Theorem undif 2397

Description: Union of complementary parts into whole.

Assertion

Ref	Expression
undif	⊢ (A ⊆ B ↔ (A ∪ (B ∖ A)) = B)

Proof of Theorem undif

Step	Hyp	Ref	Expression
1		ssequn1 2252	. 2 ⊢ (A ⊆ B ↔ (A ∪ B) = B)
2		undif2 2394	. . 3 ⊢ (A ∪ (B ∖ A)) = (A ∪ B)
3	2	eqeq1i 1525	. 2 ⊢ ((A ∪ (B ∖ A)) = B ↔ (A ∪ B) = B)
4	1, 3	bitr4i 174	1 ⊢ (A ⊆ B ↔ (A ∪ (B ∖ A)) = B)

Syntax hints:   ↔ wb 144   = wceq 992   ∖ cdif 2096   ∪ cun 2097   ⊆ wss 2099

This theorem is referenced by:  difsnid 2531  dfdom2 4525  sbthlem5 4596  sbthlem6 4597  fodomr 4628  mapdom2 4641  limensuci 4653  unfi 4697  xrsupss 6246  xrinfmss 6247  rcfpfillem6 11094  cptclsscpt 11489  dfcon2 11501  dif1en 11833  indexf 11847

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 998  ax-gen 999  ax-8 1000  ax-10 1002  ax-12 1004  ax-17 1007  ax-4 1009  ax-5o 1011  ax-6o 1014  ax-9o 1159  ax-10o 1177  ax-16 1247  ax-11o 1255  ax-ext 1500

This theorem depends on definitions:  df-bi 145  df-or 222  df-an 223  df-ex 1017  df-sb 1209  df-clab 1506  df-cleq 1511  df-clel 1514  df-ne 1630  df-v 1858  df-dif 2101  df-un 2102  df-in 2103  df-ss 2105  df-nul 2333

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