hcju/proofwikiqa huggingface dataset:
By definition, a finite propositional tableau $T$ is formed by applying the tableau extension rules a finite number of times. Each tableau extension rule extends $T$ finitely. Therefore $T$ is a finite tree. The result follows from Branch of Finite Tree is Finite. {{qed}} Category:Propositional Tableaus gg8rbya0gos7zkbxsgvc3m619tat6y7
By definition of tableau confutation, every branch of $T$ is contradictory. The result follows by definition of finished propositional tableau. {{qed}}
=== Existence === The existence of $C$ follows immediately from the definition of a basis. {{qed|lemma}} === Uniqueness === Let $C, D$ be finite subsets of $R \times B$ satisfying the requirements. Let $Q = \left\{{v: \exists r \in R: \left({r, v}\right) \in C \cup D}\right\}$. Let $V' = \operatorname{span} \left({Q}\right)$. Then $V'$ is a finite-dimensional vector space with basis $Q$ and $x \in V'$. Thus the theorem follows from Expression of Vector as Linear Combination from Basis is Unique. {{explain}} {{qed}} Category:Vector Spaces Category:Linear Algebra 0cxx8banz3dgsvjrbnh5fx843yhiku0
Define $\sequence {D_n}$ recursively: Let $D_0 = \set s$. For each $n \in \N$ let $D_{n + 1} = \map \RR {D_n}$. Now, for each $n \in \N$ let $E_n$ be the set of all enumerations of $D_n$. Then $E_n$ is non-empty and finite for each $n$. By the Axiom of Countable Choice for Finite Sets, there is a sequence $\sequence {e_n}$ such that for each $n$, $e_n \in E_n$. Now recursively define $\sequence {x_n}$: :Define $x_0$ as $s$. :Define $x_{n + 1}$ as the first element in the enumeration $e_n$ which is in $\map \RR {x_n}$. Then $\sequence {x_n}$ is the required sequence. {{qed}} {{SomeChoice|Axiom:Axiom of Countable Choice for Finite Sets}}
Let $S' = \left\{{y \in S: s \mathrel{\mathcal R^+} y}\right\}$, where $\mathcal R^+$ is the transitive closure of $\mathcal R$. Let $\mathcal R'$ be the restriction of $\mathcal R$ to $S'$.  For each $x \in S'$, there is a $y \in S$ such that $x \mathrel{\mathcal R} y$. But then $s \mathrel{\mathcal R^+} y$, so $y \in S'$, so $x \mathrel{\mathcal R'} y$. Thus $\mathcal R'$ is a left-total relation on $S'$. $S'$ is non-empty: since $\mathcal R$ is left-total, there is a $t \in S$ such that $s \mathrel{\mathcal R} t$, so $s \mathrel{\mathcal R^+} t$, so $t \in S'$. By the Axiom of Dependent Choice, there is a infinite sequence $\left\langle{ y_n}\right\rangle_{n \in \N}$ in $S'$ such that for each $n \in \N$, $y_n \mathrel{\mathcal R'} y_{n+1}$. Then by the definition of restriction, $y_n \mathrel{\mathcal R} y_{n+1}$ for each $n \in \N$. By the definition of $S'$, $s \mathrel{\mathcal R^+} y_0$. By the definition of transitive closure, there are elements $x_0, \dots, x_m$ such that $s = x_0 \mathrel{\mathcal R} x_1 \mathrel{\mathcal R} \cdots \mathrel{\mathcal R} x_m \mathrel{\mathcal R} x_m = y_0$. Then for $n > m$, define $x_n$ as $y_{n-m}$. This sequence then meets the requirements. {{qed}} {{ADC}} Category:Set Theory hncvyjtr3p4t02o9a45h5ixp2i1egkz
