hcju/proofwikiqa huggingface dataset:
{{BeginTableau|\left({\neg p \land \neg q}\right) \implies \left({\neg \left({p \lor q}\right)}\right)}} {{Assumption|1|\neg p \land \neg q}} {{SequentIntro|2|1|\neg \left({p \lor q}\right)|1|De Morgan's Laws (Logic): Conjunction of Negations: Formulation 1}} {{Implication|3||\left({\neg p \land \neg q}\right) \implies \left({\neg \left({p \lor q}\right)}\right)|1|2}} {{EndTableau}} {{qed}}
{{BeginTableau|\vdash \left({\neg \left({p \lor q}\right)}\right) \implies \left({\neg p \land \neg q}\right)}} {{Assumption|1|\neg \left({p \lor q}\right)}} {{SequentIntro|2|1|\neg p \land \neg q|1|De Morgan's Laws (Logic): Conjunction of Negations: Formulation 1}} {{Implication|3||\left({\neg \left({p \lor q}\right)}\right) \implies \left({\neg p \land \neg q}\right)|1|2}} {{EndTableau}} {{qed}} Category:De Morgan's Laws (Logic) lz2c718vtnmt72duwxve8dch6pc4scd
=== Definition $(1)$ implies Definition $(3)$ === {{transclude:Equivalence of Definitions of Generalized Ordered Space/Definition 1 implies Definition 3|section=proof}} === Definition $(3)$ implies Definition $(1)$ === {{transclude:Equivalence of Definitions of Generalized Ordered Space/Definition 3 implies Definition 1|section=proof}} === Definition $(2)$ implies Definition $(1)$ === {{:Equivalence of Definitions of Generalized Ordered Space/Definition 2 implies Definition 1}} === Definition $(3)$ implies Definition $(2)$ === This follows from GO-Space Embeds Densely into Linearly Ordered Space. {{qed}} Category:Generalized Ordered Spaces Category:Equivalence of Definitions of Generalized Ordered Space hazwxe6hpm2z8xwd4rpix6pxci8o2b8
Let $x \in U \in \tau$. Then by the definition of topological embedding: :$\map \phi U$ is an open neighborhood of $\map \phi x$ in $\map \phi S$ with the subspace topology. Thus by Basis for Topological Subspace and the definition of the order topology, there is an open interval or open ray $I' \in \tau'$ such that: :$\map \phi x \in I' \cap \map \phi S \subseteq \map \phi U$ Since $I'$ is an interval or ray, it is convex in $S'$ by Interval of Ordered Set is Convex or Ray is Convex, respectively. Then: {{begin-eqn}} {{eqn | l = x \in \phi^{-1} \sqbrk {I'}       | r = \phi^{-1} \sqbrk {I' \cap \phi \sqbrk S}       | c =  }} {{eqn | o = \subseteq       | r = \phi^{-1} \sqbrk {\phi \sqbrk U}       | c =  }} {{end-eqn}} Because $\phi$ is a topological embedding, it is injective by definition. So: :$\phi^{-1} \sqbrk {\phi \sqbrk U} = U$ Thus: :$x \in \phi^{-1} \sqbrk {I'} \subseteq U$ By Inverse Image of Convex Set under Monotone Mapping is Convex: :$\phi^{-1} \sqbrk {I'}$ is convex. Thus $\tau$ has a basis consisting of convex sets. {{qed}}
Suppose $f$ is increasing. Let $a, b, c \in X$ such that $a \le b \le c$. Let $a, c \in f^{-1} \left[{C}\right]$. By definition of inverse image: : $f \left({a}\right), f \left({c}\right) \in C$ By definition of increasing mapping: :$f \left({a}\right) \preceq f \left({b}\right) \preceq f \left({c}\right)$ Thus by definition of convex set: :$f \left({b}\right) \in C$ Then by definition of inverse image: :$b \in f^{-1} \left[{C}\right]$ Since this holds for all such triples, $f^{-1} \left[{C}\right]$ is convex in $X$. A similar argument applies if $f$ is decreasing. {{qed}} Category:Order Theory fetuvrsxewava1wwlw6q1fh3x889jej
