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Theorem difxp2 6165
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 6163 . 2 ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) = (((𝐴𝐴) × 𝐵) ∪ (𝐴 × (𝐵𝐶)))
2 difid 4370 . . . . 5 (𝐴𝐴) = ∅
32xpeq1i 5702 . . . 4 ((𝐴𝐴) × 𝐵) = (∅ × 𝐵)
4 0xp 5774 . . . 4 (∅ × 𝐵) = ∅
53, 4eqtri 2759 . . 3 ((𝐴𝐴) × 𝐵) = ∅
65uneq1i 4159 . 2 (((𝐴𝐴) × 𝐵) ∪ (𝐴 × (𝐵𝐶))) = (∅ ∪ (𝐴 × (𝐵𝐶)))
7 uncom 4153 . . 3 (∅ ∪ (𝐴 × (𝐵𝐶))) = ((𝐴 × (𝐵𝐶)) ∪ ∅)
8 un0 4390 . . 3 ((𝐴 × (𝐵𝐶)) ∪ ∅) = (𝐴 × (𝐵𝐶))
97, 8eqtri 2759 . 2 (∅ ∪ (𝐴 × (𝐵𝐶))) = (𝐴 × (𝐵𝐶))
101, 6, 93eqtrri 2764 1 (𝐴 × (𝐵𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1540  cdif 3945  cun 3946  c0 4322   × cxp 5674
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 2702  ax-sep 5299  ax-nul 5306  ax-pr 5427
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1781  df-sb 2067  df-clab 2709  df-cleq 2723  df-clel 2809  df-ral 3061  df-rex 3070  df-rab 3432  df-v 3475  df-dif 3951  df-un 3953  df-in 3955  df-ss 3965  df-nul 4323  df-if 4529  df-sn 4629  df-pr 4631  df-op 4635  df-opab 5211  df-xp 5682  df-rel 5683
This theorem is referenced by:  difxp2ss  32195  imadifxp  32266  sxbrsigalem2  33750
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