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Combined display of all available logs of ProofWiki. You can narrow down the view by selecting a log type, the username (case-sensitive), or the affected page (also case-sensitive).
- 13:41, 28 December 2022 Anjo talk contribs created page Relative Complement of Cartesian Product/Proof 1 (Created page with "== Theorem == Let $A$ and $B$ be sets. Let $X \subseteq A$ and $Y \subseteq B$. Then: :$\relcomp {A \mathop \times B} {X \times Y} = \paren {A \times \relcomp B Y} \cup \paren {\relcomp A X \times B}$ == Proof == From Set with Relative Complement forms Partition: :$A = \set {X \mid \relcomp A X}$ :$B = \set {Y \mid \relcomp B Y}$ and so by definition of partition: :$A = X \cup \relcomp A X$ :$B = Y \cup...")
- 13:41, 28 December 2022 Anjo talk contribs created page Relative Complement of Cartesian Product/Proof 2 (Created page with "== Theorem == Let $A$ and $B$ be sets. Let $X \subseteq A$ and $Y \subseteq B$. Then: :$\relcomp {A \mathop \times B} {X \times Y} $ == Proof == {{begin-eqn}} {{eqn | r = \relcomp {A \mathop \times B} {X \times Y} | o = }} {{eqn | r = A \times B \setminus X \times Y | c = {{Defof|Relative Complement}} | o = | ll= \leadstoandfrom }} {{eqn | r = \set {\tuple {x, y}: x \in A \land y \in B \land \neg (x \in X \land y \in Y)}...")
- 17:23, 24 December 2022 Anjo talk contribs created page Set is Subset of Union/Family of Sets/Proof 1 (Created page with "Let $x \in S_\beta$ for some $\beta \in I$. Then: {{begin-eqn}} {{eqn | l = x | o = \in | r = S_\beta | c = }} {{eqn | ll= \leadsto | l = x | o = \in | r = \set {x: \exists \alpha \in I: x \in S_\alpha} | c = {{Defof|Indexed Family of Sets}} }} {{eqn | ll= \leadsto | l = x | o = \in | r = \bigcup_{\alpha \mathop \in I} S_\alpha | c = {{Defof|Union of Family}} }} {{eqn | ll= \leadsto | l = S_\beta...")
- 17:18, 24 December 2022 Anjo talk contribs created page Set is Subset of Union/Family of Sets/Proof 2 (Created page with "Let a arbitrary $\beta \in I$. Then: {{begin-eqn}} {{eqn | l = \beta | o = \in | r = I | c = }} {{eqn | ll= \leadsto | l = \set {\beta} | o = \subseteq | r = I | c = Singleton of Element is Subset }} {{eqn | ll= \leadsto | l = \bigcup_{} \set {S_\beta} | o = \subseteq | r = \bigcup_{\alpha \mathop \in I} S_\alpha | c = Union of Subset of Family is Subset of Union of Family }} {{eqn | ll= \leadsto...")
- 23:48, 22 December 2022 Anjo talk contribs created page Union with Superset is Superset/Proof 1 (Created page with "Let $S \cup T = T$. Then by definition of set equality: :$S \cup T \subseteq T$ Thus: {{begin-eqn}} {{eqn | l = S | o = \subseteq | r = S \cup T | c = Subset of Union }} {{eqn | ll= \leadsto | l = S | o = \subseteq | r = T | c = Subset Relation is Transitive }} {{end-eqn}} Now let $S \subseteq T$. From Subset of Union, we have $S \cup T \supseteq T$. We also have: {{beg...")
- 23:47, 22 December 2022 Anjo talk contribs created page Union with Superset is Superset/Proof 2 (Created page with "{{begin-eqn}} {{eqn | r = S \cup T = T | o = }} {{eqn | ll= \leadstoandfrom | r = \paren {x \in S \lor x \in T \iff x \in T} | o = | c = Definition of Set Equality }} {{eqn | ll= \leadstoandfrom | r = \paren {x \in S \implies x \in T} | o = | c = Conditional iff Biconditional of Consequent with Disjunction }} {{eqn | ll= \leadstoandfrom...")
- 23:22, 22 December 2022 Anjo talk contribs created page Intersection with Subset is Subset/Proof 2 (Created page with "{{begin-eqn}} {{eqn | r = S \cap T = T | o = }} {{eqn | ll= \leadstoandfrom | r = \paren {x \in S \land x \in T \iff x \in T} | o = | c = Definition of Set Equality }} {{eqn | ll= \leadstoandfrom | r = \paren {x \in S \implies x \in T} | o = | c = Conditional iff Biconditional of Consequent with Disjunction }} {{eqn | ll= \leadstoandfrom...")
- 22:52, 22 December 2022 Anjo talk contribs created page Intersection with Subset is Subset/Proof 1 (Created page with "Let $S \cap T = S$. Then by the definition of set equality, $S \subseteq S \cap T$. Thus: {{begin-eqn}} {{eqn | l = S \cap T | o = \subseteq | r = T | c = Intersection is Subset }} {{eqn | ll= \leadsto | l = S | o = \subseteq | r = T | c = Subset Relation is Transitive }} {{end-eqn}} Now let $S \subseteq T$. From Intersection is Subset we have $S \supseteq S \cap T$. We also hav...")