Theorem undifr 4458

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 4164	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐵 ∪ 𝐴) = 𝐵)
2		undif1 4451	. . 3 ⊢ ((𝐵 ∖ 𝐴) ∪ 𝐴) = (𝐵 ∪ 𝐴)
3	2	eqeq1i 2740	. 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 1540   ∖ cdif 3923   ∪ cun 3924   ⊆ wss 3926

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2065  df-clab 2714  df-cleq 2727  df-clel 2809  df-rab 3416  df-v 3461  df-dif 3929  df-un 3931  df-in 3933  df-ss 3943  df-nul 4309

This theorem is referenced by:  difsnid  4786  f1ofvswap  7299  ralxpmap  8910  psdmullem  22103  psdmul  22104  tocyc01  33129  rprmdvdsprod  33549  aks6d1c5lem3  42150  selvvvval  42608  evlselvlem  42609  evlselv  42610  isubgr3stgrlem3  47980