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Theorem difxp2 6170
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 6168 . 2 ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) = (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶)))
2 difid 4328 . . . . 5 (𝐴 ∖ 𝐴) = ∅
32xpeq1i 5682 . . . 4 ((𝐴 ∖ 𝐴) × 𝐵) = (∅ × 𝐵)
4 0xp 5756 . . . 4 (∅ × 𝐵) = ∅
53, 4eqtri 2786 . . 3 ((𝐴 ∖ 𝐴) × 𝐵) = ∅
65uneq1i 4166 . 2 (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶)))
7 uncom 4160 . . 3 (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅)
8 un0 4347 . . 3 ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅) = (𝐴 × (𝐵 ∖ 𝐶))
97, 8eqtri 2786 . 2 (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (𝐴 × (𝐵 ∖ 𝐶))
101, 6, 93eqtrri 2791 1 (𝐴 × (𝐵 ∖ 𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   = wceq 1570   ∖ cdif 4048   ∪ cun 4152  ∅c0 4281   × cxp 5654
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735  ax-sep 5253  ax-pr 5396
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-opab 5172  df-xp 5662  df-rel 5663
This theorem is used by:  difxp2ss  33194  imadifxp  33270  sxbrsigalem2  35001
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