We first show that $V$ is countable. If $V$ is finite, then it is surely countable. Suppose instead that $V$ is infinite. Choose an arbitrary vertex $q \in V$. Recursively define a sequence $\sequence {S_n}$: :Let $S_0 = \set q$. :Let $S_{n + 1}$ be the set of all vertices that are adjacent to some element of $S_n$ but ''not'' adjacent to any element of $S_k$ for $k < n$. :That is, $S_n$ is the set of vertices whose shortest path(s) to $q$ have $n$ edges. Since $G$ is connected, $V = \displaystyle \bigcup_{n \mathop \in \N} S_n$. {{explain|Trivial, but requires talking about walks and all.}} Furthermore, $S_n$ is finite and non-empty for each $n$. {{explain}} But by Countable Union of Finite Sets is Countable, $\displaystyle \bigcup_{n \mathop \in \N} S_n$ is countable, so $V$ is countable. {{refactor|Set of Unordered Pairs of Countable Set is Countable or some such.}} By Cartesian Product of Countable Sets is Countable, $V \times V$ is countable. Let $E' = \set {\tuple {a, b}: \set {a, b} \in E}$. By Subset of Countable Set is Countable, $E'$ is countable. Let $g: E' \to E$ be defined by $\map g {a, b} = \set {a, b}$. Then $g$ is surjective. Thus by Surjection from Natural Numbers iff Countable and Composite of Surjections is Surjection, $E$ is countable as well. {{qed}} {{SomeChoice|Axiom:Axiom of Countable Choice for Finite Sets|Countable Union of Finite Sets is Countable}} Category:Graph Theory hgpkxcayhkj2ffxx4fs27ztsdkyr9nu
Let $\triangle ABC$ be a right triangle such that $\angle A$ is right. :300px {{begin-eqn}} {{eqn | l = a^2       | r = b^2 + c^2       | c = Pythagoras's Theorem }} {{eqn | l = c^2       | r = a^2 - b^2       | c = adding $-b^2$ to both sides and rearranging }} {{eqn | r = a^2 - 2 b^2 + b^2       | c = adding $0 = b^2 - b^2$ to the {{RHS}} }} {{eqn | r = a^2 - 2 a b \left({\frac b a}\right) + b^2       | c = multiplying $2 b^2$ by $\dfrac a a$ }} {{eqn | r = a^2 + b^2 - 2 a b \cos C       | c = Definition of Cosine: $\cos C = \dfrac b a$ }} {{end-eqn}} Hence the result. {{qed}}
Let $A$ be an infinite successor set. Let $B$ be another arbitrary infinite successor set. Then from Intersection is Subset, $A \cap B \subseteq A$. From Intersection of Infinite Successor Sets, $A \cap B$ is also an infinite successor set. This set $A \cap B$ is one of the subsets By the definition of $\omega$ it follows that $\omega \subseteq A \cap B$, and consequently that $\omega \subseteq B$. {{qed}}
Let $\triangle ABC$ be an acute triangle. :300px Let $BD$ be dropped perpendicular to $AC$, and let us define $h = BD$, $e = CD$ and $f = AD$. Then $\triangle CDB$ and $\triangle ADB$ are right triangles. So we have both : :$(1):\quad c^2 = h^2 + f^2 \quad$  Pythagoras's Theorem :$(2):\quad a^2 = h^2 + e^2 \quad$  Pythagoras's Theorem and also : :$(3):\quad b^2 = (e + f)^2 = e^2 + f^2 + 2ef$   :$(4):\quad e = a \cos C  \quad$ Definition:Cosine of Angle We'll start with the first equation and use the rest of them to get the desired result : {{begin-eqn}} {{eqn | l = c^2        | r = h^2 + f^2       | c = $(1)$ }} {{eqn | r = a^2 - e^2 + f^2       | c = $(2)$ }} {{eqn | r = a^2 - e^2 + f^2 + 2e^2 - 2e^2 + 2ef - 2ef       | c = adding and subtracting  $2e^2$ and $2ef$ }} {{eqn | r = a^2 + (e^2 + f^2 + 2ef) - 2e(e + f)       | c = rearanging }} {{eqn | r = a^2 + b^2 - 2 a b \cos C       | c = using $(3)$ to substitute both parentheses for $b^2$ and $b$ respectively, and $(4)$ to subst. e for $a \cos C$ }} {{end-eqn}} {{qed}}
