Theorem difuncomp 32574

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 4231	. . . . 5 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐶 ∩ 𝐴) = 𝐴)
2	1	biimpi 216	. . . 4 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∩ 𝐴) = 𝐴)
3		incom 4217	. . . 4 ⊢ (𝐶 ∩ 𝐴) = (𝐴 ∩ 𝐶)
4	2, 3	eqtr3di 2790	. . 3 ⊢ (𝐴 ⊆ 𝐶 → 𝐴 = (𝐴 ∩ 𝐶))
5	4	difeq1d 4135	. 2 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∖ 𝐵) = ((𝐴 ∩ 𝐶) ∖ 𝐵))
6		difundi 4296	. . . 4 ⊢ (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)) = ((𝐶 ∖ (𝐶 ∖ 𝐴)) ∩ (𝐶 ∖ 𝐵))
7		dfss4 4275	. . . . . 6 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
8	7	biimpi 216	. . . . 5 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
9	8	ineq1d 4227	. . . 4 ⊢ (𝐴 ⊆ 𝐶 → ((𝐶 ∖ (𝐶 ∖ 𝐴)) ∩ (𝐶 ∖ 𝐵)) = (𝐴 ∩ (𝐶 ∖ 𝐵)))
10	6, 9	eqtrid 2787	. . 3 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)) = (𝐴 ∩ (𝐶 ∖ 𝐵)))
11		indif2 4287	. . 3 ⊢ (𝐴 ∩ (𝐶 ∖ 𝐵)) = ((𝐴 ∩ 𝐶) ∖ 𝐵)
12	10, 11	eqtrdi 2791	. 2 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)) = ((𝐴 ∩ 𝐶) ∖ 𝐵))
13	5, 12	eqtr4d 2778	1 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∖ 𝐵) = (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)))

Syntax hints:   → wi 4   = wceq 1537   ∖ cdif 3960   ∪ cun 3961   ∩ cin 3962   ⊆ wss 3963

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-rab 3434  df-v 3480  df-dif 3966  df-un 3968  df-in 3970  df-ss 3980

This theorem is referenced by:  ldgenpisyslem1  34144