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Theorem difxp1 6119
Description: Difference law for Cartesian product. (Contributed by Scott Fenton, 18-Feb-2013.) (Revised by Mario Carneiro, 26-Jun-2014.)
Assertion
Ref Expression
difxp1 ((𝐴𝐵) × 𝐶) = ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶))

Proof of Theorem difxp1
StepHypRef Expression
1 difxp 6118 . 2 ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶)) = (((𝐴𝐵) × 𝐶) ∪ (𝐴 × (𝐶𝐶)))
2 difid 4325 . . . . 5 (𝐶𝐶) = ∅
32xpeq2i 5648 . . . 4 (𝐴 × (𝐶𝐶)) = (𝐴 × ∅)
4 xp0 5721 . . . 4 (𝐴 × ∅) = ∅
53, 4eqtri 2756 . . 3 (𝐴 × (𝐶𝐶)) = ∅
65uneq2i 4114 . 2 (((𝐴𝐵) × 𝐶) ∪ (𝐴 × (𝐶𝐶))) = (((𝐴𝐵) × 𝐶) ∪ ∅)
7 un0 4343 . 2 (((𝐴𝐵) × 𝐶) ∪ ∅) = ((𝐴𝐵) × 𝐶)
81, 6, 73eqtrri 2761 1 ((𝐴𝐵) × 𝐶) = ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1541  cdif 3895  cun 3896  c0 4282   × cxp 5619
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 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2705  ax-sep 5238  ax-nul 5248  ax-pr 5374
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2712  df-cleq 2725  df-clel 2808  df-ral 3049  df-rex 3058  df-rab 3397  df-v 3439  df-dif 3901  df-un 3903  df-ss 3915  df-nul 4283  df-if 4477  df-sn 4578  df-pr 4580  df-op 4584  df-opab 5158  df-xp 5627  df-rel 5628
This theorem is referenced by:  resdifdi  6190  difxp1ss  32506  sxbrsigalem2  34322
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