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Theorem difres 30265
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 5566 . . 3 (𝐶𝐵) = (𝐶 ∩ (𝐵 × V))
21difeq2i 4100 . 2 (𝐴 ∖ (𝐶𝐵)) = (𝐴 ∖ (𝐶 ∩ (𝐵 × V)))
3 difindi 4262 . . . 4 (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴𝐶) ∪ (𝐴 ∖ (𝐵 × V)))
4 ssdif 4120 . . . . . . 7 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ((𝐵 × V) ∖ (𝐵 × V)))
5 difid 4334 . . . . . . 7 ((𝐵 × V) ∖ (𝐵 × V)) = ∅
64, 5sseqtrdi 4021 . . . . . 6 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ∅)
7 ss0 4356 . . . . . 6 ((𝐴 ∖ (𝐵 × V)) ⊆ ∅ → (𝐴 ∖ (𝐵 × V)) = ∅)
86, 7syl 17 . . . . 5 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) = ∅)
98uneq2d 4143 . . . 4 (𝐴 ⊆ (𝐵 × V) → ((𝐴𝐶) ∪ (𝐴 ∖ (𝐵 × V))) = ((𝐴𝐶) ∪ ∅))
103, 9syl5eq 2873 . . 3 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴𝐶) ∪ ∅))
11 un0 4348 . . 3 ((𝐴𝐶) ∪ ∅) = (𝐴𝐶)
1210, 11syl6eq 2877 . 2 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = (𝐴𝐶))
132, 12syl5eq 2873 1 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶𝐵)) = (𝐴𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1530  Vcvv 3500  cdif 3937  cun 3938  cin 3939  wss 3940  c0 4295   × cxp 5552  cres 5556
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1904  ax-6 1963  ax-7 2008  ax-8 2109  ax-9 2117  ax-10 2138  ax-11 2153  ax-12 2169  ax-ext 2798
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 844  df-tru 1533  df-ex 1774  df-nf 1778  df-sb 2063  df-clab 2805  df-cleq 2819  df-clel 2898  df-nfc 2968  df-rab 3152  df-v 3502  df-dif 3943  df-un 3945  df-in 3947  df-ss 3956  df-nul 4296  df-res 5566
This theorem is referenced by:  qtophaus  30986
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