Users' Mathboxes Mathbox for Thierry Arnoux < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >   Mathboxes  >  difres Structured version   Visualization version   GIF version

Theorem difres 32620
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 5701 . . 3 (𝐶𝐵) = (𝐶 ∩ (𝐵 × V))
21difeq2i 4133 . 2 (𝐴 ∖ (𝐶𝐵)) = (𝐴 ∖ (𝐶 ∩ (𝐵 × V)))
3 difindi 4298 . . . 4 (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴𝐶) ∪ (𝐴 ∖ (𝐵 × V)))
4 ssdif 4154 . . . . . . 7 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ((𝐵 × V) ∖ (𝐵 × V)))
5 difid 4382 . . . . . . 7 ((𝐵 × V) ∖ (𝐵 × V)) = ∅
64, 5sseqtrdi 4046 . . . . . 6 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) ⊆ ∅)
7 ss0 4408 . . . . . 6 ((𝐴 ∖ (𝐵 × V)) ⊆ ∅ → (𝐴 ∖ (𝐵 × V)) = ∅)
86, 7syl 17 . . . . 5 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐵 × V)) = ∅)
98uneq2d 4178 . . . 4 (𝐴 ⊆ (𝐵 × V) → ((𝐴𝐶) ∪ (𝐴 ∖ (𝐵 × V))) = ((𝐴𝐶) ∪ ∅))
103, 9eqtrid 2787 . . 3 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = ((𝐴𝐶) ∪ ∅))
11 un0 4400 . . 3 ((𝐴𝐶) ∪ ∅) = (𝐴𝐶)
1210, 11eqtrdi 2791 . 2 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶 ∩ (𝐵 × V))) = (𝐴𝐶))
132, 12eqtrid 2787 1 (𝐴 ⊆ (𝐵 × V) → (𝐴 ∖ (𝐶𝐵)) = (𝐴𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1537  Vcvv 3478  cdif 3960  cun 3961  cin 3962  wss 3963  c0 4339   × cxp 5687  cres 5691
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1540  df-fal 1550  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-rab 3434  df-v 3480  df-dif 3966  df-un 3968  df-in 3970  df-ss 3980  df-nul 4340  df-res 5701
This theorem is referenced by:  qtophaus  33797
  Copyright terms: Public domain W3C validator