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Theorem difxp2 6029
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 6027 . 2 ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) = (((𝐴𝐴) × 𝐵) ∪ (𝐴 × (𝐵𝐶)))
2 difid 4285 . . . . 5 (𝐴𝐴) = ∅
32xpeq1i 5577 . . . 4 ((𝐴𝐴) × 𝐵) = (∅ × 𝐵)
4 0xp 5646 . . . 4 (∅ × 𝐵) = ∅
53, 4eqtri 2765 . . 3 ((𝐴𝐴) × 𝐵) = ∅
65uneq1i 4073 . 2 (((𝐴𝐴) × 𝐵) ∪ (𝐴 × (𝐵𝐶))) = (∅ ∪ (𝐴 × (𝐵𝐶)))
7 uncom 4067 . . 3 (∅ ∪ (𝐴 × (𝐵𝐶))) = ((𝐴 × (𝐵𝐶)) ∪ ∅)
8 un0 4305 . . 3 ((𝐴 × (𝐵𝐶)) ∪ ∅) = (𝐴 × (𝐵𝐶))
97, 8eqtri 2765 . 2 (∅ ∪ (𝐴 × (𝐵𝐶))) = (𝐴 × (𝐵𝐶))
101, 6, 93eqtrri 2770 1 (𝐴 × (𝐵𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1543  cdif 3863  cun 3864  c0 4237   × cxp 5549
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2016  ax-8 2112  ax-9 2120  ax-12 2175  ax-ext 2708  ax-sep 5192  ax-nul 5199  ax-pr 5322
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848  df-3an 1091  df-tru 1546  df-fal 1556  df-ex 1788  df-sb 2071  df-clab 2715  df-cleq 2729  df-clel 2816  df-ral 3066  df-rex 3067  df-rab 3070  df-v 3410  df-dif 3869  df-un 3871  df-in 3873  df-ss 3883  df-nul 4238  df-if 4440  df-sn 4542  df-pr 4544  df-op 4548  df-opab 5116  df-xp 5557  df-rel 5558
This theorem is referenced by:  difxp2ss  30590  imadifxp  30659  sxbrsigalem2  31965
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