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Theorem difxp1 6015
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 6014 . 2 ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶)) = (((𝐴𝐵) × 𝐶) ∪ (𝐴 × (𝐶𝐶)))
2 difid 4328 . . . . 5 (𝐶𝐶) = ∅
32xpeq2i 5575 . . . 4 (𝐴 × (𝐶𝐶)) = (𝐴 × ∅)
4 xp0 6008 . . . 4 (𝐴 × ∅) = ∅
53, 4eqtri 2842 . . 3 (𝐴 × (𝐶𝐶)) = ∅
65uneq2i 4134 . 2 (((𝐴𝐵) × 𝐶) ∪ (𝐴 × (𝐶𝐶))) = (((𝐴𝐵) × 𝐶) ∪ ∅)
7 un0 4342 . 2 (((𝐴𝐵) × 𝐶) ∪ ∅) = ((𝐴𝐵) × 𝐶)
81, 6, 73eqtrri 2847 1 ((𝐴𝐵) × 𝐶) = ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1530  cdif 3931  cun 3932  c0 4289   × cxp 5546
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1904  ax-6 1963  ax-7 2008  ax-8 2109  ax-9 2117  ax-10 2138  ax-11 2153  ax-12 2169  ax-ext 2791  ax-sep 5194  ax-nul 5201  ax-pr 5320
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1083  df-tru 1533  df-ex 1774  df-nf 1778  df-sb 2063  df-mo 2616  df-eu 2648  df-clab 2798  df-cleq 2812  df-clel 2891  df-nfc 2961  df-ral 3141  df-rex 3142  df-rab 3145  df-v 3495  df-dif 3937  df-un 3939  df-in 3941  df-ss 3950  df-nul 4290  df-if 4466  df-sn 4560  df-pr 4562  df-op 4566  df-br 5058  df-opab 5120  df-xp 5554  df-rel 5555  df-cnv 5556
This theorem is referenced by:  difxp1ss  30274  sxbrsigalem2  31532
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