Theorem undifrOLD 4438

Description: Obsolete version of undifr 4437 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 4436	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∪ (𝐵 ∖ 𝐴)) = 𝐵)
2		uncom 4112	. . 3 ⊢ (𝐴 ∪ (𝐵 ∖ 𝐴)) = ((𝐵 ∖ 𝐴) ∪ 𝐴)
3	2	eqeq1i 2742	. 2 ⊢ ((𝐴 ∪ (𝐵 ∖ 𝐴)) = 𝐵 ↔ ((𝐵 ∖ 𝐴) ∪ 𝐴) = 𝐵)
4	1, 3	bitri 275	1 ⊢ (𝐴 ⊆ 𝐵 ↔ ((𝐵 ∖ 𝐴) ∪ 𝐴) = 𝐵)

Syntax hints:   ↔ wb 206   = wceq 1542   ∖ cdif 3900   ∪ cun 3901   ⊆ wss 3903

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 2709

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 2716  df-cleq 2729  df-clel 2812  df-rab 3402  df-v 3444  df-dif 3906  df-un 3908  df-in 3910  df-ss 3920  df-nul 4288

This theorem is referenced by: (None)