Let $\triangle ABC$ be an obtuse triangle. :300px Let $AC$ be extended and $BD$ be dropped perpendicular to $AC$, and let us define $h = BD$, $e = CD$ and $f = AD$. Then $\triangle CDB$ and $\triangle ADB$ are right triangles. So we have both : :$(1):\quad c^2 = h^2 + f^2 \quad$  Pythagoras's Theorem :$(2):\quad a^2 = h^2 + e^2 \quad$  Pythagoras's Theorem and also : :$(3):\quad e^2 = (b + f)^2 = b^2 + f^2 + 2bf$   :$(4):\quad e = a \cos C  \quad$ Definition:Cosine of Angle We'll start with the first equation and use the rest of them to get the desired result : {{begin-eqn}} {{eqn | l = c^2        | r = h^2 + f^2       | c = $(1)$ }} {{eqn | r = a^2 - e^2 + f^2       | c = $(2)$ }} {{eqn | r = a^2 - b^2 - f^2 - 2bf + f^2       | c = $(3)$ }} {{eqn | r = a^2 - b^2 - 2bf + 2b^2 - 2b^2       | c = canceling out $f^2 - f^2$ and adding and subtracting  $2b^2$ }} {{eqn | r = a^2 + b^2 - 2b(b + f)       | c = rearanging }} {{eqn | r = a^2 + b^2 - 2 a b \cos C       | c = using $(4)$ to substitute $b + f = e$ with $a \cos C$ }} {{end-eqn}} {{qed}}
Let $r$ be a vertex of $G$. Recursively define a sequence $\sequence {S_n}$: Let $S_0 = \set r$. Let $S_{n + 1}$ be the set of all vertices that are adjacent to some element of $S_n$ but ''not'' adjacent to any element of $S_k$ for $k < n$. That is, $S_n$ is the set of vertices whose shortest path(s) to $r$ have $n$ edges. Since $G$ is connected: :$V = \displaystyle \bigcup_{n \mathop \in \N} S_n$ Note that $\set {S_n: n \in \N}$ is pairwise disjoint. Define a relation $\RR$ on $V$ by letting $p \mathrel \RR q$ {{iff}}: :$p$ is adjacent to $q$ and: :there exists an $n \in \N$ such that $p \in S_n$ and $q \in S_{n+1}$. Let $\RR^+$ be the transitive closure of $\RR$. Let $V'$ be the set of all $v \in V$ such that $\map {\RR^+} v$ is infinite. It will be demonstrated that $r \in V'$: Let $v \in V$. Since $G$ is connected, there is a path from $r$ to $v$. By the Well-Ordering Principle, there is a path $P$ from $r$ to $v$ of minimal length. Then the vertices along $P$ will lie in successive sets $S_k$. {{explain}} Thus $v \in \map {\RR^+} r$. As this holds for all $v \in V$, and $V$ is infinite, $\map {\RR^+} r$ is infinite. Therefore $r \in V'$. Let $\RR'$ be the restriction of $\RR$ to $V'$. It is to be shown that $\RR'$ is a left-total relation. Let $v \in V'$. Then $\map {\RR^+} v$ is infinite. Since $\map \RR v$ is finite, Finite Union of Finite Sets is Finite shows that $\map \RR v$ must have an element $u$ such that $\map {\RR^+} u$ is infinite. So $u \in V'$ and $v \mathrel {\RR'} u$. We have that $V$ is assumed to be countably infinite. {{WLOG}}, let $V = \N$. Recursively define a sequence $\sequence {v_k}$ in $V'$ thus: :$v_0 = r$ :$v_{n + 1} = \map \min {\map {\RR'} {v_n} }$ Then $\sequence {v_k}$ is the sequence of vertices along an infinite path starting at $r$. {{qed}} Category:König's Lemma 057k0nyt88igjyz7j07bbed5ivgksqv
By Locally Finite Connected Graph is Countable, $G$ has countably many vertices. Thus the result holds by König's Lemma: Countable. {{qed}}
