing}}
{{Conjunction|3|1|\paren {r \implies r} \land \paren {p \implies q}|2|1}}
{{SequentIntro|4|1|\paren {r \land p} \implies \paren {... | Factor Principles/Conjunction on Left/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Factor_Principles/Conjunction_on_Left/Formulation_1 | https://proofwiki.org/wiki/Factor_Principles/Conjunction_on_Left/Formulation_1/Proof_1 | [
"Factor Principles"
] | [] | [
"Praeclarum Theorema"
] |
proofwiki-7164 | Factor Principles/Conjunction on Left/Formulation 1 | :$p \implies q \vdash \paren {r \land p} \implies \paren {r \land q}$ | {{BeginTableau|p \implies q \vdash \paren {r \land p} \implies \paren {r \land q} }}
{{Premise|1|p \implies q}}
{{Assumption|2|r \land p}}
{{Simplification|3|2|p|2|2}}
{{ModusPonens|4|1, 2|q|1|3}}
{{Simplification|5|2|r|2|1}}
{{Conjunction|6|1, 2|r \land q|5|4}}
{{Implication|7|1|\paren {r \land p} \implies \paren {r \... | :$p \implies q \vdash \paren {r \land p} \implies \paren {r \land q}$ | {{BeginTableau|p \implies q \vdash \paren {r \land p} \implies \paren {r \land q} }}
{{Premise|1|p \implies q}}
{{Assumption|2|r \land p}}
{{Simplification|3|2|p|2|2}}
{{ModusPonens|4|1, 2|q|1|3}}
{{Simplification|5|2|r|2|1}}
{{Conjunction|6|1, 2|r \land q|5|4}}
{{Implication|7|1|\paren {r \land p} \implies \paren {r \... | Factor Principles/Conjunction on Left/Formulation 1/Proof 2 | https://proofwiki.org/wiki/Factor_Principles/Conjunction_on_Left/Formulation_1 | https://proofwiki.org/wiki/Factor_Principles/Conjunction_on_Left/Formulation_1/Proof_2 | [
"Factor Principles"
] | [] | [] |
proofwiki-7165 | Factor Principles/Conjunction on Left/Formulation 2 | :$\vdash \paren {p \implies q} \implies \paren {\paren {r \land p} \implies \paren {r \land q} }$ | {{BeginTableau|\vdash \paren {p \implies q} \implies \paren {\paren {r \land p} \implies \paren {r \land q} } }}
{{Assumption|1|p \implies q}}
{{Assumption|2|r \land p}}
{{Simplification|3|2|p|2|2}}
{{ModusPonens|4|1, 2|q|1|3}}
{{Simplification|5|2|r|2|1}}
{{Conjunction|6|1, 2|r \land q|5|4}}
{{Implication|7|1|\paren {... | :$\vdash \paren {p \implies q} \implies \paren {\paren {r \land p} \implies \paren {r \land q} }$ | {{BeginTableau|\vdash \paren {p \implies q} \implies \paren {\paren {r \land p} \implies \paren {r \land q} } }}
{{Assumption|1|p \implies q}}
{{Assumption|2|r \land p}}
{{Simplification|3|2|p|2|2}}
{{ModusPonens|4|1, 2|q|1|3}}
{{Simplification|5|2|r|2|1}}
{{Conjunction|6|1, 2|r \land q|5|4}}
{{Implication|7|1|\paren {... | Factor Principles/Conjunction on Left/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Factor_Principles/Conjunction_on_Left/Formulation_2 | https://proofwiki.org/wiki/Factor_Principles/Conjunction_on_Left/Formulation_2/Proof_1 | [
"Factor Principles"
] | [] | [] |
proofwiki-7166 | Triangle is Convex Set | The interior of a triangle embedded in $\R^2$ is a convex set. | Denote the triangle as $\triangle$, and the interior of the boundary of $\triangle$ as $\Int \triangle$.
From Boundary of Polygon is Jordan Curve, it follows that the boundary of $\triangle$ is equal to the image of a Jordan curve, so $\Int \triangle$ is well-defined.
Denote the vertices of $\triangle$ as $A_1, A_2, A_... | The [[Definition:Interior of Jordan Curve|interior]] of a [[Definition:Triangle (Geometry)|triangle]] embedded in $\R^2$ is a [[Definition:Convex Set (Vector Space)|convex set]]. | Denote the [[Definition:Triangle (Geometry)|triangle]] as $\triangle$, and the [[Definition:Interior of Jordan Curve|interior]] of the [[Definition:Boundary (Geometry)|boundary]] of $\triangle$ as $\Int \triangle$.
From [[Boundary of Polygon is Jordan Curve]], it follows that the [[Definition:Boundary (Geometry)|bound... | Triangle is Convex Set | https://proofwiki.org/wiki/Triangle_is_Convex_Set | https://proofwiki.org/wiki/Triangle_is_Convex_Set | [
"Convex Sets (Vector Spaces)"
] | [
"Definition:Jordan Curve/Interior",
"Definition:Triangle (Geometry)",
"Definition:Convex Set (Vector Space)"
] | [
"Definition:Triangle (Geometry)",
"Definition:Jordan Curve/Interior",
"Definition:Boundary (Geometry)",
"Boundary of Polygon is Jordan Curve",
"Definition:Boundary (Geometry)",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Jordan Curve",
"Definition:Polygon/Vertex",
"Definition:Angle ... |
proofwiki-7167 | Norm of Vector Cross Product | Let $\mathbf a$ and $\mathbf b$ be vectors in the real Euclidean space $\R^3$.
Let $\times$ denote the vector cross product.
Then:
:$\norm {\mathbf a \times \mathbf b} = \norm {\mathbf a} \norm {\mathbf b} \size {\sin \theta}$
where $\theta$ is the angle between $\mathbf a$ and $\mathbf b$, or an arbitrary number if $\... | Suppose either $\mathbf a$ or $\mathbf b$ is the zero vector.
Then by {{NormAxiomVector|1}}:
:$\norm {\mathbf a} = 0$
or:
:$\norm {\mathbf b} = 0$
By calculation, it follows that $\mathbf a \times \mathbf b$ is also the zero vector.
Hence:
:$\norm {\mathbf a \times \mathbf b} = 0$
and equality holds.
Now suppose that b... | Let $\mathbf a$ and $\mathbf b$ be [[Definition:Space Vector|vectors]] in the [[Definition:Real Euclidean Space|real Euclidean space]] $\R^3$.
Let $\times$ denote the [[Definition:Vector Cross Product|vector cross product]].
Then:
:$\norm {\mathbf a \times \mathbf b} = \norm {\mathbf a} \norm {\mathbf b} \size {\si... | Suppose either $\mathbf a$ or $\mathbf b$ is the [[Definition:Zero Vector|zero vector]].
