Theorem difres 32945

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 5673	. . 3 ⊢ (𝐶 ↾ 𝐵) = (𝐶 ∩ (𝐵 × V))
2	1	difeq2i 4078	. 2 ⊢ (𝐴 ∖ (𝐶 ↾ 𝐵)) = (𝐴 ∖ (𝐶 ∩ (𝐵 × V)))
3		difindi 4245	. . . 4 ⊢ (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴 ∖ 𝐶) ∪ (𝐴 ∖ (𝐵 × V)))
4		ssdif 4098	. . . . . . 7 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ((𝐵 × V) ∖ (𝐵 × V)))
5		difid 4332	. . . . . . 7 ⊢ ((𝐵 × V) ∖ (𝐵 × V)) = ∅
6	4, 5	sseqtrdi 3977	. . . . . 6 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ∅)
7		ss0 4359	. . . . . 6 ⊢ ((𝐴 ∖ (𝐵 × V)) ⊆ ∅ → (𝐴 ∖ (𝐵 × V)) = ∅)
8	6, 7	syl 18	. . . . 5 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) = ∅)
9	8	uneq2d 4122	. . . 4 ⊢ (𝐴 ⊆ (𝐵 × V) → ((𝐴 ∖ 𝐶) ∪ (𝐴 ∖ (𝐵 × V))) = ((𝐴 ∖ 𝐶) ∪ ∅))
10	3, 9	eqtrid 2810	. . 3 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴 ∖ 𝐶) ∪ ∅))
11		un0 4351	. . 3 ⊢ ((𝐴 ∖ 𝐶) ∪ ∅) = (𝐴 ∖ 𝐶)
12	10, 11	eqtrdi 2814	. 2 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = (𝐴 ∖ 𝐶))
13	2, 12	eqtrid 2810	1 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ↾ 𝐵)) = (𝐴 ∖ 𝐶))

Syntax hints:   → wi 4   = wceq 1570  Vcvv 3455   ∖ cdif 3902   ∪ cun 3903   ∩ cin 3904   ⊆ wss 3905  ∅c0 4286   × cxp 5659   ↾ cres 5663

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-rab 3417  df-v 3457  df-dif 3908  df-un 3910  df-in 3912  df-ss 3922  df-nul 4287  df-res 5673

This theorem is referenced by:  qtophaus  34226