hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cap B_\alpha}       | c =  }} {{eqn | lll=\leadsto       | ll= \forall \alpha \in I:       | l = x       | o = \in       | r = A_\alpha \cap B_\alpha        | c = {{Defof|Intersection of Family}} }} {{eqn | lll=\leadsto       | ll= \forall \alpha \in I:       | l = x       | o = \in       | r = A_\alpha       | c = {{Defof|Set Intersection}} }} {{eqn | lo= \land       | l = x       | o = \in       | r = B_\alpha       | c =  }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \bigcap_{\alpha \mathop \in I} A_\alpha       | c ={{Defof|Intersection of Family}} }} {{eqn | lo= \land       | l = x       | o = \in       | r = \bigcap_{\alpha \mathop \in I} B_\alpha       | c = {{Defof|Intersection of Family}} }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcap_{\alpha \mathop \in I} B_\alpha}       | c = {{Defof|Set Intersection}} }} {{end-eqn}} Thus by definition of subset: :$\displaystyle \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cap B_\alpha} \subseteq \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcap_{\alpha \mathop \in I} B_\alpha}$ {{qed|lemma}} {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcap_{\alpha \mathop \in I} B_\alpha}       | c =  }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \bigcap_{\alpha \mathop \in I} A_\alpha       | c = {{Defof|Set Intersection}} }} {{eqn | lo= \land       | l = x       | o = \in       | r = \bigcap_{\alpha \mathop \in I} B_\alpha       | c =  }} {{eqn | lll=\leadsto       | ll= \forall \alpha \in I:       | l = x       | o = \in       | r = A_\alpha       | c = {{Defof|Intersection of Family}} }} {{eqn | lo= \land       | l = x       | o = \in       | r = B_\alpha       | c =  }} {{eqn | lll=\leadsto       | ll= \forall \alpha \in I:       | l = x       | o = \in       | r = A_\alpha \cap B_\alpha       | c = {{Defof|Set Intersection}} }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cap B_\alpha}       | c = {{Defof|Intersection of Family}} }} {{end-eqn}} Thus by definition of subset: :$\displaystyle \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcap_{\alpha \mathop \in I} B_\alpha} \subseteq \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cap B_\alpha}$ {{qed|lemma}} By definition of set equality: :$\displaystyle \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cap B_\alpha} = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcap_{\alpha \mathop \in I} B_\alpha}$ {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = \bigcup_{\alpha \mathop \in I} \paren {A_\alpha \cap B}       | c = }} {{eqn | lll=\leadsto       | ll= \exists \alpha \in I:       | l = x       | o = \in       | r = A_\alpha \cap B       | c = {{Defof|Union of Family}} }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = A_\alpha       | c = {{Defof|Set Intersection}} }} {{eqn | lo= \land       | l = x       | o = \in       | r = B       | c =  }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha}       | c = Set is Subset of Union }} {{eqn | lo= \land       | l = x       | o = \in       | r = B       | c =  }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B       | c = {{Defof|Set Intersection}} }} {{end-eqn}} By definition of subset: :$\displaystyle \bigcup_{\alpha \mathop \in I} \paren {A_\alpha \cap B} \subseteq \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B$ {{qed|lemma}} {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B       | c =  }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha}       | c = {{Defof|Set Intersection}} }} {{eqn | lo= \land       | l = x       | o = \in       | r = B       | c =  }} {{eqn | lll=\leadsto       | ll =\exists \alpha \in I:       | l = x       | o = \in       | r = A_\alpha       | c = {{Defof|Union of Family}} }} {{eqn | lo= \land       | l = x       | o = \in       | r = B       | c =  }} {{eqn | lll=\leadsto       | ll =\exists \alpha \in I:       | l = x       | o = \in       | r = A_\alpha \cap B       | c = {{Defof|Set Intersection}} }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \bigcup_{\alpha \mathop \in I} \paren {A_\alpha \cap B}       | c = Set is Subset of Union }} {{end-eqn}} By definition of subset: :$\displaystyle \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B \subseteq \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cap B}$ {{qed|lemma}} By definition of set equality: :$\displaystyle \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cap B} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B$ {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B}       | c = }} {{eqn | lll=\leadsto       | ll =\forall \alpha \in I:       | l = x       | o = \in       | r = A_\alpha \cup B       | c = Intersection is Subset }} {{eqn | lll=\leadsto       | ll =\forall \alpha \in I:       | l = x       | o = \in       | r = A_\alpha       | c = {{Defof|Set Union}} }} {{eqn | lo= \lor       | l = x       | o = \in       | r = B       | c =  }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha}       | c = {{Defof|Intersection of Family}} }} {{eqn | lo= \lor       | l = x       | o = \in       | r = B       | c =  }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B       | c = {{Defof|Set Union}} }} {{end-eqn}} By definition of subset: :$\displaystyle \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B} \subseteq \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B$ {{qed|lemma}} {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B       | c =  }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha}       | c = {{Defof|Set Union}} }} {{eqn | lo= \lor       | l = x       | o = \in       | r = B       | c =  }} {{eqn | lll=\leadsto       | ll =\forall \alpha \in I:       | l = x       | o = \in       | r = A_\alpha       | c = Intersection is Subset }} {{eqn | lo= \lor       | l = x       | o = \in       | r = B       | c =  }} {{eqn | lll=\leadsto       | ll =\forall \alpha \in I:       | l = x       | o = \in       | r = A_\alpha \cup B       | c = {{Defof|Set Union}} }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B}       | c = {{Defof|Intersection of Family}} }} {{end-eqn}} By definition of subset: :$\displaystyle \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B \subseteq \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B}$ {{qed|lemma}} By definition of set equality: :$\displaystyle \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B} = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B$ {{qed}}