Let $\left({r}\right)$ be the $0$-length $G$-path whose only vertex is $r$.  === Connected === We will show that there is a $T$-path from $\left({r}\right)$ to each element $p$ of $P$. We proceed by induction on the length of $p$. If $p$ has length $0$, then $p = \left({r}\right)$. Thus the $0$-length $T$-path $\left({\left({r}\right)}\right)$ connects $\left({r}\right)$ to $p$. {{explain|Is $\left({\left({r}\right)}\right)$ meant here or just $\left({r}\right)$? If the former, explain the meaning of the double brackets.}} Suppose that there is a $T$-path from $\left({r}\right)$ to each element of $P$ of length $n$. Let $p \in P$ have length $n + 1$ and final vertex $z$. Let $p^*$ be the $G$-path obtained from $p$ by removing the last vertex and the last edge. Then $p^*$ is a $G$-path of length $n$. We must show that $p^* \in P$. Let $w$ be the last vertex of $p^*$. Suppose for the sake of contradiction that $p^* \notin P$. Then by the definition of $P$, there is a $G$-path $m$ from $r$ to $w$ which is shorter than $p^*$. But then appending the last edge and vertex of $p$ to $m$, one obtains a $G$-path from $r$ to $z$ which is shorter than $p$. This contradicts the supposition that $p \in P$. Thus we conclude that $p^* \in P$. By the inductive hypothesis, there is a $T$-path $b$ from $\left({r}\right)$ to $p^*$. Then by the definition of $E'$, there is an edge from $p^*$ to $p$. Appending that edge and $p$ to $b$ yields a $T$-path from $\left({r}\right)$ to $p$. As such a $T$-path exists for each $p \in P$, $T$ is connected. {{qed|lemma}} === Acyclic === {{finish}} Category:Tree Theory 585si5m2hji0ron60zdim1iqs4e4oc8
Let $G$ be an infinite planar graph. Let $C$ be a set of vertices of $G$. For each $c \in C$ let $p_c^1, p_c^2, p_c^3, p_c^4$ be propositional symbols, where : $p_c^i$ is true iff the color of vertex $c$ is $i$. Let $\mathcal P_0$ be the vocabulary defined as: :$\mathcal P_0 = \left\{{p_c^1, p_c^2, p_c^3, p_c^4: c \in C}\right\}$ Let $\mathbf H$ be the set of all statements with the following forms: :$(1): \quad$ $p_c^1 \lor p_c^2 \lor p_c^3 \lor p_c^4$ for each $c \in C$ :$(2): \quad$ $p_c^i \implies \neg p_c^j$ for each $c \in C$ and for each $i \ne j$ :$(3): \quad$ $\neg \left({p_c^i \land \neg p_{c'}^i}\right)$ for each $i$ where $c$ and $c'$ are adjacent. Let $\mathcal M$ be a model for $\mathbf H$ which corresponds to a proper vertex $4$-coloring of $G$. By hypothesis, any finite planar graph can be assigned a proper vertex $4$-coloring. Thus every finite subgraph of $G$ has a proper vertex $4$-coloring. That is, every finite subset of $\mathbf H$ has a model. By the Compactness Theorem for Boolean Interpretations, $\mathbf H$ has a model. Hence the entirety of $G$ can be assigned a proper vertex $k$-coloring. {{qed}}
We represent a walk as a sequence of vertices. However, the same argument will work for the representation as an alternating sequence of vertices and edges. The proof proceeds by induction on the length of $w$. If the length of $w$ is $0$, then it is trivially a path. Suppose that every walk of length less than $n$ has a subsequence forming a path between its endpoints. Let $w$ have length $n$. If $w$ is not a path, then it must have a cycle. That is, for some $j, k \in \Dom w$, $j < k$ and $w_j = w_k$. Then removing $w_{j + 1}, \dots, w_k$ from $w$ will form a new walk $w'$, from $x$ to $y$ which is strictly shorter than $w$. By the inductive hypothesis, $w'$ has as subsequence which is a path from $x$ to $y$. {{qed}} Category:Graph Theory srv00yb34cr37uj72p3dzo2tsy6q17b
