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Theorem difxp2 5548
Description: Difference law for Cartesian product. (Contributed by Scott Fenton, 18-Feb-2013.) (Revised by Mario Carneiro, 26-Jun-2014.)
Assertion
Ref Expression
difxp2 (𝐴 × (𝐵𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶))

Proof of Theorem difxp2
StepHypRef Expression
1 difxp 5546 . 2 ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) = (((𝐴𝐴) × 𝐵) ∪ (𝐴 × (𝐵𝐶)))
2 difid 3939 . . . . 5 (𝐴𝐴) = ∅
32xpeq1i 5125 . . . 4 ((𝐴𝐴) × 𝐵) = (∅ × 𝐵)
4 0xp 5189 . . . 4 (∅ × 𝐵) = ∅
53, 4eqtri 2642 . . 3 ((𝐴𝐴) × 𝐵) = ∅
65uneq1i 3755 . 2 (((𝐴𝐴) × 𝐵) ∪ (𝐴 × (𝐵𝐶))) = (∅ ∪ (𝐴 × (𝐵𝐶)))
7 uncom 3749 . . 3 (∅ ∪ (𝐴 × (𝐵𝐶))) = ((𝐴 × (𝐵𝐶)) ∪ ∅)
8 un0 3958 . . 3 ((𝐴 × (𝐵𝐶)) ∪ ∅) = (𝐴 × (𝐵𝐶))
97, 8eqtri 2642 . 2 (∅ ∪ (𝐴 × (𝐵𝐶))) = (𝐴 × (𝐵𝐶))
101, 6, 93eqtrri 2647 1 (𝐴 × (𝐵𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1481  cdif 3564  cun 3565  c0 3907   × cxp 5102
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1720  ax-4 1735  ax-5 1837  ax-6 1886  ax-7 1933  ax-9 1997  ax-10 2017  ax-11 2032  ax-12 2045  ax-13 2244  ax-ext 2600  ax-sep 4772  ax-nul 4780  ax-pr 4897
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1038  df-tru 1484  df-ex 1703  df-nf 1708  df-sb 1879  df-clab 2607  df-cleq 2613  df-clel 2616  df-nfc 2751  df-ral 2914  df-rex 2915  df-rab 2918  df-v 3197  df-dif 3570  df-un 3572  df-in 3574  df-ss 3581  df-nul 3908  df-if 4078  df-sn 4169  df-pr 4171  df-op 4175  df-opab 4704  df-xp 5110  df-rel 5111
This theorem is referenced by:  imadifxp  29386  sxbrsigalem2  30322
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