=== Definition 1 implies Definition 2 === {{:Equivalence of Definitions of Well-Ordering/Definition 1 implies Definition 2}} By hypothesis, every subset of $S$ has a smallest element. By Smallest Element is Minimal it follows that every subset of $S$ has a minimal element. Thus it follows that $\preceq$ is a well-ordering on $S$ by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\preceq$ be a well-ordering on $S$ by definition 2. That is: :$\preceq$ is a well-founded total ordering. By definition of well-founded, every $T \subseteq S$ has a minimal element. By Minimal Element in Toset is Unique and Smallest, every $T \subseteq S$ has a smallest element. The result follows. {{qed}} Category:Well-Orderings aexj9bpqs6iubuikzd3vbbt2y6mzbcv
Let $A$ be a class. Let $B$ be a non-empty subset of $A$. Let $\prec$ be a relation on $A$. === Definition 1 implies Definition 2 === Suppose that $\prec$ is a foundational strict total ordering of $A$. By the definition of strict total ordering, $\prec$ connects $A$. {{qed|lemma}} === Definition 2 implies Definition 1 === Suppose that $\prec$ connects $A$ and that $\prec$ is a well-founded relation. That is: : for any $x, y \in A$, either $x = y$, $x \prec y$, or $y \prec x$ : whenever $b$ is a non-empty subset of $A$, $b$ has a $\prec$-minimal element. $\prec$ is transitive: Let $x, y, z \in A$. Let $x \prec y$ and $y \prec z$. Since $\prec$ connects $A$, either $x \prec z$ or $z \prec x$. By Foundational Relation has no Relational Loops it is not the case that $x \prec y$, $y \prec z$ and $z \prec x$. Thus we conclude that $x \prec z$. As this holds for all such $x, y, z$, $\prec$ is transitive. $\prec$ is antireflexive by Foundational Relation is Antireflexive. Thus $\prec$ is a strict total ordering. {{qed}} Category:Order Theory dcuz4f7v1y4s3wbk1dsir2i2d3zu2n3
==== $\RR^+$ is transitive ==== Let $x,y,z \in S$. Let $x \mathrel {\RR^+} y$ and $y \mathrel {\RR^+} z$. Then for some $m, n \in \N_{>0}$ there are $s_0, s_1, \dots, s_m$ and $t_0, t_1, \dots, t_n$ such that $s_0 = x$, $s_m = y$, $t_0 = y$, $t_n = z$, and the following hold: :$\forall k \in \N_m: s_k \mathrel {\RR^+} s_{k + 1}$ :$\forall k \in \N_n: t_k \mathrel {\RR^+} t_{k + 1}$ Let $\sequence {u_k}_{k \mathop \in \N_{m + n} }$ be defined thus: :$u_k = \cases {s_k & \text{if $k \le m$} \\ t_{k - m} & \text {if $k > m$}}$ Then clearly $u_k \mathrel {\RR^+} u_{k+1}$ whenever $k < m$ and whenever $k > m$. But $u_m = s_m = y = t_0 \mathrel {\RR^+} t_1 = u_{m+1}$, so this holds also for $k = m$. Furthermore, $u_0 = s_0 = x$ and $u_{m+n} = t_n = z$. Therefore $x \mathrel {\RR^+} z$. As this holds for all such $x$ and $z$, $\RR^+$ is transitive. {{qed|lemma}} ==== $\RR^+$ is smallest ==== Let $\QQ$ be any transitive relation on $S$ such that $\RR \subseteq \QQ$. For any $x, y \in S$ such that $x \mathrel {\RR^+} y$, let $d \left({x, y}\right)$ be the smallest natural number $n > 0$ such that there exist $s_0, s_1, \dots, s_n \in S$ such that $s_0 = x$, $s_n = y$, and: :$\forall k \in \N_n: s_k \mathrel \RR s_{k + 1}$ Such an $n$ always exists by the definition of $\RR^+$ and the fact that $\N$ is well-ordered by $\le$. We will show by induction on $n$ that for every $x, y$ such that $x \mathrel {\RR^+} y$ and $\map d {x, y} = n$, $x \mathrel \QQ y$. This will show that $\RR^+ \subseteq \QQ$. If $\map d {x, y} = 1$ then $x \mathrel \RR y$, so $x \mathrel \QQ y$. Suppose that the result holds for $n$. Let $\map d {x, y} = n + 1$. Then there exist $s_0, s_1, \dots, s_{n + 1}$ such that $s_0 = x$, $s_{n + 1} = y$, and: :$\forall k \in \N_{n + 1}: s_k \mathrel \RR s_{k + 1}$ Then dropping the last term: :$\forall k \in \N_n: s_k \mathrel \RR s_{k + 1}$ so $x \mathrel {\RR^+} s_n$. {{explain|Explain better: either show that minimality implies this or replace standard induction with complete induction}} It should be clear, then, that $\map d {x, s_n} = n$. Thus by the inductive hypothesis, $x \mathrel \QQ s_n$. Since $\RR \subseteq \QQ$, $s_n \mathrel \QQ s_{n + 1} = y$. Since $x \mathrel \QQ s_n$, $s_n \mathrel \QQ y$, and $\QQ$ is transitive: :$x \mathrel \QQ y$ As this holds for all such $x$ and $y$, $\RR^+ \subseteq \QQ$. {{qed}} Category:Equivalence of Definitions of Transitive Closure (Relation Theory) t1gd2ly0jh4kzcqqzcse4obdyhx8c6i
A well-ordered set is a totally ordered set by definition. A bijection is a surjection by definition. The result follows from Order Isomorphism iff Strictly Increasing Surjection. {{qed}} Category:Well-Orderings Category:Order Isomorphisms bd9m2xcftcxo5aowtn8z62ncldgdsm8
Since $\RR$ is set-like, $\map {\RR^{-1} } {\set x}$ is a set for each $x$ in $A$. As $B \subseteq A$, this holds also for each $x \in B$. {{explain|Explain better.}} But then $\displaystyle \map {\RR^{-1} } B = \bigcup_{x \mathop \in B} \map {\RR^{-1} } {\set x}$, which is a set by the Axiom of Unions. {{qed}} Category:Set Theory a0lkqyt765j7yboaj8pvu4luf5ejni3
Let $g: U \to \C$ be such that $\map g x = 1 / \map f x$. Since $\map f x$ is nonzero for $x \in U$, $g$ is well-defined. By Quotient Rule for Continuous Functions, $g$ is continuous. Let $z_0 \in U$. As $g$ is continuous: :$\displaystyle \lim_{h \mathop \to 0} \frac 1 {\map f {z_0 + h} } = \frac 1 {\map f {z_0} }$ As $f$ is holomorphic: :$\displaystyle \lim_{h \mathop \to 0} \frac {\map f {z_0 + h} - \map f {z_0} } h = \map {f'} {z_0}$ 	 By the Combination Theorem for Limits of Functions,  {{begin-eqn}} {{eqn | l = \lim_{h \mathop \to 0} \frac {\map g {z_0 + h} - \map g {z_0} } h       | r = \lim_{h \mathop \to 0} \frac {\frac 1 {\map f {z_0 + h} } - \frac 1 {\map f {z_0} } } h }} {{eqn | r = -\frac 1 {\map f {z_0} } \lim_{h \mathop \to 0} \paren {\paren {\frac 1 {\map f {z_0 + h} } } \paren {\frac {\map f {z_0 + h} - \map f {z_0} } h} } }} {{eqn | r = -\frac 1 {\map f {z_0}^2} \cdot \map {f'} {z_0} }}	 {{end-eqn}} It follows that $g$ is holomorphic. {{qed}} Category:Complex Analysis Category:Reciprocals 901s09w466k3i1akdzp6afbndcxrcez
Let $x \in A$. Let $A'$ be the class of all subsets of $A$. For each $s \in A'$, $\RR^{-1}$ is a subset of $A$. Hence by Inverse Image of Set under Set-Like Relation is Set and the definition of endorelation: :$\RR^{-1} \in A'$ Define a mapping $G: A' \to A'$ as: :$\forall s \in A': \map G s = \map {\RR^{-1} } s$ Recursively define a mapping $f: \N \to A'$ as follows: :$\map f n = \begin {cases} \set x & : n = 0 \\ \map G {\map f {n - 1} } & : n > 0 \end {cases}$ By the Axiom of Infinity and the Axiom of Replacement: :$\map f \N$ is a set. Thus by the Axiom of Unions: :$\displaystyle \bigcup \map f \N$ is a set. Let $y \in \map {\paren {\RR^+}^{-1} } x$. By the definition of transitive closure: :for some $n \in \N_{>0}$ there are $a_0, a_1, \dots, a_n$ such that $y = a_0 \mathrel \RR a_1 \mathrel \RR \cdots \mathrel \RR a_n = x$. {{explain|it's a finite sort of induction, and a simple and common pattern.}} Then by induction (working from $n$ to $0$), $\displaystyle a_n, a_{n - 1}, \dots, a_0 \in \bigcup \map f \N$. As this holds for all such $y$: :$\displaystyle \map {\paren {\RR^+}^{-1} } x \subseteq \bigcup \map f \N$ By the Axiom of Specification: :$\map {\paren {\RR^+}^{-1} } x$ is a set. As this holds for all $x \in A$: :$\RR^+$ is a set-like relation. {{qed}} Category:Set Theory Category:Transitive Closures 8wsogx7xf6uebkx6imqtrrgno2wptdm