Recall that a tree is connected and has no cycles. Thus there is exactly one path from each node of $T$ to each other node of $T$. {{explain|This is in fact a result that already exists and can be quoted directly.}} Let $A$ be the set of all ordered pairs $x, y \in V$ such that: :$\tuple {x, y} \in E$ and :The unique path from $r$ to $y$ passes through $x$. {{finish}}
Let $G = \left({V, A}\right)$ be a directed graph. Let $u, v \in V$. Let $\mathcal R$ be the reachability relation for $G$, defined as the transitive closure of $A$. === Definition 1 implies Definition 2 === Suppose that $v$ is reachable from $u$ by definition 1. Then there exists a directed walk $\left({u = x_0, \dots, x_n = v}\right)$ from $u$ to $v$. Then by the definition of directed walk: : $x_0 \mathrel {\mathcal R} x_1 \mathrel {\mathcal R} \cdots \mathrel {\mathcal R} x_n$ Then by the definition of transitive closure: : $u \mathrel {\mathcal R} v$ Thus $v$ is reachable from $u$ by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Suppose that $v$ is reachable from $u$ by definition 2. That is: : $u \mathrel {\mathcal R} v$ Thus by the definition of transitive closure: :$\exists x_0, \ldots, x_n \in V: x_0 \mathrel {\mathcal R} x_1 \mathrel {\mathcal R} \cdots \mathrel {\mathcal R} x_n$ Then $\left({x_0 \mathrel {\mathcal R} x_1 \mathrel {\mathcal R} \cdots \mathrel {\mathcal R} x_n}\right)$ is a directed walk from $u$ to $v$. Therefore, $v$ is reachable from $u$ by definition 1. {{qed}} Category:Digraphs a1pj5naddponrz3u27bnzrkhlmg8bjq
This follows immediately from Equivalence of Definitions of Reachable. {{qed}} Category:Digraphs sz7vf78mh5hkfdrmrmsm4divqhwezlq
{{BeginTableau|\left({p \implies q}\right) \implies p \vdash p}} {{Premise|1|\left({p \implies q}\right) \implies p}} {{Assumption|2|\neg p}} {{SequentIntro|3|2|p \implies q|2|False Statement implies Every Statement}} {{ModusPonens|4|1,2|p|1|3}} {{NonContradiction|5|1,2|2|4}} {{Reductio|6|1|p|2|5}} {{EndTableau}} {{qed}}
{{BeginTableau|\left({p \implies q}\right) \implies p \vdash p}} {{Premise|1|\left({p \implies q}\right) \implies p}} {{ExcludedMiddle|2|p \lor \neg p}} {{Assumption|3|\neg p}} {{SequentIntro|4|3|p \implies q|3|False Statement implies Every Statement}} {{ModusPonens|5|1, 3|p|1|4}} {{Assumption|6|p}} {{ProofByCases|7|1|p|2|3|5|6|6}} {{EndTableau}} {{qed}}
{{BeginTableau|\neg p \implies p \vdash p}} {{Premise|1|\neg p \implies p}} {{Assumption|2|p \implies \bot}} {{SequentIntro|3|2|\neg p|2|Negation as Implication of Bottom}} {{ModusPonens|4|1,2|p|1|3}} {{Implication|5|1|(p \implies \bot) \implies p|2|4}} {{SequentIntro|6|1|p|5|Peirce's Law}} {{EndTableau}} {{qed}}
{{BeginTableau|\vdash p \lor \neg p}} {{Assumption|1|\neg (p \lor \neg p)}} {{SequentIntro|2|1|\bot|1|Negation of Excluded Middle is False}} {{Explosion|3|1|p \lor \neg p|2}} {{Implication|4||\neg(p \lor \neg p) \implies p \lor \neg p|1|3}} {{SequentIntro|5||p \lor \neg p|4|Clavius's Law}} {{EndTableau}} {{qed}}
Let $r = 0$, so: : $p = 2 k$ Then: :$p^2 = \left({2 k}\right)^2 = 4 k^2 = 2 \left({2 k^2}\right)$ and so $p^2$ is even.
Let $r = 1$, so: :$p = 2 k + 1$ Then: :$p^2 = \paren {2 k + 1}^2 = 4 k^2 + 4 k + 1 = 2 \paren {2 k^2 + 2 k} + 1$ and so $p^2$ is odd.
