Theorem difuncomp 30317

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		incom 4128	. . . 4 ⊢ (𝐶 ∩ 𝐴) = (𝐴 ∩ 𝐶)
2		sseqin2 4142	. . . . 5 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐶 ∩ 𝐴) = 𝐴)
3	2	biimpi 219	. . . 4 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∩ 𝐴) = 𝐴)
4	1, 3	syl5reqr 2848	. . 3 ⊢ (𝐴 ⊆ 𝐶 → 𝐴 = (𝐴 ∩ 𝐶))
5	4	difeq1d 4049	. 2 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∖ 𝐵) = ((𝐴 ∩ 𝐶) ∖ 𝐵))
6		difundi 4206	. . . 4 ⊢ (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)) = ((𝐶 ∖ (𝐶 ∖ 𝐴)) ∩ (𝐶 ∖ 𝐵))
7		dfss4 4185	. . . . . 6 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
8	7	biimpi 219	. . . . 5 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
9	8	ineq1d 4138	. . . 4 ⊢ (𝐴 ⊆ 𝐶 → ((𝐶 ∖ (𝐶 ∖ 𝐴)) ∩ (𝐶 ∖ 𝐵)) = (𝐴 ∩ (𝐶 ∖ 𝐵)))
10	6, 9	syl5eq 2845	. . 3 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)) = (𝐴 ∩ (𝐶 ∖ 𝐵)))
11		indif2 4197	. . 3 ⊢ (𝐴 ∩ (𝐶 ∖ 𝐵)) = ((𝐴 ∩ 𝐶) ∖ 𝐵)
12	10, 11	eqtrdi 2849	. 2 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)) = ((𝐴 ∩ 𝐶) ∖ 𝐵))
13	5, 12	eqtr4d 2836	1 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∖ 𝐵) = (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)))

Syntax hints:   → wi 4   = wceq 1538   ∖ cdif 3878   ∪ cun 3879   ∩ cin 3880   ⊆ wss 3881

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 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2770

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-rab 3115  df-v 3443  df-dif 3884  df-un 3886  df-in 3888  df-ss 3898

This theorem is referenced by:  ldgenpisyslem1  31532