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6 May 2024
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08:47 | User:Leigh.Samphier/Matroids/Corollary of Set Difference Then Union Equals Union Then Set Difference 2 changes history +52 [Leigh.Samphier (2×)] | |||
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08:47 (cur | prev) 0 Leigh.Samphier talk contribs | ||||
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08:47 (cur | prev) +52 Leigh.Samphier talk contribs | ||||
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N 08:36 | User:Leigh.Samphier/Matroids/Set Difference Then Union Equals Union Then Set Difference 4 changes history +793 [Leigh.Samphier (4×)] | |||
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08:36 (cur | prev) −48 Leigh.Samphier talk contribs | ||||
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08:34 (cur | prev) +225 Leigh.Samphier talk contribs | ||||
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08:22 (cur | prev) −67 Leigh.Samphier talk contribs | ||||
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08:21 (cur | prev) +683 Leigh.Samphier talk contribs (Created page with "{{Proofread}} == Theorem == Let $S, A, B$ be sets. Let $A \subseteq S$. Let $A \cap B = \O$. Then: :$\paren{S \setminus A} \cup B = \paren{S \cup B} \setminus A$ == Proof == From Subsets of Disjoint Sets are Disjoint: :$A \cap B = \O$ From Set Difference with Disjoint Set: :$(1) \quad B \setminus A = B$ We have: {{begin-eqn}} {{eqn | l = \paren{S \cup B} \setminus A | r = \paren {S \setminus A} \cup \paren {B \setminus...") | |||
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08:28 | User:Leigh.Samphier/Matroids/Set Difference Then Union Equals Union Then Set Difference/Corollary 2 changes history −6 [Leigh.Samphier (2×)] | |||
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08:28 (cur | prev) +27 Leigh.Samphier talk contribs | ||||
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08:27 (cur | prev) −33 Leigh.Samphier talk contribs | ||||