Then by {{NormAxiomVector|1}}:
:$\norm {\mathbf a} = 0$
or:
:$\norm {\mathbf b} = 0$
By calculation, it follows that $\mathbf a \times \mathbf b$ is also the [[Definition:Zero Vector|zero vector]].
Hence:
:$\norm {\mathbf a \ti... | Norm of Vector Cross Product | https://proofwiki.org/wiki/Norm_of_Vector_Cross_Product | https://proofwiki.org/wiki/Norm_of_Vector_Cross_Product | [
"Vector Cross Product"
] | [
"Definition:Vector/Real Euclidean Space/Space Vector",
"Definition:Euclidean Space/Real",
"Definition:Vector Cross Product",
"Definition:Angle between Vectors",
"Definition:Real Number",
"Definition:Zero Vector"
] | [
"Definition:Zero Vector",
"Definition:Zero Vector",
"Definition:Zero Vector",
"Definition:Vector",
"Square of Norm of Vector Cross Product",
"Cosine Formula for Dot Product",
"Sum of Squares of Sine and Cosine",
"Definition:Square Root/Complex Number/Principal Square Root"
] |
proofwiki-7168 | Subset Relation is Compatible with Subset Product/Corollary 1 | Let $A, B, C, D \in \powerset S$.
Let $A \subseteq B$ and $C \subseteq D$.
Then:
:$A \circ_\PP C \subseteq B \circ_\PP D$ | By Subset Relation is Compatible with Subset Product, $\subseteq$ is compatible with $\circ_\PP$.
By Subset Relation is Transitive, $\subseteq$ is transitive.
Thus the theorem holds by Operating on Transitive Relationships Compatible with Operation.
{{qed}}
Category:Subset Products
Category:Compatible Relations
9ik95eg... | Let $A, B, C, D \in \powerset S$.
Let $A \subseteq B$ and $C \subseteq D$.
Then:
:$A \circ_\PP C \subseteq B \circ_\PP D$ | By [[Subset Relation is Compatible with Subset Product]], $\subseteq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ_\PP$.
By [[Subset Relation is Transitive]], $\subseteq$ is [[Definition:Transitive Relation|transitive]].
Thus the theorem holds by [[Operating on Transitive Relationships ... | Subset Relation is Compatible with Subset Product/Corollary 1 | https://proofwiki.org/wiki/Subset_Relation_is_Compatible_with_Subset_Product/Corollary_1 | https://proofwiki.org/wiki/Subset_Relation_is_Compatible_with_Subset_Product/Corollary_1 | [
"Subset Products",
"Compatible Relations"
] | [] | [
"Subset Relation is Compatible with Subset Product",
"Definition:Relation Compatible with Operation",
"Subset Relation is Transitive",
"Definition:Transitive Relation",
"Operating on Transitive Relationships Compatible with Operation",
"Category:Subset Products",
"Category:Compatible Relations"
] |
proofwiki-7169 | Subset Relation is Compatible with Subset Product/Corollary 2 | Let $A, B \subseteq S$.
Let $A \subseteq B$.
Then:
{{begin-eqn}}
{{eqn | q = \forall x \in S
| l = x \circ A
| o = \subseteq
| r = x \circ B
}}
{{eqn | l = A \circ x
| o = \subseteq
| r = B \circ x
}}
{{end-eqn}} | This follows from Subset Relation is Compatible with Subset Product and the definition of the subset product with a singleton.
{{qed}}
Category:Subset Products
axp9ujgrok4hgprkv48upqjov76lvmf | Let $A, B \subseteq S$.
Let $A \subseteq B$.
Then:
{{begin-eqn}}
{{eqn | q = \forall x \in S
| l = x \circ A
| o = \subseteq
| r = x \circ B
}}
{{eqn | l = A \circ x
| o = \subseteq
| r = B \circ x
}}
{{end-eqn}} | This follows from [[Subset Relation is Compatible with Subset Product]] and the definition of the [[Definition:Subset Product with Singleton|subset product with a singleton]].
{{qed}}
[[Category:Subset Products]]
axp9ujgrok4hgprkv48upqjov76lvmf | Subset Relation is Compatible with Subset Product/Corollary 2 | https://proofwiki.org/wiki/Subset_Relation_is_Compatible_with_Subset_Product/Corollary_2 | https://proofwiki.org/wiki/Subset_Relation_is_Compatible_with_Subset_Product/Corollary_2 | [
"Subset Products"
] | [] | [
"Subset Relation is Compatible with Subset Product",
"Definition:Subset Product/Singleton",
"Category:Subset Products"
] |
proofwiki-7170 | Equivalence of Definitions of Normal Subset/1 iff 2 | Let $\struct {G, \circ}$ be a group.
Let $S$ be a subset of $G$.
Then Normal Subset/Definition 1 is equivalent to Normal Subset/Definition 2.
That is, the following three statements are equivalent:
:$(1): \quad \forall g \in G: g \circ S = S \circ g$
:$(2): \quad \forall g \in G: g \circ S \circ g^{-1} = S$
:$(3): \qua... | Let $e$ be the identity of $G$.
First note that:
:$(4): \quad \paren {\forall g \in G: g \circ S \circ g^{-1} = S} \iff \paren {\forall g \in G: g^{-1} \circ S \circ g = S}$
which is shown by, for example, setting $h := g^{-1}$ and substituting.
=== Necessary Condition ===
Suppose that $S$ satisfies $(1)$.
Let $g \in G... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $S$ be a [[Definition:Subset|subset]] of $G$.
Then [[Definition:Normal Subset/Definition 1|Normal Subset/Definition 1]] is equivalent to [[Definition:Normal Subset/Definition 2|Normal Subset/Definition 2]].
That is, the following three statements are eq... | Let $e$ be the [[Definition:Identity Element|identity]] of $G$.
First note that:
:$(4): \quad \paren {\forall g \in G: g \circ S \circ g^{-1} = S} \iff \paren {\forall g \in G: g^{-1} \circ S \circ g = S}$
which is shown by, for example, setting $h := g^{-1}$ and substituting.
=== Necessary Condition ===
Suppose th... | Equivalence of Definitions of Normal Subset/1 iff 2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Normal_Subset/1_iff_2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Normal_Subset/1_iff_2 | [
"Normal Subsets"
] | [
"Definition:Group",
"Definition:Subset",
"Definition:Normal Subset/Definition 1",
"Definition:Normal Subset/Definition 2"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Subset Product within Semigroup is Associative/Corollary",
"Subset Product with Identity",
"Subset Product within Semigroup is Associative/Corollary",
"Subset Product with Identity"
] |
proofwiki-7171 | Subset Product with Identity | Let $\struct {S, \circ}$ be a magma.
Let $\struct {S, \circ}$ have an identity element $e$.
