Theorem difres 32572

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 5623	. . 3 ⊢ (𝐶 ↾ 𝐵) = (𝐶 ∩ (𝐵 × V))
2	1	difeq2i 4068	. 2 ⊢ (𝐴 ∖ (𝐶 ↾ 𝐵)) = (𝐴 ∖ (𝐶 ∩ (𝐵 × V)))
3		difindi 4237	. . . 4 ⊢ (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴 ∖ 𝐶) ∪ (𝐴 ∖ (𝐵 × V)))
4		ssdif 4089	. . . . . . 7 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ((𝐵 × V) ∖ (𝐵 × V)))
5		difid 4321	. . . . . . 7 ⊢ ((𝐵 × V) ∖ (𝐵 × V)) = ∅
6	4, 5	sseqtrdi 3970	. . . . . 6 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ∅)
7		ss0 4347	. . . . . 6 ⊢ ((𝐴 ∖ (𝐵 × V)) ⊆ ∅ → (𝐴 ∖ (𝐵 × V)) = ∅)
8	6, 7	syl 17	. . . . 5 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) = ∅)
9	8	uneq2d 4113	. . . 4 ⊢ (𝐴 ⊆ (𝐵 × V) → ((𝐴 ∖ 𝐶) ∪ (𝐴 ∖ (𝐵 × V))) = ((𝐴 ∖ 𝐶) ∪ ∅))
10	3, 9	eqtrid 2778	. . 3 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴 ∖ 𝐶) ∪ ∅))
11		un0 4339	. . 3 ⊢ ((𝐴 ∖ 𝐶) ∪ ∅) = (𝐴 ∖ 𝐶)
12	10, 11	eqtrdi 2782	. 2 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = (𝐴 ∖ 𝐶))
13	2, 12	eqtrid 2778	1 ⊢ (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ↾ 𝐵)) = (𝐴 ∖ 𝐶))

Syntax hints:   → wi 4   = wceq 1541  Vcvv 3436   ∖ cdif 3894   ∪ cun 3895   ∩ cin 3896   ⊆ wss 3897  ∅c0 4278   × cxp 5609   ↾ cres 5613

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-rab 3396  df-v 3438  df-dif 3900  df-un 3902  df-in 3904  df-ss 3914  df-nul 4279  df-res 5623

This theorem is referenced by:  qtophaus  33841