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Theorem difxp2 5777
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 5775 . 2 ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) = (((𝐴𝐴) × 𝐵) ∪ (𝐴 × (𝐵𝐶)))
2 difid 4149 . . . . 5 (𝐴𝐴) = ∅
32xpeq1i 5338 . . . 4 ((𝐴𝐴) × 𝐵) = (∅ × 𝐵)
4 0xp 5404 . . . 4 (∅ × 𝐵) = ∅
53, 4eqtri 2821 . . 3 ((𝐴𝐴) × 𝐵) = ∅
65uneq1i 3961 . 2 (((𝐴𝐴) × 𝐵) ∪ (𝐴 × (𝐵𝐶))) = (∅ ∪ (𝐴 × (𝐵𝐶)))
7 uncom 3955 . . 3 (∅ ∪ (𝐴 × (𝐵𝐶))) = ((𝐴 × (𝐵𝐶)) ∪ ∅)
8 un0 4163 . . 3 ((𝐴 × (𝐵𝐶)) ∪ ∅) = (𝐴 × (𝐵𝐶))
97, 8eqtri 2821 . 2 (∅ ∪ (𝐴 × (𝐵𝐶))) = (𝐴 × (𝐵𝐶))
101, 6, 93eqtrri 2826 1 (𝐴 × (𝐵𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1653  cdif 3766  cun 3767  c0 4115   × cxp 5310
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-9 2166  ax-10 2185  ax-11 2200  ax-12 2213  ax-13 2377  ax-ext 2777  ax-sep 4975  ax-nul 4983  ax-pr 5097
This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-3an 1110  df-tru 1657  df-ex 1876  df-nf 1880  df-sb 2065  df-clab 2786  df-cleq 2792  df-clel 2795  df-nfc 2930  df-ral 3094  df-rex 3095  df-rab 3098  df-v 3387  df-dif 3772  df-un 3774  df-in 3776  df-ss 3783  df-nul 4116  df-if 4278  df-sn 4369  df-pr 4371  df-op 4375  df-opab 4906  df-xp 5318  df-rel 5319
This theorem is referenced by:  imadifxp  29931  sxbrsigalem2  30864
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