Theorem difuncomp 32566

Description: Express a class difference using unions and class complements. (Contributed by Thierry Arnoux, 21-Jun-2020.)

Assertion

Ref	Expression
difuncomp	⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∖ 𝐵) = (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)))

Proof of Theorem difuncomp

Step	Hyp	Ref	Expression
1		sseqin2 4223	. . . . 5 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐶 ∩ 𝐴) = 𝐴)
2	1	biimpi 216	. . . 4 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∩ 𝐴) = 𝐴)
3		incom 4209	. . . 4 ⊢ (𝐶 ∩ 𝐴) = (𝐴 ∩ 𝐶)
4	2, 3	eqtr3di 2792	. . 3 ⊢ (𝐴 ⊆ 𝐶 → 𝐴 = (𝐴 ∩ 𝐶))
5	4	difeq1d 4125	. 2 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∖ 𝐵) = ((𝐴 ∩ 𝐶) ∖ 𝐵))
6		difundi 4290	. . . 4 ⊢ (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)) = ((𝐶 ∖ (𝐶 ∖ 𝐴)) ∩ (𝐶 ∖ 𝐵))
7		dfss4 4269	. . . . . 6 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
8	7	biimpi 216	. . . . 5 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
9	8	ineq1d 4219	. . . 4 ⊢ (𝐴 ⊆ 𝐶 → ((𝐶 ∖ (𝐶 ∖ 𝐴)) ∩ (𝐶 ∖ 𝐵)) = (𝐴 ∩ (𝐶 ∖ 𝐵)))
10	6, 9	eqtrid 2789	. . 3 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)) = (𝐴 ∩ (𝐶 ∖ 𝐵)))
11		indif2 4281	. . 3 ⊢ (𝐴 ∩ (𝐶 ∖ 𝐵)) = ((𝐴 ∩ 𝐶) ∖ 𝐵)
12	10, 11	eqtrdi 2793	. 2 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)) = ((𝐴 ∩ 𝐶) ∖ 𝐵))
13	5, 12	eqtr4d 2780	1 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∖ 𝐵) = (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)))

Syntax hints:   → wi 4   = wceq 1540   ∖ cdif 3948   ∪ cun 3949   ∩ cin 3950   ⊆ wss 3951

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 2007  ax-8 2110  ax-9 2118  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1543  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-rab 3437  df-v 3482  df-dif 3954  df-un 3956  df-in 3958  df-ss 3968

This theorem is referenced by:  ldgenpisyslem1  34164