Then $e \circ S = S \circ e = S$, where $\circ$ is understood to be the subset product with singleton. | {{begin-eqn}}
{{eqn | l = e \circ S
| r = \set e \circ S
| c = {{Defof|Subset Product with Singleton}}
}}
{{eqn | r = \set {x \circ y: x \in \set e, \, y \in S}
| c = {{Defof|Subset Product}}
}}
{{eqn | r = \set {e \circ y: y \in S}
| c =
}}
{{eqn | r = \set {y: y \in S}
| c = {{Defof|Ide... | Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
Let $\struct {S, \circ}$ have an [[Definition:Identity Element|identity element]] $e$.
Then $e \circ S = S \circ e = S$, where $\circ$ is understood to be the [[Definition:Subset Product with Singleton|subset product with singleton]]. | {{begin-eqn}}
{{eqn | l = e \circ S
| r = \set e \circ S
| c = {{Defof|Subset Product with Singleton}}
}}
{{eqn | r = \set {x \circ y: x \in \set e, \, y \in S}
| c = {{Defof|Subset Product}}
}}
{{eqn | r = \set {e \circ y: y \in S}
| c =
}}
{{eqn | r = \set {y: y \in S}
| c = {{Defof|Ide... | Subset Product with Identity | https://proofwiki.org/wiki/Subset_Product_with_Identity | https://proofwiki.org/wiki/Subset_Product_with_Identity | [
"Subset Products"
] | [
"Definition:Magma",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subset Product/Singleton"
] | [
"Category:Subset Products"
] |
proofwiki-7172 | Equivalence of Definitions of Normal Subset/2 implies 3 | Let $\left({G, \circ}\right)$ be a group.
Let $S \subseteq G$.
Let $S$ be a normal subset of $G$ by Definition 2.
Then $S$ is a normal subset of $G$ by Definition 3.
That is, if:
:$\forall g \in G: g \circ S \circ g^{-1} = S$
or:
:$\forall g \in G: g^{-1} \circ S \circ g = S$
then:
:$\forall g \in G: g \circ S \circ g^... | We have that:
:$\left({\forall g \in G: g \circ S \circ g^{-1} = S}\right) \iff \left({\forall g \in G: g^{-1} \circ S \circ g = S}\right)$
The result follows by definition of set equality.
{{qed}} | Let $\left({G, \circ}\right)$ be a [[Definition:group|group]].
Let $S \subseteq G$.
Let $S$ be a [[Definition:Normal Subset/Definition 2|normal subset of $G$ by Definition 2]].
Then $S$ is a [[Definition:Normal Subset/Definition 3|normal subset of $G$ by Definition 3]].
That is, if:
:$\forall g \in G: g \circ S \c... | We have that:
:$\left({\forall g \in G: g \circ S \circ g^{-1} = S}\right) \iff \left({\forall g \in G: g^{-1} \circ S \circ g = S}\right)$
The result follows by definition of [[Definition:Set Equality/Definition 2|set equality]].
{{qed}} | Equivalence of Definitions of Normal Subset/2 implies 3 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Normal_Subset/2_implies_3 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Normal_Subset/2_implies_3 | [
"Normal Subsets"
] | [
"Definition:group",
"Definition:Normal Subset/Definition 2",
"Definition:Normal Subset/Definition 3"
] | [
"Definition:Set Equality/Definition 2"
] |
proofwiki-7173 | Superset Relation is Compatible with Subset Product | Let $\struct {S, \circ}$ be a magma.
Let $\circ_\PP$ be the subset product on $\powerset S$, the power set of $S$.
Then the superset relation $\supseteq$ is compatible with $\circ_\PP$. | By Subset Relation is Compatible with Subset Product, the subset relation $\subseteq$ is compatible with $\circ_\PP$.
From Inverse of Subset Relation is Superset, the inverse of $\subseteq$ is $\supseteq$.
The result follows from Inverse of Relation Compatible with Operation is Compatible.
{{qed}}
Category:Compatible R... | Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
Let $\circ_\PP$ be the [[Definition:Subset Product|subset product]] on $\powerset S$, the [[Definition:Power Set|power set]] of $S$.
Then the [[Definition:Superset|superset]] relation $\supseteq$ is [[Definition:Relation Compatible with Operation|compatible]]... | By [[Subset Relation is Compatible with Subset Product]], the [[Definition:Subset|subset]] relation $\subseteq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ_\PP$.
From [[Inverse of Subset Relation is Superset]], the [[Definition:Inverse Relation|inverse]] of $\subseteq$ is $\supseteq$.
... | Superset Relation is Compatible with Subset Product | https://proofwiki.org/wiki/Superset_Relation_is_Compatible_with_Subset_Product | https://proofwiki.org/wiki/Superset_Relation_is_Compatible_with_Subset_Product | [
"Compatible Relations"
] | [
"Definition:Magma",
"Definition:Subset Product",
"Definition:Power Set",
"Definition:Subset/Superset",
"Definition:Relation Compatible with Operation"
] | [
"Subset Relation is Compatible with Subset Product",
"Definition:Subset",
"Definition:Relation Compatible with Operation",
"Inverse of Subset Relation is Superset",
"Definition:Inverse Relation",
"Inverse of Relation Compatible with Operation is Compatible",
"Category:Compatible Relations"
] |
proofwiki-7174 | Equivalence of Definitions of Normal Subset/3 iff 4 | Let $\struct {G,\circ}$ be a group.
Let $S \subseteq G$.
Then:
: $S$ is a normal subset of $G$ by Definition 3
{{iff}}:
: $S$ is a normal subset of $G$ by Definition 4.
That is, the following conditions are equivalent:
: $(1) \quad \forall g \in G: g \circ S \circ g^{-1} \subseteq S$
: $(2) \quad \forall g \in G: g^{-1... | First note that:
:$(5): \quad \paren {\forall g \in G: g \circ S \circ g^{-1} \subseteq S} \iff \paren {\forall g \in G: g^{-1} \circ S \circ g \subseteq S}$
:$(6): \quad \paren {\forall g \in G: S \subseteq g \circ S \circ g^{-1}} \iff \paren {\forall g \in G: S \subseteq g^{-1} \circ S \circ g}$
which is shown by, fo... | Let $\struct {G,\circ}$ be a [[Definition:group|group]].
Let $S \subseteq G$.
Then:
: $S$ is a [[Definition:Normal Subset/Definition 3|normal subset of $G$ by Definition 3]]
{{iff}}:
: $S$ is a [[Definition:Normal Subset/Definition 4|normal subset of $G$ by Definition 4]].
