Theorem undifr 4482

Description: Union of complementary parts into whole. (Contributed by Thierry Arnoux, 21-Nov-2023.) (Proof shortened by SN, 11-Mar-2025.)

Assertion

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

Proof of Theorem undifr

Step	Hyp	Ref	Expression
1		ssequn2 4183	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐵 ∪ 𝐴) = 𝐵)
2		undif1 4475	. . 3 ⊢ ((𝐵 ∖ 𝐴) ∪ 𝐴) = (𝐵 ∪ 𝐴)
3	2	eqeq1i 2736	. 2 ⊢ (((𝐵 ∖ 𝐴) ∪ 𝐴) = 𝐵 ↔ (𝐵 ∪ 𝐴) = 𝐵)
4	1, 3	bitr4i 278	1 ⊢ (𝐴 ⊆ 𝐵 ↔ ((𝐵 ∖ 𝐴) ∪ 𝐴) = 𝐵)

Syntax hints:   ↔ wb 205   = wceq 1540   ∖ cdif 3945   ∪ cun 3946   ⊆ wss 3948

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2702

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1543  df-fal 1553  df-ex 1781  df-sb 2067  df-clab 2709  df-cleq 2723  df-clel 2809  df-rab 3432  df-v 3475  df-dif 3951  df-un 3953  df-in 3955  df-ss 3965  df-nul 4323

This theorem is referenced by:  difsnid  4813  f1ofvswap  7307  ralxpmap  8896  tocyc01  32713  selvvvval  41620  evlselvlem  41621  evlselv  41622