Theorem undifr 4436

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 4142	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐵 ∪ 𝐴) = 𝐵)
2		undif1 4429	. . 3 ⊢ ((𝐵 ∖ 𝐴) ∪ 𝐴) = (𝐵 ∪ 𝐴)
3	2	eqeq1i 2734	. 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 3902   ∪ cun 3903   ⊆ wss 3905

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 2008  ax-8 2111  ax-9 2119  ax-ext 2701

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 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-rab 3397  df-v 3440  df-dif 3908  df-un 3910  df-in 3912  df-ss 3922  df-nul 4287

This theorem is referenced by:  difsnid  4764  f1ofvswap  7247  ralxpmap  8830  psdmullem  22068  psdmul  22069  tocyc01  33073  rprmdvdsprod  33484  aks6d1c5lem3  42113  selvvvval  42561  evlselvlem  42562  evlselv  42563  isubgr3stgrlem3  47956