That is, the following conditions are equ... | First note that:
:$(5): \quad \paren {\forall g \in G: g \circ S \circ g^{-1} \subseteq S} \iff \paren {\forall g \in G: g^{-1} \circ S \circ g \subseteq S}$
:$(6): \quad \paren {\forall g \in G: S \subseteq g \circ S \circ g^{-1}} \iff \paren {\forall g \in G: S \subseteq g^{-1} \circ S \circ g}$
which is shown by,... | Equivalence of Definitions of Normal Subset/3 iff 4 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Normal_Subset/3_iff_4 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Normal_Subset/3_iff_4 | [
"Normal Subsets"
] | [
"Definition:group",
"Definition:Normal Subset/Definition 3",
"Definition:Normal Subset/Definition 4"
] | [
"Subset Relation is Compatible with Subset Product/Corollary 2",
"Subset Product within Semigroup is Associative/Corollary",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Subset Relation is Compatible with Subset Product/Corollary 2",
"Subset Product within Semigroup is Associative/Corollary",
"Defini... |
proofwiki-7175 | Equivalence of Definitions of Normal Subset/3 and 4 imply 2 | Let $\struct {G, \circ}$ be a group.
Let $S \subseteq G$.
Let $S$ be a normal subset of $G$ by Definition 3 and Definition 4.
Then $S$ is a normal subset of $G$ by Definition 2. | By Equivalence of Definitions of Normal Subset: 3 iff 4, $S$ being a normal subset of $G$ by Definition 3 and Definition 4 implies that the following hold:
:$(1)\quad \forall g \in G: g \circ S \circ g^{-1} \subseteq S$
:$(2)\quad \forall g \in G: g^{-1} \circ S \circ g \subseteq S$
:$(3)\quad \forall g \in G: S \subse... | Let $\struct {G, \circ}$ be a [[Definition:group|group]].
Let $S \subseteq G$.
Let $S$ be a [[Definition:Normal Subset/Definition 3|normal subset of $G$ by Definition 3]] and [[Definition:Normal Subset/Definition 4|Definition 4]].
Then $S$ is a [[Definition:Normal Subset/Definition 2|normal subset of $G$ by Definit... | By [[Equivalence of Definitions of Normal Subset/3 iff 4|Equivalence of Definitions of Normal Subset: 3 iff 4]], $S$ being a normal subset of $G$ by [[Definition:Normal Subset/Definition 3|Definition 3]] and [[Definition:Normal Subset/Definition 4|Definition 4]] implies that the following hold:
:$(1)\quad \forall g \in... | Equivalence of Definitions of Normal Subset/3 and 4 imply 2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Normal_Subset/3_and_4_imply_2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Normal_Subset/3_and_4_imply_2 | [
"Normal Subsets"
] | [
"Definition:group",
"Definition:Normal Subset/Definition 3",
"Definition:Normal Subset/Definition 4",
"Definition:Normal Subset/Definition 2"
] | [
"Equivalence of Definitions of Normal Subset/3 iff 4",
"Definition:Normal Subset/Definition 3",
"Definition:Normal Subset/Definition 4",
"Definition:Set Equality/Definition 2",
"Definition:Set Equality/Definition 2"
] |
proofwiki-7176 | Equivalence of Definitions of Normal Subset/3 iff 5 | Let $\struct {G, \circ}$ be a group.
Let $S \subseteq G$.
Then:
: $S$ is a normal subset of $G$ by Definition 3
{{iff}}:
: $S$ is a normal subset of $G$ by Definition 5. | ==== 3 implies 5 ====
Suppose that $S$ is a normal subset of $G$ by Definition 3.
That is:
:$\forall g \in G: g^{-1} \circ S \circ g \subseteq S$.
Let $x, y \in G$ such that $x \circ y \in S$.
Then:
{{begin-eqn}}
{{eqn | l = y \circ x
| r = e \circ \paren {y \circ x}
| c = {{Group-axiom|2}}
}}
{{eqn | r = \... | Let $\struct {G, \circ}$ be a [[Definition:group|group]].
Let $S \subseteq G$.
Then:
: $S$ is a [[Definition:Normal Subset/Definition 3|normal subset of $G$ by Definition 3]]
{{iff}}:
: $S$ is a [[Definition:Normal Subset/Definition 5|normal subset of $G$ by Definition 5]]. | ==== 3 implies 5 ====
Suppose that $S$ is a normal subset of $G$ by [[Definition:Normal Subset/Definition 3|Definition 3]].
That is:
:$\forall g \in G: g^{-1} \circ S \circ g \subseteq S$.
Let $x, y \in G$ such that $x \circ y \in S$.
Then:
{{begin-eqn}}
{{eqn | l = y \circ x
| r = e \circ \paren {y \circ x... | Equivalence of Definitions of Normal Subset/3 iff 5 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Normal_Subset/3_iff_5 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Normal_Subset/3_iff_5 | [
"Normal Subsets"
] | [
"Definition:group",
"Definition:Normal Subset/Definition 3",
"Definition:Normal Subset/Definition 5"
] | [
"Definition:Normal Subset/Definition 3",
"Definition:Normal Subset/Definition 5"
] |
proofwiki-7177 | Subgroup is Normal iff Normal Subset | Let $\left({G, \circ}\right)$ be a group.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ (by definition 1) {{iff}} it is a normal subset of $G$. | === Necessary Condition ===
Let $N$ be normal in $G$ (by definition 1):
Thus for each $g \in G$:
:$\forall g \in G: g \circ N = N \circ g$
where $g \circ N$ denotes the subset product of $g$ with $N$.
Thus $N$ is a normal subset of $G$ (by definition 1):
:$\forall g \in G: g \circ N = N \circ g$
=== Sufficient Conditio... | Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]].
Let $N$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $N$ is [[Definition:Normal Subgroup/Definition 1|normal in $G$ (by definition 1)]] {{iff}} it is a [[Definition:Normal Subset|normal subset]] of $G$. | === Necessary Condition ===
Let $N$ be [[Definition:Normal Subgroup/Definition 1|normal in $G$ (by definition 1)]]:
Thus for each $g \in G$:
:$\forall g \in G: g \circ N = N \circ g$
where $g \circ N$ denotes the [[Definition:Subset Product with Singleton|subset product of $g$ with $N$]].