=== $B'$ is $\RR$-transitive === Let $x \in B'$ and $y \in A$, and let $y \mathrel \RR x$. If $x \in B$, then by the definition of transitive closure: :$y \mathrel {\RR^+} x$ so: :$y \in B'$ Let $x \in \map {\paren {\RR^+}^{-1} } B$. Then: :$x \mathrel {\RR^+} b$ for some $b \in B$. Since $\RR \subseteq \RR^+$, it follows that: :$y \mathrel {\RR^+} x$ Since $\RR^+$ is transitive: :$y \mathrel {\RR^+} b$ That is: :$y \in \map {\paren {\RR^+}^{-1} } B$ so $y \in B'$. As this holds for all such $x$ and $y$, $B'$ is $\RR$-transitive. {{qed|lemma}} === $B' \subseteq C$ === Let $x \in B'$. Then $x \in B$ or $x \in \map {\paren {\RR^+}^{-1} } B$. Let $x \in B$. Then because $B \subseteq C$: :$x \in C$ Suppose that $x \in \map {\paren {\RR^+}^{-1} } B$. Then for some $b \in B$: :$x \mathrel \RR b$ By the definition of transitive closure: :for some $n \in \N_{>0}$ there exist $a_0, a_1, \dots, a_n$ such that: ::$x = a_0 \mathrel \RR a_1 \mathrel \RR \cdots \mathrel \RR a_n = b$ {{explain|Expand to full argument.}} Thus by the Principle of Mathematical Induction: :$x \in C$ {{qed|lemma}} === Set-like implies set === Let $B$ be a set. Let $\RR$ be set-like. Then by Inverse Image of Set under Set-Like Relation is Set: :$\paren {\RR^+}^{-1}$ is a set. Thus $B'$ is a set by the Axiom of Unions. {{qed}} Category:Relational Closures a96d1ujc5xam9piqm5vf9dw2yfd416c
=== Sufficient Condition === Let $m$ be $\mathcal R$-minimal in $B$. Let $x$ be any element of $B$. Suppose for the sake of contradiction that $x \mathrel {\mathcal R'} m$. Then since $\mathcal R' \subseteq \mathcal R$: :$x \mathrel{\mathcal R} m$ contradicting the fact that $m$ is $\mathcal R$-minimal in $B$. Thus: :$\lnot \left({x \mathrel{\mathcal R'} m}\right)$ As this holds for all $x \in B$, $m$ is $\mathcal R'$-minimal in $B$. {{qed|lemma}} === Necessary Condition === Let $m$ be $\mathcal R'$-minimal in $B$. Let $x \in B$. Suppose for the sake of contradiction that $x \mathrel{\mathcal R} m$. Then $x, m \in B$. Therefore: :$\left({x, m}\right) \in B \times B$ Thus: :$\left({x, m}\right) \in \mathcal R \cap \left({B \times B}\right) = \mathcal R'$ so $x \mathrel{\mathcal R'} m$ This contradicts the fact that $m$ is $\mathcal R'$-minimal in $B$. Thus: :$\lnot \left({x \mathrel{\mathcal R} m}\right)$ As this holds for all $x \in B$, it follows that $m$ is $\mathcal R$-minimal in $B$. {{qed}} Category:Restrictions 7apan44wwcbfxgntcgd88jt1jz9mpw0
