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Theorem difxp2 6091
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 6089 . 2 ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) = (((𝐴𝐴) × 𝐵) ∪ (𝐴 × (𝐵𝐶)))
2 difid 4314 . . . . 5 (𝐴𝐴) = ∅
32xpeq1i 5633 . . . 4 ((𝐴𝐴) × 𝐵) = (∅ × 𝐵)
4 0xp 5703 . . . 4 (∅ × 𝐵) = ∅
53, 4eqtri 2764 . . 3 ((𝐴𝐴) × 𝐵) = ∅
65uneq1i 4103 . 2 (((𝐴𝐴) × 𝐵) ∪ (𝐴 × (𝐵𝐶))) = (∅ ∪ (𝐴 × (𝐵𝐶)))
7 uncom 4097 . . 3 (∅ ∪ (𝐴 × (𝐵𝐶))) = ((𝐴 × (𝐵𝐶)) ∪ ∅)
8 un0 4334 . . 3 ((𝐴 × (𝐵𝐶)) ∪ ∅) = (𝐴 × (𝐵𝐶))
97, 8eqtri 2764 . 2 (∅ ∪ (𝐴 × (𝐵𝐶))) = (𝐴 × (𝐵𝐶))
101, 6, 93eqtrri 2769 1 (𝐴 × (𝐵𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1540  cdif 3893  cun 3894  c0 4266   × cxp 5605
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2707  ax-sep 5237  ax-nul 5244  ax-pr 5366
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1781  df-sb 2067  df-clab 2714  df-cleq 2728  df-clel 2814  df-ral 3062  df-rex 3071  df-rab 3404  df-v 3442  df-dif 3899  df-un 3901  df-in 3903  df-ss 3913  df-nul 4267  df-if 4471  df-sn 4571  df-pr 4573  df-op 4577  df-opab 5149  df-xp 5613  df-rel 5614
This theorem is referenced by:  difxp2ss  31002  imadifxp  31071  sxbrsigalem2  32389
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