Theorem undifrOLD 4478

Description: Obsolete version of undifr 4477 as of 11-Mar-2025. (Contributed by Thierry Arnoux, 21-Nov-2023.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

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

Proof of Theorem undifrOLD

Step	Hyp	Ref	Expression
1		undif 4476	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∪ (𝐵 ∖ 𝐴)) = 𝐵)
2		uncom 4148	. . 3 ⊢ (𝐴 ∪ (𝐵 ∖ 𝐴)) = ((𝐵 ∖ 𝐴) ∪ 𝐴)
3	2	eqeq1i 2731	. 2 ⊢ ((𝐴 ∪ (𝐵 ∖ 𝐴)) = 𝐵 ↔ ((𝐵 ∖ 𝐴) ∪ 𝐴) = 𝐵)
4	1, 3	bitri 275	1 ⊢ (𝐴 ⊆ 𝐵 ↔ ((𝐵 ∖ 𝐴) ∪ 𝐴) = 𝐵)

Syntax hints:   ↔ wb 205   = wceq 1533   ∖ cdif 3940   ∪ cun 3941   ⊆ wss 3943

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2697

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1536  df-fal 1546  df-ex 1774  df-sb 2060  df-clab 2704  df-cleq 2718  df-clel 2804  df-rab 3427  df-v 3470  df-dif 3946  df-un 3948  df-in 3950  df-ss 3960  df-nul 4318

This theorem is referenced by: (None)