{{BeginTableau|\vdash p \iff \paren {p \land q} \lor \paren {p \land \neg q} }} {{Assumption|1|p}} {{ExcludedMiddle|2|q \lor \neg q}} {{Assumption|3|q}} {{Conjunction|4|1, 3|p \land q|1|2}} {{Addition|5|1, 3|\paren {p \land q} \lor \paren {p \land \neg q}|4|1}} {{Assumption|6|\neg q}} {{Conjunction|7|1, 6|p \land \neg q|1|6}} {{Addition|8|1, 6|\paren {p \land q} \lor \paren {p \land \neg q}|7|2}} {{ProofByCases|9|1|\paren {p \land q} \lor \paren {p \land \neg q}|2|3|5|6|8}} {{Implication|10||p \implies \paren {p \land q} \lor \paren {p \land \neg q}|1|9}} {{Assumption|11|\paren {p \land q} \lor \paren {p \land \neg q} }} {{SequentIntro|12|11|p \land \paren {q \lor \neg q}|11|Conjunction Distributes over Disjunction}} {{Simplification|13|11|p|11|2}} {{Implication|14||\paren {p \land q} \lor \paren {p \land \neg q} \implies p|11|13}} {{BiconditionalIntro|15||p \iff \paren {p \land q} \lor \paren {p \land \neg q}|12|14}} {{EndTableau}} {{qed}}
{{BeginTableau|\vdash p \iff \left({p \lor q}\right) \land \left({p \lor \neg q}\right)}} {{Assumption|1|p}} {{Addition|2|1|p \lor q|1|1}} {{Addition|3|1|p \lor \neg q|1|2}} {{Conjunction|4|1|\left({p \lor q}\right) \land \left({p \lor \neg q}\right)|2|3}} {{Implication|5||p \implies \left({p \lor q}\right) \land \left({p \lor \neg q}\right)|1|4}} {{Assumption|6|\left({p \lor q}\right) \land \left({p \lor \neg q}\right)}} {{SequentIntro|7|6|p \lor \left({q \land \neg q}\right)|6|Disjunction Distributes over Conjunction}} {{Assumption|8|p}} {{TheoremIntro|9|\neg \left({q \land \neg q}\right)|Principle of Non-Contradiction: Formulation 2}} {{ModusTollendoPonens|10|6|p|7|9|2}} {{ProofByCases|11|6|p|6|8|8|9|10}} {{Implication|12||\left({p \lor q}\right) \land \left({p \lor \neg q}\right) \implies p|6|11}} {{BiconditionalIntro|13||p \iff \left({p \lor q}\right) \land \left({p \lor \neg q}\right)|5|12}} {{EndTableau}} {{qed}}
The cases for upward-pointing and downward-pointing rays are equivalent. {{explain|"Dual of convex is convex" and duality, upon which the above statement depends.}} {{WLOG}}, suppose that $U$ is an upward-pointing ray. By the definition of a ray, there exists an $a \in S$ such that: :$I = a^\succ$ or; :$I = a^\succeq$ according to whether $U$ is open or closed. Let $x, y, z \in S$ such that $x \prec y \prec z$ and $x, z \in I$. Then: :$a \preceq x \prec y$ so: :$a \prec y$ Therefore: :$y \in a^\succ \subseteq I$ Thus $I$ is convex. {{qed}}
Let $a, b, c \in S$. Let $a, c \in \displaystyle \bigcap \mathcal C$. Let $a \prec b \prec c$. By the definition of intersection: :$\forall C \in \mathcal C$: $a, c \in C$ Since each $C \in C$ is convex: :$\forall C \in \mathcal C$: $b \in C$. Thus by the definition of intersection: :$b \in \displaystyle \bigcap \mathcal C$ Thus $\displaystyle \bigcap \mathcal C$ is convex. {{qed}} Category:Convex Sets 8dmimjz3fi4bkv3mxh4bpiye8k4tqi5
Any interval can be represented as the intersection of two rays. {{explain|Obvious though it is, the above needs to be stated as a theorem in its own right.}} Thus by Ray is Convex and Intersection of Convex Sets is Convex Set (Order Theory), $I$ is convex. {{qed}} Category:Order Theory kob8akch5iwfbh8kkzmoembt3sost1s
