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Theorem difxp1 6176
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 6175 . 2 ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶)) = (((𝐴𝐵) × 𝐶) ∪ (𝐴 × (𝐶𝐶)))
2 difid 4375 . . . . 5 (𝐶𝐶) = ∅
32xpeq2i 5709 . . . 4 (𝐴 × (𝐶𝐶)) = (𝐴 × ∅)
4 xp0 6169 . . . 4 (𝐴 × ∅) = ∅
53, 4eqtri 2754 . . 3 (𝐴 × (𝐶𝐶)) = ∅
65uneq2i 4160 . 2 (((𝐴𝐵) × 𝐶) ∪ (𝐴 × (𝐶𝐶))) = (((𝐴𝐵) × 𝐶) ∪ ∅)
7 un0 4395 . 2 (((𝐴𝐵) × 𝐶) ∪ ∅) = ((𝐴𝐵) × 𝐶)
81, 6, 73eqtrri 2759 1 ((𝐴𝐵) × 𝐶) = ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1534  cdif 3944  cun 3945  c0 4325   × cxp 5680
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-11 2147  ax-12 2167  ax-ext 2697  ax-sep 5304  ax-nul 5311  ax-pr 5433
This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-3an 1086  df-tru 1537  df-fal 1547  df-ex 1775  df-sb 2061  df-clab 2704  df-cleq 2718  df-clel 2803  df-ral 3052  df-rex 3061  df-rab 3420  df-v 3464  df-dif 3950  df-un 3952  df-ss 3964  df-nul 4326  df-if 4534  df-sn 4634  df-pr 4636  df-op 4640  df-br 5154  df-opab 5216  df-xp 5688  df-rel 5689  df-cnv 5690
This theorem is referenced by:  resdifdi  6247  difxp1ss  32449  sxbrsigalem2  34120
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