{{BeginTableau|\neg (p \lor \neg p) \vdash \bot}} {{Assumption|1|\neg (p \lor \neg p)}} {{SequentIntro|2|1|\neg p \land \neg \neg p|1|De Morgan's Laws}} {{Simplification|3|1|\neg p|2|1}} {{Simplification|4|1|\neg\neg p|2|2}} {{NonContradiction|5|1|3|4}} {{EndTableau}} {{qed}}
{{BeginTableau|\neg\neg (p \lor \neg p)}} {{Assumption|1|\neg (p \lor \neg p)}} {{SequentIntro|2|1|\bot|1|Negation of Excluded Middle is False: Form 1}} {{Implication|3||\neg (p \lor \neg p) \implies \bot|1|2}} {{SequentIntro|4||\neg \neg (p \lor \neg p)|3|Negation as Implication of Bottom}} {{EndTableau}} {{qed}} Category:Propositional Logic 9inhesbm7f2swudqcr1zrzyh8j4zygg
Let $f: Y \to Z$ be an injection. Let $X$ be a set Let $g_1: X \to Y, g_2: X \to Y$ be mappings such that: : $f \circ g_1 = f \circ g_2$ Then $\forall x \in X$: {{begin-eqn}} {{eqn | l = \map f {g_1 \paren x}       | r = \map {f \circ g_1} x       | c = {{Defof|Composite Mapping}} }} {{eqn | r = \map {f \circ g_2} x       | c = By Hypothesis }} {{eqn | r = \map f {\map {g_2} x}       | c = {{Defof|Composite Mapping}} }} {{end-eqn}} As $f$ is an injection, $\map {g_1} x = \map {g_2} x$ and thus the condition for left cancellability holds.
We use a Proof by Contraposition. That is, we show that if $f: Y \to Z$ is not injective, then $f$ is not left cancellable. Hence, suppose $f: Y \to Z$ is not injective. Then: : $\exists y_1 \ne y_2 \in Y: f \left({y_1}\right) = f \left({y_2}\right)$ Let the two mappings $g_1: Y \to Y, g_2: Y \to Y$ be defined as follows: :$\forall y \in Y: g_1 \left({y}\right) = y$ :$\forall y \in Y: g_2 \left({y}\right) = \begin{cases} y_2 & : y = y_1 \\ y & : y \ne y_1 \end{cases}$ Thus we have $g_1 \ne g_2$ such that $f \circ g_1 = f \circ g_2$. That is, $f$ is not left cancellable. From Rule of Transposition it follows that if $f$ is left cancellable, it is injective.
{{BeginTableau|\vdash p \lor \neg p}} {{Assumption|1|(p \lor \neg p) \implies \bot}} {{SequentIntro|2|1|\neg(p \lor \neg p)|1|Negation as Implication of Bottom}} {{SequentIntro|3|1|\bot|2|Negation of Excluded Middle is False}} {{Explosion|4|1|p \lor \neg p|3}} {{Implication|5||((p \lor \neg p) \implies \bot) \implies p \lor \neg p|1|4}} {{SequentIntro|6||p \lor \neg p|5|Peirce's Law}} {{EndTableau}} {{qed}}
By the definition of pseudocomplemented lattice, $L$ has a smallest element $\bot$. Let $x \in L$. Then $x \wedge \bot = \bot$. {{explain}} By the definition of pseudocomplemented lattice, there is a greatest element $\bot^*$ such that $\bot \wedge \bot^* = \bot$. But then by the definition of greatest element, $x \preceq \bot^*$ for each $x \in L$, so $\bot^*$ is the greatest element of $L$. Since $L$ has a smallest element and a greatest element, it is a bounded lattice. {{qed}} Category:Lattice Theory exrizkezr9qwqnbcbf1nsm83ffdwezu
To show that $\struct {\tau, \subseteq}$ is a complete lattice, we must show that every subset of $\tau$ has a supremum and an infimum. Let $S \subseteq \tau$. By the definition of a topology: :$\displaystyle \bigcup S \in \tau$ By Union is Smallest Superset, $\displaystyle \bigcup S$ is the supremum of $S$. Let $I$ be the interior of $\displaystyle \bigcap S$, where by Intersection of Empty Set $\displaystyle \bigcap \O$ is conventionally taken to be $X$. Then by the definition of interior and Intersection is Largest Subset: :$I \in \tau$ and :$I \subseteq U$ for each $U \in S$. Let $V \in \tau$ with $V \subseteq U$ for each $U \in S$. By Intersection is Largest Subset: :$\displaystyle V \subseteq \bigcap S$. Then by the definition of interior: :$V \subseteq I$ Thus $I$ is the infimum of $S$. So each subset of $\tau$ has a supremum and an infimum. Thus, by definition, $\struct {\tau, \subseteq}$ is a complete lattice. {{qed}} Category:Topology Category:Complete Lattices e80gtf0e1vr5oz1zjtffa3nnnfmzk5z