The cases for upper and lower closures are dual, so we need only prove the case for upper closures. Suppose, then, that $C = a^\succeq$ or $C = a^\succ$. Suppose that $x, y, z \in S$, $x \prec y \prec z$, and $x, z \in C$. Then $a \preceq x \prec y$, so $a \prec y$ by Extended Transitivity. Therefore $y \in a^\succ \subseteq C$. Thus $C$ is convex. {{qed}} Category:Lower Closures Category:Upper Closures Category:Convex Sets 1ubqkqahcv5dlic8bk0xfbemucj22th
Let $a, b \in S$ with $a \mathrel{\mathcal T} b$. By the definition of transitive closure, there is an $n \in \N$ such that $a \mathrel{\mathcal R^n} b $. Thus there are $x_0, x_1, \dots x_n \in S$ such that: : $x_0 = a$ : $x_n = b$ : For $k = 0, \dots, n-1$: $x_k \mathrel{\mathcal R} x_{k+1}$ For $k = 0, \dots, n$, let $y_k = x_{n-k}$. Then: : $y_0 = x_n = b$ : $y_n = x_0 = a$ For $k = 0, \dots, n-1$: $x_{n-k-1} \mathrel{\mathcal R} x_{n-k}$. Thus $y_{k+1} \mathrel{\mathcal R} y_k$. Since $\mathcal R$ is symmetric: : For $k = 0, \dots, n-1$: $y_k \mathrel{\mathcal R} y_{k+1}$ Thus $b \mathrel{\mathcal R^n} a$, so $b \mathrel{\mathcal T} a$. {{qed}} Category:Symmetric Relations Category:Transitive Closures kw5062bup14w60mar3g1krnuhq44vfa
Let $a \in S$. Since $\mathcal R$ is reflexive: :$\left({a, a}\right) \in \mathcal R$ By the definition of transitive closure: :$\mathcal R \subseteq \mathcal T$ Thus by the definition of subset: :$\left({a, a}\right) \in \mathcal T$ Since this holds for all $a \in S, \mathcal T$ is reflexive. {{qed}} Category:Reflexive Relations Category:Transitive Closures ivk48ffpye1ebp4yjc0cc6v3qja04eb
Let $\sim$ be the transitive closure of $\mathcal R$. Checking in turn each of the criteria for equivalence: === Reflexivity === By Transitive Closure of Reflexive Relation is Reflexive: : $\sim$ is reflexive. {{qed|lemma}} === Symmetry === By Transitive Closure of Symmetric Relation is Symmetric: : $\sim$ is symmetric. {{qed|lemma}} === Transitivity === By the definition of transitive closure: : $\sim$ is transitive. {{qed|lemma}} $\sim$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}} Category:Transitive Closures Category:Equivalence Relations o62uk3c5ms86bq9v4zgnbs4mju6yw3n
Let $a,b,c \in S$ Let $a,c \in U \cup V$. Let $a \prec b \prec c$. If $a,c \in U$ then $b \in U$ because $U$ is convex. Thus $b \in U \cup V$ by the definition of union. Similarly, if $a,c \in V$ then $b \in U \cup V$. Otherwise, suppose WLOG that $a \in U$ and $c \in V$. Since $U \cap V$ is nonempty by the premise, it has an element $p$. Since $\preceq$ is a total ordering: :$b \preceq p$ or $p \preceq b$. If $b \preceq p$, then since $a \prec b$, $a,p \in U$, and $U$ is convex, we can conclude that :$b \in U$ so $b \in U \cup V$. A similar argument shows that it $p \preceq b$ then $b \in V$, so $b \in U \cup V$. Thus in all cases we can conclude that $b \in U \cup V$, so $U \cup V$ is convex. {{qed}} Category:Total Orderings phpaeazpqhrvjvlm7bqfm97jnvw0amk
Let $a, c \in \bigcup \mathcal A$. Let $b \in S$. Let $a \prec b \prec c$. Since $a, c \in \bigcup \mathcal A$, there are $P, Q \in \mathcal A$ such that $a \in P$ and $c \in Q$. By the premise, there are elements $C_0, \dots, C_n \in \mathcal A$ such that: :$C_0 = P$ :$C_n = Q$ :For $k = 0, \dots, n-1: C_k \cap C_{k+1} \ne \varnothing$ {{explain|details of induction. Consider putting the finite chain case into a separate lemma.}} Applying Union of Overlapping Intervals is Interval inductively: :$\displaystyle \bigcup_{k \mathop = 0}^n C_k$ is convex. Since $\displaystyle a, c \in \bigcup_{k \mathop = 0}^n C_k$, by the definition of convexity: :$\displaystyle b \in \bigcup_{k \mathop = 0}^n C_k$ Thus: :$\displaystyle b \in \bigcup \mathcal A$ Since this holds for all such triples $a, b, c$, it follows that $\mathcal A$ is convex. {{qed}} Category:Total Orderings 0zqc4uvgtwv16acj56off7wr2zf6zy0
