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Theorem difxp2 6058
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 6056 . 2 ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) = (((𝐴𝐴) × 𝐵) ∪ (𝐴 × (𝐵𝐶)))
2 difid 4301 . . . . 5 (𝐴𝐴) = ∅
32xpeq1i 5606 . . . 4 ((𝐴𝐴) × 𝐵) = (∅ × 𝐵)
4 0xp 5675 . . . 4 (∅ × 𝐵) = ∅
53, 4eqtri 2766 . . 3 ((𝐴𝐴) × 𝐵) = ∅
65uneq1i 4089 . 2 (((𝐴𝐴) × 𝐵) ∪ (𝐴 × (𝐵𝐶))) = (∅ ∪ (𝐴 × (𝐵𝐶)))
7 uncom 4083 . . 3 (∅ ∪ (𝐴 × (𝐵𝐶))) = ((𝐴 × (𝐵𝐶)) ∪ ∅)
8 un0 4321 . . 3 ((𝐴 × (𝐵𝐶)) ∪ ∅) = (𝐴 × (𝐵𝐶))
97, 8eqtri 2766 . 2 (∅ ∪ (𝐴 × (𝐵𝐶))) = (𝐴 × (𝐵𝐶))
101, 6, 93eqtrri 2771 1 (𝐴 × (𝐵𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1539  cdif 3880  cun 3881  c0 4253   × cxp 5578
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-12 2173  ax-ext 2709  ax-sep 5218  ax-nul 5225  ax-pr 5347
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-ral 3068  df-rex 3069  df-rab 3072  df-v 3424  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4254  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-opab 5133  df-xp 5586  df-rel 5587
This theorem is referenced by:  difxp2ss  30772  imadifxp  30841  sxbrsigalem2  32153
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