Thus $N$ is a [[Definition... | Subgroup is Normal iff Normal Subset | https://proofwiki.org/wiki/Subgroup_is_Normal_iff_Normal_Subset | https://proofwiki.org/wiki/Subgroup_is_Normal_iff_Normal_Subset | [
"Conjugacy",
"Normal Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Normal Subgroup/Definition 1",
"Definition:Normal Subset"
] | [
"Definition:Normal Subgroup/Definition 1",
"Definition:Subset Product/Singleton",
"Definition:Normal Subset/Definition 1",
"Definition:Normal Subset/Definition 1",
"Definition:Normal Subgroup/Definition 1"
] |
proofwiki-7178 | Rule of Explosion/Sequent Form | :$\bot \vdash \phi$ | {{BeginTableau|\bot \vdash \phi}}
{{Premise|1|\bot}}
{{Explosion|2|1|\phi|1}}
{{EndTableau}}
{{Qed}}
Category:Rule of Explosion
qvl3j0govct9cz4i06dono4n38594tf | :$\bot \vdash \phi$ | {{BeginTableau|\bot \vdash \phi}}
{{Premise|1|\bot}}
{{Explosion|2|1|\phi|1}}
{{EndTableau}}
{{Qed}}
[[Category:Rule of Explosion]]
qvl3j0govct9cz4i06dono4n38594tf | Rule of Explosion/Sequent Form | https://proofwiki.org/wiki/Rule_of_Explosion/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Explosion/Sequent_Form | [
"Rule of Explosion"
] | [] | [
"Category:Rule of Explosion"
] |
proofwiki-7179 | Law of Excluded Middle/Sequent Form | The '''Law of Excluded Middle''' can be symbolised by the sequent:
:$\vdash p \lor \neg p$ | {{BeginTableau|\vdash p \lor \neg p}}
{{ExcludedMiddle|1|p \lor \neg p}}
{{EndTableau}}
{{Qed}} | The '''[[Law of Excluded Middle]]''' can be symbolised by the [[Definition:Sequent|sequent]]:
:$\vdash p \lor \neg p$ | {{BeginTableau|\vdash p \lor \neg p}}
{{ExcludedMiddle|1|p \lor \neg p}}
{{EndTableau}}
{{Qed}} | Law of Excluded Middle/Sequent Form/Proof 1 | https://proofwiki.org/wiki/Law_of_Excluded_Middle/Sequent_Form | https://proofwiki.org/wiki/Law_of_Excluded_Middle/Sequent_Form/Proof_1 | [
"Law of Excluded Middle"
] | [
"Law of Excluded Middle",
"Definition:Sequent"
] | [] |
proofwiki-7180 | Law of Excluded Middle/Sequent Form | The '''Law of Excluded Middle''' can be symbolised by the sequent:
:$\vdash p \lor \neg p$ | {{BeginTableau|\vdash p \lor \neg p|Instance 2 of the Hilbert-style systems}}
{{TheoremIntro|1|\paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }|Hypothetical Syllogism}}
{{TableauLine
| n = 2
| f = \paren {\paren {p \lor p} \implies p} \implies \paren {\paren {p \implies \p... | The '''[[Law of Excluded Middle]]''' can be symbolised by the [[Definition:Sequent|sequent]]:
:$\vdash p \lor \neg p$ | {{BeginTableau|\vdash p \lor \neg p|[[Definition:Hilbert Proof System/Instance 2|Instance 2 of the Hilbert-style systems]]}}
{{TheoremIntro|1|\paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }|[[Hypothetical Syllogism/Formulation 5/Proof 2|Hypothetical Syllogism]]}}
{{TableauL... | Law of Excluded Middle/Sequent Form/Proof 2 | https://proofwiki.org/wiki/Law_of_Excluded_Middle/Sequent_Form | https://proofwiki.org/wiki/Law_of_Excluded_Middle/Sequent_Form/Proof_2 | [
"Law of Excluded Middle"
] | [
"Law of Excluded Middle",
"Definition:Sequent"
] | [
"Definition:Hilbert Proof System/Instance 2",
"Hypothetical Syllogism/Formulation 5/Proof 2"
] |
proofwiki-7181 | Law of Excluded Middle/Sequent Form | The '''Law of Excluded Middle''' can be symbolised by the sequent:
:$\vdash p \lor \neg p$ | We apply the Method of Truth Tables to the proposition $\vdash p \lor \neg p$.
As can be seen by inspection, the truth value of the main connective, that is $\lor$, is $\T$ for each boolean interpretation for $p$.
:<nowiki>$\begin{array}{|c|c|cc|} \hline
p & \lor & \neg & p \\
\hline
\F & \T & \T & \F \\
\T & \T & \F &... | The '''[[Law of Excluded Middle]]''' can be symbolised by the [[Definition:Sequent|sequent]]:
:$\vdash p \lor \neg p$ | We apply the [[Method of Truth Tables]] to the proposition $\vdash p \lor \neg p$.
As can be seen by inspection, the [[Definition:Truth Value|truth value]] of the [[Definition:Main Connective (Propositional Logic)|main connective]], that is $\lor$, is $\T$ for each [[Definition:Boolean Interpretation|boolean interpret... | Law of Excluded Middle/Sequent Form/Proof by Truth Table | https://proofwiki.org/wiki/Law_of_Excluded_Middle/Sequent_Form | https://proofwiki.org/wiki/Law_of_Excluded_Middle/Sequent_Form/Proof_by_Truth_Table | [
"Law of Excluded Middle"
] | [
"Law of Excluded Middle",
"Definition:Sequent"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7182 | Power Structure of Monoid is Monoid | Let $\struct {G, \circ}$ be a monoid with identity $e$.
Let $\struct {\powerset G, \circ_\PP}$ be the power structure of $\struct {G, \circ}$.
Then $\struct {\powerset G, \circ_\PP}$ is a monoid with identity $\set e$. | By definition of a monoid, $\struct {G, \circ}$ is a semigroup.
By Power Structure of Semigroup is Semigroup, $\struct {\powerset G, \circ_\PP}$ is a semigroup.
By Subset Product by Identity Singleton, $\set e$ is an identity for $\struct {\powerset G, \circ_\PP}$.
Thus $\struct {\powerset G, \circ_\PP}$ is a monoid wi... | Let $\struct {G, \circ}$ be a [[Definition:Monoid|monoid]] with [[Definition:Identity Element|identity]] $e$.
Let $\struct {\powerset G, \circ_\PP}$ be the [[Definition:Power Structure|power structure]] of $\struct {G, \circ}$.
Then $\struct {\powerset G, \circ_\PP}$ is a [[Definition:Monoid|monoid]] with [[Definiti... | By definition of a [[Definition:Monoid|monoid]], $\struct {G, \circ}$ is a [[Definition:Semigroup|semigroup]].
By [[Power Structure of Semigroup is Semigroup]], $\struct {\powerset G, \circ_\PP}$ is a [[Definition:Semigroup|semigroup]].