From Involution is Permutation, $f$ is a permutation and so a bijection. Since $f$ is continuous, it suffices to verify that its inverse is also continuous. Now recall $f$ is an involution, that is, $f^{-1} = f$. Thus $f^{-1}$ is also continuous. Hence $f$ is a homeomorphism. {{qed}} Category:Continuous Mappings Category:Homeomorphisms Category:Involutions 2djq8sagezscmf2b3kgfi523p99kpm4
By definition, a permutation is a bijection from a set to itself. Thus it is sufficient to show that $f$ is a bijection. By definition of involution, for each $x \in S$: :$\map f {\map f x} = x$ By Equality of Mappings: :$f \circ f = I_S$ where $I_S$ is the identity mapping on $S$. Thus $f$ is both a left inverse and a right inverse of itself. The result follows from Bijection iff Left and Right Inverse. {{qed}} Category:Involutions Category:Permutation Theory kpxtxhpxzu12jovs9dcovzqnrd9bn2g
{{BeginTableau|\neg \left ({p \iff q}\right) \vdash \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)}} {{Premise|1|\neg \left ({p \iff q}\right)}} {{SequentIntro|2|1|\neg \left({\left ({p \implies q}\right) \land \left ({q \implies p}\right)}\right)|1   |Rule of Material Equivalence}} {{SequentIntro|3|1|\neg \left({\left ({\neg p \lor q}\right) \land \left ({\neg q \lor p}\right)}\right)|2   |Rule of Material Implication (twice)}} {{DeMorgan|4|1|\neg \left({\neg p \lor q}\right) \lor \neg \left ({\neg q \lor p}\right)|3|Disjunction of Negations}} {{DeMorgan|5|1|\left ({\neg \neg  p \land \neg q}\right) \lor \left ({\neg \neg q \land \neg p}\right)|4|Conjunction of Negations}} {{DoubleNegElimination|6|1|\left ({p \land \neg q}\right) \lor \left ({q \land \neg p}\right)|5}} {{Commutation|7|1|\left ({p \land \neg q}\right) \lor \left ({\neg p \land q}\right)|6|Conjunction}} {{Commutation|8|1|\left ({\neg p \land q}\right) \lor \left ({p \land \neg q}\right)|7|Disjunction}} {{EndTableau}} {{qed}} {{LEM|Double Negation Elimination}}
{{BeginTableau|\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right) \vdash \neg \left ({p \iff q}\right)}} {{Premise|1|\left ({\neg p \land q}\right) \lor \left ({p \land \neg q}\right)}} {{Commutation|2|1|\left ({p \land \neg q}\right) \lor \left ({\neg p \land q}\right)|1|Disjunction}} {{Commutation|3|1|\left ({p \land \neg q}\right) \lor \left ({q \land \neg p}\right)|2|Conjunction}} {{DoubleNegIntro|4|1|\left ({\neg \neg  p \land \neg q}\right) \lor \left ({\neg \neg q \land \neg p}\right)|3}} {{DeMorgan|5|1|\neg \left({\neg p \lor q}\right) \lor \neg \left ({\neg q \lor p}\right)|4|Conjunction of Negations}} {{DeMorgan|6|1|\neg \left({\left ({\neg p \lor q}\right) \land \left ({\neg q \lor p}\right)}\right)|5|Disjunction of Negations}} {{SequentIntro|7|1  |\neg \left({\left ({p \implies q}\right) \land \left ({q \implies p}\right)}\right)|6   |Rule of Material Implication (twice)}} {{SequentIntro|8|1|\neg \left ({p \iff q}\right)|7|Rule of Material Equivalence}} {{EndTableau}} {{qed}}
