Theorem undifr 4423

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 4129	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐵 ∪ 𝐴) = 𝐵)
2		undif1 4416	. . 3 ⊢ ((𝐵 ∖ 𝐴) ∪ 𝐴) = (𝐵 ∪ 𝐴)
3	2	eqeq1i 2741	. 2 ⊢ (((𝐵 ∖ 𝐴) ∪ 𝐴) = 𝐵 ↔ (𝐵 ∪ 𝐴) = 𝐵)
4	1, 3	bitr4i 278	1 ⊢ (𝐴 ⊆ 𝐵 ↔ ((𝐵 ∖ 𝐴) ∪ 𝐴) = 𝐵)

Syntax hints:   ↔ wb 206   = wceq 1542   ∖ cdif 3886   ∪ cun 3887   ⊆ wss 3889

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2715  df-cleq 2728  df-clel 2811  df-rab 3390  df-v 3431  df-dif 3892  df-un 3894  df-in 3896  df-ss 3906  df-nul 4274

This theorem is referenced by:  difsnid  4753  f1ofvswap  7261  ralxpmap  8844  psdmullem  22131  psdmul  22132  tocyc01  33179  rprmdvdsprod  33594  evlextv  33686  esplyind  33719  esplyindfv  33720  vietalem  33723  aks6d1c5lem3  42576  selvvvval  43018  evlselvlem  43019  evlselv  43020  isubgr3stgrlem3  48444