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Theorem difxp2 6156
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 × (𝐵 ∖ 𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶))

Proof of Theorem difxp2
StepHypRef Expression
1 difxp 6154 . 2 ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) = (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶)))
2 difid 4324 . . . . 5 (𝐴 ∖ 𝐴) = ∅
32xpeq1i 5676 . . . 4 ((𝐴 ∖ 𝐴) × 𝐵) = (∅ × 𝐵)
4 0xp 5749 . . . 4 (∅ × 𝐵) = ∅
53, 4eqtri 2783 . . 3 ((𝐴 ∖ 𝐴) × 𝐵) = ∅
65uneq1i 4163 . 2 (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶)))
7 uncom 4157 . . 3 (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅)
8 un0 4343 . . 3 ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅) = (𝐴 × (𝐵 ∖ 𝐶))
97, 8eqtri 2783 . 2 (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (𝐴 × (𝐵 ∖ 𝐶))
101, 6, 93eqtrri 2788 1 (𝐴 × (𝐵 ∖ 𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   = wceq 1569   ∖ cdif 4045   ∪ cun 4149  ∅c0 4278   × cxp 5648
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732  ax-sep 5248  ax-pr 5390
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-ss 3896  df-dif 4047  df-in 4091  df-un 4151  df-nul 4279  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-opab 5167  df-xp 5656  df-rel 5657
This theorem is used by:  difxp2ss  33166  imadifxp  33242  sxbrsigalem2  34964
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