{{BeginTableau|\vdash \paren {\neg \paren {p \iff q} } \iff \paren {\paren {\neg p \land q} \lor \paren {p \land \neg q} } }} {{Assumption|1|\neg \paren {p \iff q} }} {{SequentIntro|2|1|\paren {\neg p \land q} \lor \paren {p \land \neg q}|1|Non-Equivalence as Disjunction of Conjunctions: Formulation 1}} {{Implication|3||\paren {\neg \paren {p \iff q} } \implies \paren {\paren {\neg p \land q} \lor \paren {p \land \neg q} }|1|2}} {{Assumption|4|\paren {\neg p \land q} \lor \paren {p \land \neg q} }} {{SequentIntro|5|4|\neg \paren {p \iff q}|4|Non-Equivalence as Disjunction of Conjunctions: Formulation 1}} {{Implication|6||\paren {\paren {\neg p \land q} \lor \paren {p \land \neg q} } \implies \paren {\neg \paren {p \iff q} }|4|5}} {{BiconditionalIntro|7||\paren {\neg \paren {p \iff q} } \iff \paren {\paren {\neg p \land q} \lor \paren {p \land \neg q} }|3|6}} {{EndTableau|qed}}
{{BeginTableau|\neg \paren {p \iff q} \vdash \paren {p \iff \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}} {{DeMorgan            | 3|1|\neg \paren {p \implies q} \lor \neg \paren {q \implies p} |2|Disjunction of Negations}} {{Assumption          | 4|p}} {{IdentityLaw         | 5|4|p|4}} {{Assumption          | 6|q}} {{Implication         | 7|6|p \implies q|4|6}} {{DoubleNegIntro      | 8|6|\neg \neg \paren {p \implies q}|7}} {{ModusTollendoPonens | 9|1, 6|\neg \paren {q \implies p}|3|8|1}} {{SequentIntro        |10|1, 6|q \land \neg p|9|Conjunction with Negative Equivalent to Negation of Implication}} {{Simplification      |11|1, 6|\neg p|10|2}} {{NonContradiction    |12|1, 4, 6|4|11}} {{Contradiction       |13|1, 4|\neg q|6|12}} {{Implication         |14|1|p \implies \neg q|4|13}} {{Assumption          |15|\neg q}} {{IdentityLaw         |16|15|\neg q|15}} {{Assumption          |17|\neg p}} {{Implication         |18|17|\neg q \implies \neg p|16|17}} {{SequentIntro        |19|17|p \implies q|18|Rule of Transposition}} {{ModusTollendoPonens |20|1, 17|\neg \paren {q \implies p} |3|19|1}} {{SequentIntro        |21|1, 17|q \land \neg p|20|Conjunction with Negative Equivalent to Negation of Implication}} {{Simplification      |22|1, 17|q|21|1}} {{NonContradiction    |23|1, 15, 17|15|22}} {{Contradiction       |24|1, 15|p|17|23}} {{Implication         |25|1|\neg q \implies p|15|24}} {{BiconditionalIntro  |26|1|\paren {p \iff \neg q}|14|25}} {{EndTableau|qed}} {{LEM||3}}
{{BeginTableau | \paren {p \iff \neg q} \vdash \neg \paren {p \iff q} }} {{Premise      | 1 | p \iff \neg q}} {{SequentIntro | 2 | 1 |\neg \paren {p \iff \neg \neg q} | 1 | Non-Equivalence as Equivalence with Negation: Forward Implication}} {{TheoremIntro | 3 | \neg \neg q \iff q | Double Negation}} {{SequentIntro | 4 | 1 |\neg \paren {p \iff q} |2, 3| Biconditional is Transitive}} {{EndTableau}}
=== Forward Implication: Proof === {{:Non-Equivalence as Equivalence with Negation/Formulation 1/Forward Implication/Proof}} === Reverse Implication: Proof === {{:Non-Equivalence as Equivalence with Negation/Formulation 1/Reverse Implication/Proof}}
