Theorem difuncomp 33246

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 4123	. . . . 5 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐶 ∩ 𝐴) = 𝐴)
2	1	biimpi 219	. . . 4 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∩ 𝐴) = 𝐴)
3		incom 4109	. . . 4 ⊢ (𝐶 ∩ 𝐴) = (𝐴 ∩ 𝐶)
4	2, 3	eqtr3di 2813	. . 3 ⊢ (𝐴 ⊆ 𝐶 → 𝐴 = (𝐴 ∩ 𝐶))
5	4	difeq1d 4065	. 2 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∖ 𝐵) = ((𝐴 ∩ 𝐶) ∖ 𝐵))
6		difundi 4238	. . . 4 ⊢ (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)) = ((𝐶 ∖ (𝐶 ∖ 𝐴)) ∩ (𝐶 ∖ 𝐵))
7		dfss4 4217	. . . . . 6 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
8	7	biimpi 219	. . . . 5 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴)
9	8	ineq1d 4119	. . . 4 ⊢ (𝐴 ⊆ 𝐶 → ((𝐶 ∖ (𝐶 ∖ 𝐴)) ∩ (𝐶 ∖ 𝐵)) = (𝐴 ∩ (𝐶 ∖ 𝐵)))
10	6, 9	eqtrid 2810	. . 3 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)) = (𝐴 ∩ (𝐶 ∖ 𝐵)))
11		indif2 4229	. . 3 ⊢ (𝐴 ∩ (𝐶 ∖ 𝐵)) = ((𝐴 ∩ 𝐶) ∖ 𝐵)
12	10, 11	eqtrdi 2814	. 2 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)) = ((𝐴 ∩ 𝐶) ∖ 𝐵))
13	5, 12	eqtr4d 2801	1 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∖ 𝐵) = (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)))

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ⊆ wss 3900   ∖ cdif 4048   ∩ cin 4092   ∪ cun 4152

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154

This theorem is used by:  ldgenpisyslem1  34907