By [[Subset Product by Identity Singleton]], $\set e$ is an [[Definition:Identity... | Power Structure of Monoid is Monoid | https://proofwiki.org/wiki/Power_Structure_of_Monoid_is_Monoid | https://proofwiki.org/wiki/Power_Structure_of_Monoid_is_Monoid | [
"Monoids",
"Power Structures"
] | [
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Power Structure",
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Monoid",
"Definition:Semigroup",
"Power Structure of Semigroup is Semigroup",
"Definition:Semigroup",
"Subset Product with Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Monoid",
"Category:Monoids",
"Category:Power Structures"
] |
proofwiki-7183 | Principle of Non-Contradiction/Sequent Form/Formulation 2 | :$\vdash \neg \paren {p \land \neg p}$ | {{BeginTableau|\vdash \neg \left({p \land \neg p}\right)}}
{{Assumption|1|p \land \neg p}}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|\neg p|1|2}}
{{NonContradiction|4|1|2|3}}
{{Contradiction|5||\neg \left({p \land \neg p}\right)|1|4}}
{{EndTableau|qed}} | :$\vdash \neg \paren {p \land \neg p}$ | {{BeginTableau|\vdash \neg \left({p \land \neg p}\right)}}
{{Assumption|1|p \land \neg p}}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|\neg p|1|2}}
{{NonContradiction|4|1|2|3}}
{{Contradiction|5||\neg \left({p \land \neg p}\right)|1|4}}
{{EndTableau|qed}} | Principle of Non-Contradiction/Sequent Form/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Principle_of_Non-Contradiction/Sequent_Form/Formulation_2 | https://proofwiki.org/wiki/Principle_of_Non-Contradiction/Sequent_Form/Formulation_2/Proof_1 | [
"Principle of Non-Contradiction"
] | [] | [] |
proofwiki-7184 | Principle of Non-Contradiction/Sequent Form/Formulation 2 | :$\vdash \neg \paren {p \land \neg p}$ | We apply the Method of Truth Tables to the proposition $\neg \paren {p \land \neg p}$.
As can be seen by inspection, the truth value of the main connective, that is $\neg$, is $T$ for each boolean interpretation for $p$.
:<nowiki>$\begin {array} {|ccccc|} \hline
\neg & (p & \land & \neg & p)\\
\hline
\T & \F & \F & \T ... | :$\vdash \neg \paren {p \land \neg p}$ | We apply the [[Method of Truth Tables]] to the proposition $\neg \paren {p \land \neg p}$.
As can be seen by inspection, the [[Definition:Truth Value|truth value]] of the [[Definition:Main Connective (Propositional Logic)|main connective]], that is $\neg$, is $T$ for each [[Definition:Boolean Interpretation|boolean in... | Principle of Non-Contradiction/Sequent Form/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Principle_of_Non-Contradiction/Sequent_Form/Formulation_2 | https://proofwiki.org/wiki/Principle_of_Non-Contradiction/Sequent_Form/Formulation_2/Proof_by_Truth_Table | [
"Principle of Non-Contradiction"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7185 | Power Structure of Semigroup is Semigroup | Let $\struct {S, \circ}$ be a semigroup.
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.
Then $\struct {\powerset S, \circ_\PP}$ is a semigroup. | From Power Structure of Magma is Magma we conclude that $\struct {\powerset S, \circ_\PP}$ is a magma.
It follows from Subset Product within Semigroup is Associative that $\circ_\PP$ is associative in $\struct {\powerset S, \circ_\PP}$.
Thus $\struct {\powerset S, \circ_\PP}$ is a semigroup.
{{qed}}
Category:Semigroups... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $\struct {\powerset S, \circ_\PP}$ be the [[Definition:Power Structure|power structure]] of $\struct {S, \circ}$.
Then $\struct {\powerset S, \circ_\PP}$ is a [[Definition:Semigroup|semigroup]]. | From [[Power Structure of Magma is Magma]] we conclude that $\struct {\powerset S, \circ_\PP}$ is a [[Definition:Magma|magma]].
It follows from [[Subset Product within Semigroup is Associative]] that $\circ_\PP$ is [[Definition:Associative Operation|associative]] in $\struct {\powerset S, \circ_\PP}$.
Thus $\struct {... | Power Structure of Semigroup is Semigroup | https://proofwiki.org/wiki/Power_Structure_of_Semigroup_is_Semigroup | https://proofwiki.org/wiki/Power_Structure_of_Semigroup_is_Semigroup | [
"Semigroups",
"Power Structures"
] | [
"Definition:Semigroup",
"Definition:Power Structure",
"Definition:Semigroup"
] | [
"Power Structure of Magma is Magma",
"Definition:Magma",
"Subset Product within Semigroup is Associative",
"Definition:Associative Operation",
"Definition:Semigroup",
"Category:Semigroups",
"Category:Power Structures"
] |
proofwiki-7186 | Power Structure of Group is Monoid | Let $\struct {G, \circ}$ be a group with identity $e$.
Let $\struct {\powerset G, \circ_\PP}$ be the power structure of $\struct {G, \circ}$.
Then $\struct {\powerset G, \circ_\PP}$ is a monoid with identity $\set e$. | By definition of a group, $\struct {G, \circ}$ is a monoid.
The result follows from Power Structure of Monoid is Monoid.
{{qed}}
Category:Power Structures
Category:Monoids
Category:Group Theory
4mllzcji74j6twhcs98ka2qafla6gy9 | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] with [[Definition:Identity Element|identity]] $e$.
Let $\struct {\powerset G, \circ_\PP}$ be the [[Definition:Power Structure|power structure]] of $\struct {G, \circ}$.
Then $\struct {\powerset G, \circ_\PP}$ is a [[Definition:Monoid|monoid]] with [[Definition... | By definition of a [[Definition:Group|group]], $\struct {G, \circ}$ is a [[Definition:Monoid|monoid]].
The result follows from [[Power Structure of Monoid is Monoid]].
{{qed}}
[[Category:Power Structures]]
[[Category:Monoids]]
[[Category:Group Theory]]
4mllzcji74j6twhcs98ka2qafla6gy9 | Power Structure of Group is Monoid | https://proofwiki.org/wiki/Power_Structure_of_Group_is_Monoid | https://proofwiki.org/wiki/Power_Structure_of_Group_is_Monoid | [
"Power Structures",
"Monoids",
"Group Theory"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Power Structure",
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Group",
"Definition:Monoid",
"Power Structure of Monoid is Monoid",
"Category:Power Structures",
"Category:Monoids",
"Category:Group Theory"
] |
proofwiki-7187 | Power Structure of Magma is Magma | Let $\struct {S, \circ}$ be a magma.
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.
Then $\struct {\powerset S, \circ_\PP}$ is a magma.
That is, $\circ_\PP$ is closed in $\powerset S$. | Let $\struct {S, \circ}$ be a magma.
Let $A, B \subseteq S$.
{{begin-eqn}}
{{eqn | q = \forall a \in A, b \in B
| l = a \circ b
| o = \in
| r = S
| c = {{Defof|Magma}}
}}
{{eqn | ll= \leadsto
| l = A \circ B
| o = \subseteq
| r = S
| c = {{Defof|Subset Product}}
}}
{{eqn ... | Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
Let $\struct {\powerset S, \circ_\PP}$ be the [[Definition:Power Structure|power structure]] of $\struct {S, \circ}$.
Then $\struct {\powerset S, \circ_\PP}$ is a [[Definition:Magma|magma]].
That is, $\circ_\PP$ is [[Definition:Closed Operation|closed]] in $... | Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
Let $A, B \subseteq S$.