From Continuous Real Function is Darboux Integrable, the definite integrals under discussion are guaranteed to exist. Consider the case where $\forall x \in \closedint a b: \map f x \ge 0$. Define a constant mapping: :$f_0: \closedint a b \to \R$: :$\map {f_0} x = 0$ Then: {{begin-eqn}} {{eqn | l = \map {f_0} x       | o = \le       | r = \map f x       | c = for any $x \in \closedint a b$: recall $\map f x \ge 0$ }} {{eqn | ll= \leadsto       | l = \int_a^b \map {f_0} x \rd x       | o = \le       | r = \int_a^b \map f x \rd x       | c = Relative Sizes of Definite Integrals }} {{eqn | ll= \leadsto       | l = 0 \paren {b - a}       | o = \le       | r = \int_a^b \map f x \rd x       | c = Integral of Constant }} {{eqn | ll= \leadsto       | l = \int_a^b \map f x \rd x       | o = \ge       | r = 0 }} {{end-eqn}} The proofs of the other cases are similar. {{qed}} Category:Integral Calculus 2b6zbio87ls85qmqge23xxn4rety4a6
{{BeginTableau|\vdash \neg \left ({p \iff q}\right) \iff \left({p \iff \neg q}\right)}} {{Assumption|1|\neg \left ({p \iff q}\right)}} {{SequentIntro|2|1|p \iff \neg q|1|Non-Equivalence as Equivalence with Negation: Formulation 1}} {{Implication|3||\left({\neg \left ({p \iff q}\right)}\right) \implies \left({p \iff \neg q}\right)|1|2}} {{Assumption|4|p \iff \neg q}} {{SequentIntro|5|4|\neg \left ({p \iff q}\right)|4|Non-Equivalence as Equivalence with Negation: Formulation 1}} {{Implication|6||\left({p \iff \neg q}\right) \implies \left({\neg \left ({p \iff q}\right)}\right)|4|5}} {{BiconditionalIntro|7||\left({\neg \left ({p \iff q}\right)}\right) \iff \left({p \iff \neg q}\right)|3|6}} {{EndTableau}} {{qed}}
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. $\begin{array}{|cccc||cccc|} \hline \neg & (p & \iff & q) & (p & \iff & \neg & q) \\ \hline F & F & T & F & T & F & T & F \\ T & F & F & T & T & T & F & T  \\ T & T & F & F & F & T & T & F  \\ F & T & T & T & F & F & F & T  \\ \hline \end{array}$ {{qed}}
$\RR$ is vacuously transitive because there are no elements $a, b, c \in S$ such that $a \mathrel \RR b$ and $b \mathrel \RR c$. Let $\RR^\leftrightarrow$ be the symmetric closure of $\RR$. Then $\RR^\leftrightarrow = \RR \cup \RR^{-1} = \set {\tuple {p, q}, \tuple {q, p} }$. Then: :$p \mathrel {\RR^\leftrightarrow} q$ and $q \mathrel {\RR^\leftrightarrow} p$ but: :$p \not \mathrel {\RR^\leftrightarrow} p$ Therefore $\RR^\leftrightarrow$ is not transitive. {{qed}} Category:Transitive Relations qqk8rl27ob84fq93bz3e62s112gjkzt
By the definition of symmetric closure: :$\RR^\leftrightarrow = \RR \cup \RR^{-1}$. Here $\RR^{-1}$ is the inverse of $\RR$. By Inverse of Relation Compatible with Operation is Compatible, $\RR^{-1}$ is compatible with $\circ$. Thus by Union of Relations Compatible with Operation is Compatible: :$\RR^\leftrightarrow = \RR \cup \RR^{-1}$ is compatible with $\circ$. {{qed}} Category:Compatible Relations b6pdw9g2zqye1clef2ior1iq7fzjwqb
Let: :$a, c \in \left\{{x}\right\}$ :$b \in S$ :$a \preceq b \preceq c$ Then $a = c = x$. Thus $x \preceq b \preceq x$. Since $\preceq$ is a ordering, it is antisymmetric. Thus $b = x$, so $b \in \left\{{x}\right\}$. Since this holds for the only such triple, $\left\{{x}\right\}$ is convex. {{qed}} Category:Convex Sets Category:Singletons 8yo4pko9er4prextiv9qlaxo5auvz4g
=== Proof of Forward Implication === {{:Rule of Transposition/Variant 1/Formulation 2/Forward Implication/Proof}} === Proof of Reverse Implication === {{:Rule of Transposition/Variant 1/Formulation 2/Reverse Implication/Proof}} {{BeginTableau|\vdash \paren {p \implies \neg q} \iff \paren {q \implies \neg p} }} {{TheoremIntro|1|\paren {p \implies \neg q} \implies \paren {q \implies \neg p}|Rule of Transposition: Forward Implication}} {{TheoremIntro|2|\paren {q \implies \neg p} \implies \paren {p \implies \neg q}|Rule of Transposition: Reverse Implication}} {{BiconditionalIntro|3||\paren {p \implies \neg q} \iff \paren {q \implies \neg p}|1|2}} {{EndTableau}} {{qed}}
