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Theorem difres 29258
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 5086 . . 3 (𝐶𝐵) = (𝐶 ∩ (𝐵 × V))
21difeq2i 3703 . 2 (𝐴 ∖ (𝐶𝐵)) = (𝐴 ∖ (𝐶 ∩ (𝐵 × V)))
3 difindi 3857 . . . 4 (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴𝐶) ∪ (𝐴 ∖ (𝐵 × V)))
4 ssdif 3723 . . . . . . 7 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ((𝐵 × V) ∖ (𝐵 × V)))
5 difid 3922 . . . . . . 7 ((𝐵 × V) ∖ (𝐵 × V)) = ∅
64, 5syl6sseq 3630 . . . . . 6 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ∅)
7 ss0 3946 . . . . . 6 ((𝐴 ∖ (𝐵 × V)) ⊆ ∅ → (𝐴 ∖ (𝐵 × V)) = ∅)
86, 7syl 17 . . . . 5 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) = ∅)
98uneq2d 3745 . . . 4 (𝐴 ⊆ (𝐵 × V) → ((𝐴𝐶) ∪ (𝐴 ∖ (𝐵 × V))) = ((𝐴𝐶) ∪ ∅))
103, 9syl5eq 2667 . . 3 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴𝐶) ∪ ∅))
11 un0 3939 . . 3 ((𝐴𝐶) ∪ ∅) = (𝐴𝐶)
1210, 11syl6eq 2671 . 2 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = (𝐴𝐶))
132, 12syl5eq 2667 1 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶𝐵)) = (𝐴𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1480  Vcvv 3186  cdif 3552  cun 3553  cin 3554  wss 3555  c0 3891   × cxp 5072  cres 5076
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1878  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2750  df-ral 2912  df-rab 2916  df-v 3188  df-dif 3558  df-un 3560  df-in 3562  df-ss 3569  df-nul 3892  df-res 5086
This theorem is referenced by:  qtophaus  29685
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