Theorem difres 33073

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 5667	. . 3 ⊢ (𝐶 ↾ 𝐵) = (𝐶 ∩ (𝐵 × V))
2	1	difeq2i 4071	. 2 ⊢ (𝐴 ∖ (𝐶 ↾ 𝐵)) = (𝐴 ∖ (𝐶 ∩ (𝐵 × V)))
3		difindi 4238	. . . 4 ⊢ (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴 ∖ 𝐶) ∪ (𝐴 ∖ (𝐵 × V)))
4		ssdif 4091	. . . . . . 7 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ((𝐵 × V) ∖ (𝐵 × V)))
5		difid 4325	. . . . . . 7 ⊢ ((𝐵 × V) ∖ (𝐵 × V)) = ∅
6	4, 5	sseqtrdi 3971	. . . . . 6 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ∅)
7		ss0 4352	. . . . . 6 ⊢ ((𝐴 ∖ (𝐵 × V)) ⊆ ∅ → (𝐴 ∖ (𝐵 × V)) = ∅)
8	6, 7	syl 18	. . . . 5 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) = ∅)
9	8	uneq2d 4115	. . . 4 ⊢ (𝐴 ⊆ (𝐵 × V) → ((𝐴 ∖ 𝐶) ∪ (𝐴 ∖ (𝐵 × V))) = ((𝐴 ∖ 𝐶) ∪ ∅))
10	3, 9	eqtrid 2807	. . 3 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴 ∖ 𝐶) ∪ ∅))
11		un0 4344	. . 3 ⊢ ((𝐴 ∖ 𝐶) ∪ ∅) = (𝐴 ∖ 𝐶)
12	10, 11	eqtrdi 2811	. 2 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = (𝐴 ∖ 𝐶))
13	2, 12	eqtrid 2807	1 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ↾ 𝐵)) = (𝐴 ∖ 𝐶))

This proof depends on syntax axioms:   → wi 4   = wceq 1570  Vcvv 3450   ∖ cdif 3896   ∪ cun 3897   ∩ cin 3898   ⊆ wss 3899  ∅c0 4279   × cxp 5653   ↾ cres 5657

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-res 5667

This theorem is used by:  qtophaus  34346