{{begin-eqn}}
{{eqn | q = \forall a \in A, b \in B
| l = a \circ b
| o = \in
| r = S
| c = {{Defof|Magma}}
}}
{{eqn | ll= \leadsto
| l = A \circ B
| o = \subseteq
| r = S
| c = {{Defof|Sub... | Power Structure of Magma is Magma | https://proofwiki.org/wiki/Power_Structure_of_Magma_is_Magma | https://proofwiki.org/wiki/Power_Structure_of_Magma_is_Magma | [
"Power Structures",
"Magmas"
] | [
"Definition:Magma",
"Definition:Power Structure",
"Definition:Magma",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Definition:Magma",
"Definition:Magma",
"Category:Power Structures",
"Category:Magmas"
] |
proofwiki-7188 | Element in Left Coset iff Product with Inverse in Subgroup | Let $y H$ denote the left coset of $H$ by $y$.
Then:
:$x \in y H \iff x^{-1} y \in H$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = y H
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \exists h \in H
| l = x
| r = y h
| c = {{Defof|Left Coset}}
}}
{{eqn | ll= \leadstoandfrom
| q = \exists h \in H
| l = x^{-1}
| r = h^{-1} y^{-1}
| c = Inverse of... | Let $y H$ denote the [[Definition:Left Coset|left coset]] of $H$ by $y$.
Then:
:$x \in y H \iff x^{-1} y \in H$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = y H
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \exists h \in H
| l = x
| r = y h
| c = {{Defof|Left Coset}}
}}
{{eqn | ll= \leadstoandfrom
| q = \exists h \in H
| l = x^{-1}
| r = h^{-1} y^{-1}
| c = [[Inverse ... | Element in Left Coset iff Product with Inverse in Subgroup | https://proofwiki.org/wiki/Element_in_Left_Coset_iff_Product_with_Inverse_in_Subgroup | https://proofwiki.org/wiki/Element_in_Left_Coset_iff_Product_with_Inverse_in_Subgroup | [
"Cosets"
] | [
"Definition:Coset/Left Coset"
] | [
"Inverse of Group Product",
"Definition:Subgroup"
] |
proofwiki-7189 | Element in Right Coset iff Product with Inverse in Subgroup | Let $H \circ y$ denote the right coset of $H$ by $y$.
Then:
:$x \in H y \iff x y^{-1} \in H$ | Let $\struct {G, *}$ be the opposite group of $G$.
Then:
:$x \in H y \iff x \in y * H$
:$x y^{-1} \in H \iff y^{-1} * x \in H$
Since $H$ is closed under inverses:
:$x y^{-1} \in H \iff x^{-1} * y \in H$
By Element in Left Coset iff Product with Inverse in Subgroup:
:$x \in y * H \iff x^{-1} * y \in H$
Hence the resul... | Let $H \circ y$ denote the [[Definition:Right Coset|right coset]] of $H$ by $y$.
Then:
:$x \in H y \iff x y^{-1} \in H$ | Let $\struct {G, *}$ be the [[Definition:Opposite Group|opposite group]] of $G$.
Then:
:$x \in H y \iff x \in y * H$
:$x y^{-1} \in H \iff y^{-1} * x \in H$
Since $H$ is closed under inverses:
:$x y^{-1} \in H \iff x^{-1} * y \in H$
By [[Element in Left Coset iff Product with Inverse in Subgroup]]:
:$x \in y * H \... | Element in Right Coset iff Product with Inverse in Subgroup | https://proofwiki.org/wiki/Element_in_Right_Coset_iff_Product_with_Inverse_in_Subgroup | https://proofwiki.org/wiki/Element_in_Right_Coset_iff_Product_with_Inverse_in_Subgroup | [
"Cosets"
] | [
"Definition:Coset/Right Coset"
] | [
"Definition:Opposite Group",
"Element in Left Coset iff Product with Inverse in Subgroup"
] |
proofwiki-7190 | Inverse of Inverse of Subset of Group | Let $\struct {G, \circ}$ be a group.
Let $X \subseteq G$.
Then:
:$\paren {X^{-1} }^{-1} = X$.
where $X^{-1}$ denotes the inverse of $X$. | {{begin-eqn}}
{{eqn | l = \paren {X^{-1} }^{-1}
| r = \set {x^{-1}: x \in X^{-1} }
| c = {{Defof|Inverse of Subset of Group}}
}}
{{eqn | r = \set {\paren {x^{-1} }^{-1}: x \in X}
| c = {{Defof|Inverse of Subset of Group}}
}}
{{eqn | r = \set {x: x \in X}
| c = Inverse of Group Inverse
}}
{{eqn |... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $X \subseteq G$.
Then:
:$\paren {X^{-1} }^{-1} = X$.
where $X^{-1}$ denotes the [[Definition:Inverse of Subset of Group|inverse of $X$]]. | {{begin-eqn}}
{{eqn | l = \paren {X^{-1} }^{-1}
| r = \set {x^{-1}: x \in X^{-1} }
| c = {{Defof|Inverse of Subset of Group}}
}}
{{eqn | r = \set {\paren {x^{-1} }^{-1}: x \in X}
| c = {{Defof|Inverse of Subset of Group}}
}}
{{eqn | r = \set {x: x \in X}
| c = [[Inverse of Group Inverse]]
}}
{{e... | Inverse of Inverse of Subset of Group | https://proofwiki.org/wiki/Inverse_of_Inverse_of_Subset_of_Group | https://proofwiki.org/wiki/Inverse_of_Inverse_of_Subset_of_Group | [
"Group Theory"
] | [
"Definition:Group",
"Definition:Inverse of Subset/Group"
] | [
"Inverse of Group Inverse",
"Category:Group Theory"
] |
proofwiki-7191 | Exclusive Or is Commutative | : $p \oplus q \dashv \vdash q \oplus p$ | {{BeginTableau|p \oplus q \vdash q \oplus p}}
{{Premise|1|p \oplus q}}
{{SequentIntro|2|1|\left({p \lor q} \right) \land \neg \left({p \land q}\right)|1|Definition of Exclusive Or}}
{{Commutation|3|1|\left({q \lor p} \right) \land \neg \left({p \land q}\right)|2|Disjunction}}
{{Commutation|4|1|\left({q \lor p} \right) ... | : $p \oplus q \dashv \vdash q \oplus p$ | {{BeginTableau|p \oplus q \vdash q \oplus p}}
{{Premise|1|p \oplus q}}
{{SequentIntro|2|1|\left({p \lor q} \right) \land \neg \left({p \land q}\right)|1|Definition of [[Non-Equivalence|Exclusive Or]]}}
{{Commutation|3|1|\left({q \lor p} \right) \land \neg \left({p \land q}\right)|2|Disjunction}}
{{Commutation|4|1|\left... | Exclusive Or is Commutative/Proof 1 | https://proofwiki.org/wiki/Exclusive_Or_is_Commutative | https://proofwiki.org/wiki/Exclusive_Or_is_Commutative/Proof_1 | [
"Exclusive Or",
"Exclusive Or is Commutative"
] | [] | [
"Non-Equivalence",
"Non-Equivalence",
"Non-Equivalence",
"Non-Equivalence"
] |
proofwiki-7192 | Exclusive Or is Commutative | : $p \oplus q \dashv \vdash q \oplus p$ | 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 & \oplus & q & q & \oplus & p \\
\hline
F & F & F & F & F & F \\
F & T & T & T & T & F \\
T & T & F & F & T & T \\
T & F & T & T & F &... | : $p \oplus q \dashv \vdash q \oplus p$ | 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]].
