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Theorem difres 32614
Description: Case when class difference in unaffected by restriction. (Contributed by Thierry Arnoux, 1-Jan-2020.)
Assertion
Ref Expression
difres (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶𝐵)) = (𝐴𝐶))

Proof of Theorem difres
StepHypRef Expression
1 df-res 5696 . . 3 (𝐶𝐵) = (𝐶 ∩ (𝐵 × V))
21difeq2i 4122 . 2 (𝐴 ∖ (𝐶𝐵)) = (𝐴 ∖ (𝐶 ∩ (𝐵 × V)))
3 difindi 4291 . . . 4 (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴𝐶) ∪ (𝐴 ∖ (𝐵 × V)))
4 ssdif 4143 . . . . . . 7 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ((𝐵 × V) ∖ (𝐵 × V)))
5 difid 4375 . . . . . . 7 ((𝐵 × V) ∖ (𝐵 × V)) = ∅
64, 5sseqtrdi 4023 . . . . . 6 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ∅)
7 ss0 4401 . . . . . 6 ((𝐴 ∖ (𝐵 × V)) ⊆ ∅ → (𝐴 ∖ (𝐵 × V)) = ∅)
86, 7syl 17 . . . . 5 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) = ∅)
98uneq2d 4167 . . . 4 (𝐴 ⊆ (𝐵 × V) → ((𝐴𝐶) ∪ (𝐴 ∖ (𝐵 × V))) = ((𝐴𝐶) ∪ ∅))
103, 9eqtrid 2788 . . 3 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴𝐶) ∪ ∅))
11 un0 4393 . . 3 ((𝐴𝐶) ∪ ∅) = (𝐴𝐶)
1210, 11eqtrdi 2792 . 2 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = (𝐴𝐶))
132, 12eqtrid 2788 1 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶𝐵)) = (𝐴𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1539  Vcvv 3479  cdif 3947  cun 3948  cin 3949  wss 3950  c0 4332   × cxp 5682  cres 5686
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-ext 2707
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1542  df-fal 1552  df-ex 1779  df-sb 2064  df-clab 2714  df-cleq 2728  df-clel 2815  df-rab 3436  df-v 3481  df-dif 3953  df-un 3955  df-in 3957  df-ss 3967  df-nul 4333  df-res 5696
This theorem is referenced by:  qtophaus  33836
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