=== Proof of Forward Implication === {{:Rule of Transposition/Variant 1/Formulation 1/Forward Implication/Proof}} === Proof of Reverse Implication === {{:Rule of Transposition/Variant 1/Formulation 1/Reverse Implication/Proof}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|cccc||cccc|} \hline p & \implies & \neg & q & q & \implies & \neg & p \\ \hline F & T & T & F & F & T & T & F \\ F & T & F & T & T & T & T & F \\ T & F & T & F & F & F & F & T \\ T & T & F & T & T & T & F & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|cccc||cccc|} \hline \neg & p & \implies & q & \neg & q & \implies & p \\ \hline T & F & T & F & T & F & T & F \\ T & F & T & T & F & T & T & F \\ F & T & F & F & T & F & F & T \\ F & T & T & T & F & T & T & T \\ \hline \end{array}$ {{qed}}
As: :$\Int \gamma$ and $\Img \gamma$ are disjoint by the Jordan Curve Theorem and: :$\map \sigma {\openint c d} \subseteq \Int \gamma$ it follows that: :$\Img \gamma \cap \Img \sigma = \set {\map \sigma c, \map \sigma d}$. As $\gamma$ is a Jordan curve, it follows that $\gamma {\restriction_{\closedint a {t_1} } }$ and $\gamma {\restriction_{\closedint {t_2} b} }$ intersect only in $\map \gamma a$. It follows that $\gamma_1$ is a Jordan arc. As the initial point of $\gamma_1$ is $\map \gamma a$, and the final point of $\gamma_1$ is $\map \gamma b = \map \gamma a$, it follows that $\gamma_1$ is a Jordan curve. As $\Img {-\sigma} = \Img \sigma$, it follows that $\gamma_2$ is a Jordan arc. As $\map \gamma {t_1} = \map \sigma c = -\map \sigma d$, it follows that $\gamma_2$ is a Jordan curve. {{qed|lemma}} Denote the exterior of $\gamma$ as $\Ext \gamma$. Let $q_0 \in \Ext \gamma$ be determined. Let $q \in \Ext \gamma$. By the Jordan Curve Theorem, $\Ext \gamma$ is unbounded. Hence for all $N \in \N$ we can choose $q \in \Ext \gamma$ such that: :$\map d {\mathbf 0, q} > N$ where $d$ denotes the Euclidean metric on $\R^2$. {{explain|The word "chooses" suggests that some choice function may be involved here, possibly Axiom:Axiom of Countable Choice? Someone knowledgeable may need to be consulted.}} The Jordan Curve Theorem also shows that $\Ext \gamma$ is open and connected. From Connected Open Subset of Euclidean Space is Path-Connected, there exists a path: :$\rho: \closedint 0 1 \to \Ext \gamma$ joining $q$ and $q_0$. We have: :$\Img {\gamma_1} \subseteq \Int \gamma \cup \Img \gamma$ This is disjoint with $\Ext \gamma$ Thus it follows that $\rho$ is a path in either $\Ext {\gamma_1}$ or $\Int {\gamma_1}$. We have that $\map d {\mathbf 0, q}$ can be arbitrary large. Also, $\Int {\gamma_1}$ is bounded. So follows that $\rho$ is a path in $\Ext {\gamma_1}$. In particularly: :$q \in \Ext {\gamma_1}$ Therefore: :$\Ext \gamma \subseteq \Ext {\gamma_1}$ Let $q_1 \in \Int {\gamma_1}$. Then: :$q_1 \notin \Ext \gamma$ as $\Int {\gamma_1}$ and $\Ext {\gamma_1}$ are disjoint. It follows that: :$q_1 \in \Int {\gamma_1}$ Therefore: :$\Int {\gamma_1} \subseteq \Int \gamma$ Similarly, it follows that: :$\Int {\gamma_2} \subseteq \Int \gamma$ {{qed}} Category:Jordan Curves Category:Jordan Arcs tca9jwpxhq4l9f8frelckm12c0k0px3