$\begin{array}{|ccc||ccc|} \hline
p & \oplus ... | Exclusive Or is Commutative/Proof 2 | https://proofwiki.org/wiki/Exclusive_Or_is_Commutative | https://proofwiki.org/wiki/Exclusive_Or_is_Commutative/Proof_2 | [
"Exclusive Or",
"Exclusive Or is Commutative"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7193 | Exclusive Or is Associative | Exclusive or is associative:
:$p \oplus \paren {q \oplus r} \dashv \vdash \paren {p \oplus q} \oplus 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.
$\begin{array}{|ccccc||ccccc|} \hline
p & \oplus & (q & \oplus & r) & (p & \oplus & q) & \oplus & r \\
\hline
\F & \F & \F & \F & \F & \F & \F & \F & \F & \F \\
\F & \T &... | [[Definition:Exclusive Or|Exclusive or]] is [[Definition:Associative Operation|associative]]:
:$p \oplus \paren {q \oplus r} \dashv \vdash \paren {p \oplus q} \oplus 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]].
$\begin{array}{|ccccc||ccccc|} \hline
p & \op... | Exclusive Or is Associative | https://proofwiki.org/wiki/Exclusive_Or_is_Associative | https://proofwiki.org/wiki/Exclusive_Or_is_Associative | [
"Exclusive Or",
"Truth Table Proofs"
] | [
"Definition:Exclusive Or",
"Definition:Associative Operation"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7194 | Exclusive Or with Itself | Exclusive or destroys copies of itself:
:$p \oplus p \dashv \vdash \bot$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective match for each boolean interpretations.
$\begin{array}{|ccc|} \hline
p & \oplus & p \\
\hline
\F & \F & \F \\
\T & \F & \T \\
\hline
\end{array}$
{{qed}} | [[Definition:Exclusive Or|Exclusive or]] destroys copies of itself:
:$p \oplus p \dashv \vdash \bot$ | 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 each [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|ccc|} \hline
p & \oplus & p \\... | Exclusive Or with Itself | https://proofwiki.org/wiki/Exclusive_Or_with_Itself | https://proofwiki.org/wiki/Exclusive_Or_with_Itself | [
"Exclusive Or",
"Truth Table Proofs"
] | [
"Definition:Exclusive Or"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7195 | Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Proof 1 | : $\neg \left ({p \iff q}\right) \dashv \vdash \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$ | === Forward Implication: Proof ===
{{:Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof}}
=== Reverse Implication: Proof ===
{{:Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof}} | : $\neg \left ({p \iff q}\right) \dashv \vdash \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$ | === [[Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof|Forward Implication: Proof]] ===
{{:Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof}}
=== [[Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof|Reverse I... | Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1/Proof_1 | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1/Proof_1 | [
"Non-Equivalence as Disjunction of Conjunctions"
] | [] | [
"Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof",
"Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof"
] |
proofwiki-7196 | Non-Equivalence as Disjunction of Conjunctions/Formulation 1 | :$\neg \paren {p \iff q} \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$ | {{BeginTableau|\neg \paren {p \iff q} \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q} }}
{{Premise|1|\neg \paren {p \iff q} }}
{{SequentIntro|2|1|\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }|1
|Rule of Material Equivalence}}
{{SequentIntro|3|1|\neg \paren {\paren {\neg p \lor q} \land ... | :$\neg \paren {p \iff q} \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$ | {{BeginTableau|\neg \paren {p \iff q} \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q} }}
{{Premise|1|\neg \paren {p \iff q} }}
{{SequentIntro|2|1|\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }|1
|[[Rule of Material Equivalence]]}}
{{SequentIntro|3|1|\neg \paren {\paren {\neg p \lor q} \l... | Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1 | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1/Forward_Implication/Proof | [
"Non-Equivalence as Disjunction of Conjunctions"
] | [] | [
"Rule of Material Equivalence",
"Rule of Material Implication"
] |
proofwiki-7197 | Non-Equivalence as Disjunction of Conjunctions/Formulation 1 | :$\neg \paren {p \iff q} \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$ | === Forward Implication: Proof ===
{{:Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof}}
=== Reverse Implication: Proof ===
{{:Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof}} | :$\neg \paren {p \iff q} \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$ | === [[Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof|Forward Implication: Proof]] ===
{{:Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof}}
=== [[Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof|Reverse I... | Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1 | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1/Proof_1 | [
"Non-Equivalence as Disjunction of Conjunctions"
] | [] | [
"Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof",
"Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof"
] |
proofwiki-7198 | Non-Equivalence as Disjunction of Conjunctions/Formulation 1 | :$\neg \paren {p \iff q} \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$ | 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>$\begin{array}{|cccc||ccccccccc|} \hline
\neg & (p & \iff & q) & (\neg & p & \land & q) & \lor & (p & \land & \neg & q) \\
\hline
\F & \F & \T & \F & \T & \F & \... | :$\neg \paren {p \iff q} \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$ | 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>$\begin{array}{|cccc||ccccccccc|} \hl... | Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1 | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1/Proof_by_Truth_Table | [
"Non-Equivalence as Disjunction of Conjunctions"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7199 | Non-Equivalence as Disjunction of Conjunctions/Formulation 1 | :$\neg \paren {p \iff q} \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$ | {{BeginTableau|\paren {\neg p \land q} \lor \paren {p \land \neg q} \vdash \neg \paren {p \iff q} }}
{{Premise|1|\paren {\neg p \land q} \lor \paren {p \land \neg q} }}
{{Commutation|2|1|\paren {p \land \neg q} \lor \paren {\neg p \land q}|1|Disjunction}}
{{Commutation|3|1|\paren {p \land \neg q} \lor \paren {q \land \... | :$\neg \paren {p \iff q} \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$ | {{BeginTableau|\paren {\neg p \land q} \lor \paren {p \land \neg q} \vdash \neg \paren {p \iff q} }}
{{Premise|1|\paren {\neg p \land q} \lor \paren {p \land \neg q} }}
{{Commutation|2|1|\paren {p \land \neg q} \lor \paren {\neg p \land q}|1|Disjunction}}
{{Commutation|3|1|\paren {p \land \neg q} \lor \paren {q \land \... | Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1 | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1/Reverse_Implication/Proof | [
"Non-Equivalence as Disjunction of Conjunctions"
] | [] | [
"Rule of Material Implication",
"Rule of Material Equivalence"
] |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.