Theorem undifrOLD 4434

Description: Obsolete version of undifr 4433 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 4432	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∪ (𝐵 ∖ 𝐴)) = 𝐵)
2		uncom 4108	. . 3 ⊢ (𝐴 ∪ (𝐵 ∖ 𝐴)) = ((𝐵 ∖ 𝐴) ∪ 𝐴)
3	2	eqeq1i 2736	. 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 1541   ∖ cdif 3899   ∪ cun 3900   ⊆ wss 3902

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-rab 3396  df-v 3438  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4284

This theorem is referenced by: (None)