Theorem undifr 4449

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 4150	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐵 ∪ 𝐴) = 𝐵)
2		undif1 4442	. . 3 ⊢ ((𝐵 ∖ 𝐴) ∪ 𝐴) = (𝐵 ∪ 𝐴)
3	2	eqeq1i 2774	. 2 ⊢ (((𝐵 ∖ 𝐴) ∪ 𝐴) = 𝐵 ↔ (𝐵 ∪ 𝐴) = 𝐵)
4	1, 3	bitr4i 281	1 ⊢ (𝐴 ⊆ 𝐵 ↔ ((𝐵 ∖ 𝐴) ∪ 𝐴) = 𝐵)

Syntax hints:   ↔ wb 209   = wceq 1567   ∖ cdif 3910   ∪ cun 3911   ⊆ wss 3913

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1570  df-fal 1580  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-rab 3424  df-v 3465  df-dif 3916  df-un 3918  df-in 3920  df-ss 3930  df-nul 4295

This theorem is referenced by:  difsnid  4780  f1ofvswap  7305  ralxpmap  8893  selvvvval  22261  psdmullem  22296  psdmul  22297  tocyc01  33378  rprmdvdsprod  33768  evlextv  33876  esplyind  33909  esplyindfv  33910  vietalem  33913  aks6d1c5lem3  42793  evlselvlem  43211  evlselv  43212  isubgr3stgrlem3  48621