{{BeginTableau|\left({p \lor \neg p}\right) \vdash \left({\left({p \implies q}\right) \implies p}\right) \implies p}} {{Premise|1|p \lor \neg p}} {{Assumption|2|p}} {{SequentIntro|3|2|\left({\left({p \implies q}\right) \implies p}\right) \implies p|2|True Statement is implied by Every Statement}} {{Assumption|4|\neg p}} {{SequentIntro|5|4|p \implies q|4|False Statement implies Every Statement}} {{Assumption|6|\left({p \implies q}\right) \implies p}} {{ModusPonens|7|6, 5|p|6|5}} {{Implication|8|5|\left({\left({p \implies q}\right) \implies p}\right) \implies p|6|7}} {{ProofByCases|9|1|\left({\left({p \implies q}\right) \implies p}\right) \implies p|1|2|3|4|8}} {{EndTableau}}
=== Definition 1 implies Definition 2 === Let $\Bbb S$ be a set of subsets $\Bbb S$ of $S$ such that: :$(1): \quad$ $\Bbb S$ is pairwise disjoint: $\forall S_1, S_2 \in \Bbb S: S_1 \cap S_2 = \O$ when $S_1 \ne S_2$ :$(2): \quad$ The union of $\Bbb S$ forms the whole set $S$: $\displaystyle \bigcup \Bbb S = S$ :$(3): \quad$ None of the elements of $\Bbb S$ is empty: $\forall T \in \Bbb S: T \ne \O$. By $(3)$ all sets in $\Bbb S$ are non-empty. Suppose there exists $x \in S$ in more than one set of $\Bbb S$. That is: : $\exists S_1, S_2 \in \Bbb S, S_1 \ne S_2: x \in S_1 \land x \in S_2$ Then by definition of set intersection: :$x \in S_1 \cap S_2$ and so $S_1 \cap S_2 \ne \varnothing$. This contradicts condition $(1)$. So no element of $S$ can be in more than one set of $\Bbb S$. Suppose there exists $x \in S$ in none of the sets of $\Bbb S$. Then $\displaystyle x \notin \bigcup \Bbb S$ and so $\Bbb S \notin S$. This contradicts condition $(2)$. So every element of $S$ is in at least one set of $\Bbb S$. Thus $x$ lies in exactly one element of $\Bbb S$. Hence it follows that $\Bbb S$ is a set of non-empty subsets of $S$ such that each element of $S$ lies in exactly one element of $\Bbb S$. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\Bbb S$ be a set of non-empty subsets of $S$ such that each element of $S$ lies in exactly one element of $\Bbb S$. $(3)$ is fulfilled automatically by definition. Consider $\Bbb S$. From Union of Subsets is Subset: Subset of Power Set: :$\displaystyle \bigcup \Bbb S \subseteq S$ Let $x \in T$. By hypothesis: :$\exists S_1 \in \Bbb S: x \in S_1$ That is: :$\displaystyle x \in \bigcup \Bbb S$ and so: :$\displaystyle S \subseteq \bigcup \Bbb S$ By definition of set equality: :$\displaystyle \bigcup \Bbb S = S$ thus demonstrating that $(2)$ holds. Suppose $\Bbb S$ were not pairwise disjoint, and that: :$\exists S_1, S_2 \in \Bbb S: S_1 \cap S_2 \ne \O$ Then: :$\exists x \in S: x \in S_1 \cap S_2$ By definition of set intersection: :$x \in S_1$ and $x \in S_2$ thus contradicting the supposition that each element of $S$ lies in exactly one element of $\Bbb S$. So $\Bbb S$ is pairwise disjoint and so $(1)$ holds. Hence the result. {{qed}} Category:Set Partitions h0k7s366quubd64lntdwns05gzts9fy