Let $i = \bigcap A$. === Set === By Intersection of Non-Empty Class is Set, $i$ is a set. {{qed|lemma}} === Transitive === Let $n \in i$ and let $m \in n$. Let $a \in A$. By the definition of intersection, $n \in a$. Since $a$ is an ordinal, it is transitive. Thus $m \in a$. As this holds for all $a \in A$, $m \in i$. Thus $i$ is transitive. {{qed|lemma}} === $\in$-connected === Let $x,y \in i$. Since $A$ is non-empty, it has an element $a$. By the definition of intersection, $x, y \in a$. Since $a$ is an ordinal, it is $\in$-connected. Thus $x \in y$ or $y \in x$. As this holds for all $x, y \in i$, $i$ is $\in$-connected. {{qed|lemma}} === Well-founded === Let $b$ be a non-empty subset of $i$. Let $a \in A$ (such exists because $A$ is non-empty). By Intersection is Largest Subset, $b \subseteq a$. Since $a$ is an ordinal it is well-founded. Thus $b$ has an element $x$ such that $x \cap b = \varnothing$. As this holds for all such $b$, $i$ is well-founded. {{qed}} Category:Ordinals i2frgc6mpmaln0n9etgjv69mtis7d0u
Let $\exists r, s \in S: r \wedge \neg r = a, \ s \wedge \neg s = b$ Then: {{begin-eqn}} {{eqn | l = a       | r = r \wedge \neg r       | c = by hypothesis }} {{eqn | r = \paren {s \wedge \neg s} \vee \paren {r \wedge \neg r}       | c = Boolean Algebra: Axiom $(BA_2 \ 5)$ }} {{eqn | r = \paren {r \wedge \neg r} \vee \paren {s \wedge \neg s}       | c = Boolean Algebra: Axiom $(BA_2 \ 1)$ }} {{eqn | r = s \wedge \neg s       | c = Boolean Algebra: Axiom $(BA_2 \ 5)$ }} {{eqn | r = b       | c = by hypothesis }} {{end-eqn}} Thus, whatever $r$ and $s$ may be: : $r \wedge \neg r = s \wedge \neg s$ This unique element can be assigned the symbol $\bot$ and named '''bottom''' as required. {{qed}}
Let $\exists r, s \in S: r \vee \neg r = a, \ s \vee \neg s = b$ Then: {{begin-eqn}} {{eqn | l = a       | r = r \vee \neg r       | c = by hypothesis }} {{eqn | r = \paren {s \vee \neg s} \wedge \paren {r \vee \neg r}       | c = Boolean Algebra: Axiom $(\text {BA}_2 \ 5)$ }} {{eqn | r = \paren {r \vee \neg r} \wedge \paren {s \vee \neg s}       | c = Boolean Algebra: Axiom $(\text {BA}_2 \ 1)$ }} {{eqn | r = s \vee \neg s       | c = Boolean Algebra: Axiom $(\text {BA}_2 \ 5)$ }} {{eqn | r = b       | c = by hypothesis }} {{end-eqn}} Thus, whatever $r$ and $s$ may be: :$r \vee \neg r = s \vee \neg s$ This unique element can be assigned the symbol $\top$ and named '''top''' as required. {{qed}}
Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be ordered sets. === Definition 1 implies Definition 2 === Let $\phi:S \to T$ be an order isomorphism by Definition 1. Then $\phi$ is bijective, and thus trivially surjective. Let $x, y \in S$. Then by Definition 1: :$x \preceq_1 y \implies \phi \left({x}\right) \preceq_2 \phi \left({y}\right)$ Suppose that $\phi \left({x}\right) \preceq_2 \phi \left({y}\right)$. Then by Definition 1: :$\phi^{-1}\left({\phi \left({x}\right)}\right) \preceq_1 \phi^{-1} \left({\phi \left({y}\right)}\right)$ By the definition of inverse: :$x \preceq_1 y$. Thus by Rule of Implication: :$\phi \left({x}\right) \preceq_2 \phi \left({y}\right) \implies x \preceq_1 y$. So: :$x \preceq_1 y \iff \phi \left({x}\right) \preceq_2 \phi \left({y}\right)$ As this holds for all $x, y \in S$ and $\phi$ is surjective, $\phi$ is an order isomorphism by Definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\phi:S \to T$ be an order isomorphism by Definition 2. Then $\phi$ is a surjective order embedding. By Order Embedding is Injection, $\phi$ is injective. As it is also surjective, $\phi$ is bijective, and thus satisfies the first condition of Definition 1. Since $\phi$ is an order embedding: : $\forall x, y \in S: \left({x \preceq_1 y \iff \phi\left({x}\right) \preceq_2 \phi\left({y}\right)}\right)$ Thus : $\forall x, y \in S: \left({x \preceq_1 y \implies \phi\left({x}\right) \preceq_2 \phi\left({y}\right)}\right)$ This satisfies the second condition of Definition 1. Furthermore: : $\forall x, y \in S: \left({\phi\left({x}\right) \preceq_2 \phi\left({y}\right) \implies x \preceq_1 y}\right)$ Let $p, q \in T$ and let $p \preceq_2 q$. Then since $\phi$ is bijective, it has an inverse $\phi^{-1}$ such that :$\phi\left({\phi^{-1}\left({p}\right)}\right) = p$ :$\phi\left({\phi^{-1}\left({q}\right)}\right) = q$ Thus $\phi\left({\phi^{-1}\left({p}\right)}\right) \preceq_2 \phi\left({\phi^{-1}\left({q}\right)}\right)$. Therefore: :$\phi^{-1}\left({p}\right) \preceq_1 \phi^{-1}\left({q}\right)$. Thus we see that: : $\forall p, q \in T: \left({p \preceq_2 q \implies \phi^{-1}\left({p}\right) \preceq_2 \phi^{-1}\left({q}\right)}\right)$ Thus we have satisfied the final condition of Definition 1. {{qed}} Category:Order Isomorphisms h8ibzos6s2og12e98tpm0o22rrfyihy
Let $a_1 \preceq a_2$. Then by the definition of lower closure: :$a_1 \in {a_2}^\preceq$ Let $a_3 \in {a_1}^\preceq$. Then by definition: :$a_3 \preceq a_1$ As an ordering is transitive, it follows that: :$a_3 \preceq a_2$ and so: :$a_3 \in {a_2}^\preceq$ This holds for all $a_3 \in {a_1}^\preceq$. Thus by definition of subset: :${a_1}^\preceq \subseteq {a_2}^\preceq$ {{qed}} Category:Order Theory lut06i1dlqz5c71uz5dlnzd7qpxmw5f
Follows from: : Restriction of Transitive Relation is Transitive : Restriction of Antireflexive Relation is Antireflexive : Restriction of Connected Relation is Connected {{qed}} Category:Order Theory 5xe55ko8u2svkmxnc3vzi33rnk4odei
By Restriction of Ordering is Ordering, $\preceq'$ is an ordering. Let $A$ be a non-empty subset of $T$. {{explain}} Then $A$ is a non-empty subset of $S$. Since $\preceq$ is well-founded, $A$ has a minimal element $m$ with respect to $\preceq$. Let $x \in A$ and suppose $x \preceq' m$. Then by the definition of restriction, $x \preceq m$. Thus by the definition of a minimal element, $x = m$. As this holds for all $x \in A$, $m$ is minimal in $A$ with respect to $\preceq'$. As this holds for all subsets $A$ of $T$, $\preceq'$ is a well-founded ordering of $T$. {{qed}} Category:Order Theory gxz77g5woviyhcod6vl0y0nmzpwee3y
By the definition of well-ordering, $\preceq$ is a well-founded total ordering. By Restriction of Total Ordering is Total Ordering, $\preceq'$ is a total ordering. By Restriction of Well-Founded Ordering, $\preceq'$ is well-founded. Thus $\preceq'$ is a well-ordering. {{qed}} Category:Well-Orderings 2ao0noil1turi1fr4eb23cc72f8hj8v
{{begin-eqn}} {{eqn | l = \Sigma       | o = \ne       | r = \O       | c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 1)$}} }} {{eqn | ll= \leadsto       | lo= \exists A:        | l = A        | o = \in       | r = \Sigma       | c = {{Defof|Empty Set}} }} {{eqn | ll= \leadsto       | l = \Omega \setminus A       | o = \in       | r = \Sigma       | c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 2)$}} }} {{eqn | ll= \leadsto       | l = A \cup \paren {\Omega \setminus A}       | o = \in       | r = \Sigma       | c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 3)$}} }} {{eqn | ll= \leadsto       | l = \Omega       | o = \in       | r = \Sigma       | c = Union with Relative Complement }} {{end-eqn}} {{qed}}