Theorem undifrOLD 4450

Description: Obsolete version of undifr 4449 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 4448	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∪ (𝐵 ∖ 𝐴)) = 𝐵)
2		uncom 4124	. . 3 ⊢ (𝐴 ∪ (𝐵 ∖ 𝐴)) = ((𝐵 ∖ 𝐴) ∪ 𝐴)
3	2	eqeq1i 2735	. 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 1540   ∖ cdif 3914   ∪ cun 3915   ⊆ wss 3917

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 2702

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 2709  df-cleq 2722  df-clel 2804  df-rab 3409  df-v 3452  df-dif 3920  df-un 3922  df-in 3924  df-ss 3934  df-nul 4300

This theorem is referenced by: (None)