Theorem difres 32688

Description: Case when class difference in unaffected by restriction. (Contributed by Thierry Arnoux, 1-Jan-2020.)

Assertion

Ref	Expression
difres	⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ↾ 𝐵)) = (𝐴 ∖ 𝐶))

Proof of Theorem difres

Step	Hyp	Ref	Expression
1		df-res 5637	. . 3 ⊢ (𝐶 ↾ 𝐵) = (𝐶 ∩ (𝐵 × V))
2	1	difeq2i 4064	. 2 ⊢ (𝐴 ∖ (𝐶 ↾ 𝐵)) = (𝐴 ∖ (𝐶 ∩ (𝐵 × V)))
3		difindi 4233	. . . 4 ⊢ (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴 ∖ 𝐶) ∪ (𝐴 ∖ (𝐵 × V)))
4		ssdif 4085	. . . . . . 7 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ((𝐵 × V) ∖ (𝐵 × V)))
5		difid 4317	. . . . . . 7 ⊢ ((𝐵 × V) ∖ (𝐵 × V)) = ∅
6	4, 5	sseqtrdi 3963	. . . . . 6 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ∅)
7		ss0 4343	. . . . . 6 ⊢ ((𝐴 ∖ (𝐵 × V)) ⊆ ∅ → (𝐴 ∖ (𝐵 × V)) = ∅)
8	6, 7	syl 17	. . . . 5 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) = ∅)
9	8	uneq2d 4109	. . . 4 ⊢ (𝐴 ⊆ (𝐵 × V) → ((𝐴 ∖ 𝐶) ∪ (𝐴 ∖ (𝐵 × V))) = ((𝐴 ∖ 𝐶) ∪ ∅))
10	3, 9	eqtrid 2784	. . 3 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴 ∖ 𝐶) ∪ ∅))
11		un0 4335	. . 3 ⊢ ((𝐴 ∖ 𝐶) ∪ ∅) = (𝐴 ∖ 𝐶)
12	10, 11	eqtrdi 2788	. 2 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = (𝐴 ∖ 𝐶))
13	2, 12	eqtrid 2784	1 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ↾ 𝐵)) = (𝐴 ∖ 𝐶))

Syntax hints:   → wi 4   = wceq 1542  Vcvv 3430   ∖ cdif 3887   ∪ cun 3888   ∩ cin 3889   ⊆ wss 3890  ∅c0 4274   × cxp 5623   ↾ cres 5627

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 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-rab 3391  df-v 3432  df-dif 3893  df-un 3895  df-in 3897  df-ss 3907  df-nul 4275  df-res 5637

This theorem is referenced by:  qtophaus  33999