Theorem difres 29413

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 5126	. . 3 ⊢ (𝐶 ↾ 𝐵) = (𝐶 ∩ (𝐵 × V))
2	1	difeq2i 3725	. 2 ⊢ (𝐴 ∖ (𝐶 ↾ 𝐵)) = (𝐴 ∖ (𝐶 ∩ (𝐵 × V)))
3		difindi 3881	. . . 4 ⊢ (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴 ∖ 𝐶) ∪ (𝐴 ∖ (𝐵 × V)))
4		ssdif 3745	. . . . . . 7 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ((𝐵 × V) ∖ (𝐵 × V)))
5		difid 3948	. . . . . . 7 ⊢ ((𝐵 × V) ∖ (𝐵 × V)) = ∅
6	4, 5	syl6sseq 3651	. . . . . 6 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ∅)
7		ss0 3974	. . . . . 6 ⊢ ((𝐴 ∖ (𝐵 × V)) ⊆ ∅ → (𝐴 ∖ (𝐵 × V)) = ∅)
8	6, 7	syl 17	. . . . 5 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) = ∅)
9	8	uneq2d 3767	. . . 4 ⊢ (𝐴 ⊆ (𝐵 × V) → ((𝐴 ∖ 𝐶) ∪ (𝐴 ∖ (𝐵 × V))) = ((𝐴 ∖ 𝐶) ∪ ∅))
10	3, 9	syl5eq 2668	. . 3 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴 ∖ 𝐶) ∪ ∅))
11		un0 3967	. . 3 ⊢ ((𝐴 ∖ 𝐶) ∪ ∅) = (𝐴 ∖ 𝐶)
12	10, 11	syl6eq 2672	. 2 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = (𝐴 ∖ 𝐶))
13	2, 12	syl5eq 2668	1 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ↾ 𝐵)) = (𝐴 ∖ 𝐶))

Syntax hints:   → wi 4   = wceq 1483  Vcvv 3200   ∖ cdif 3571   ∪ cun 3572   ∩ cin 3573   ⊆ wss 3574  ∅c0 3915   × cxp 5112   ↾ cres 5116

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ral 2917  df-rab 2921  df-v 3202  df-dif 3577  df-un 3579  df-in 3581  df-ss 3588  df-nul 3916  df-res 5126

This theorem is referenced by:  qtophaus  29903