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Theorem difxp1 6121
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 6120 . 2 ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶)) = (((𝐴𝐵) × 𝐶) ∪ (𝐴 × (𝐶𝐶)))
2 difid 4334 . . . . 5 (𝐶𝐶) = ∅
32xpeq2i 5664 . . . 4 (𝐴 × (𝐶𝐶)) = (𝐴 × ∅)
4 xp0 6114 . . . 4 (𝐴 × ∅) = ∅
53, 4eqtri 2761 . . 3 (𝐴 × (𝐶𝐶)) = ∅
65uneq2i 4124 . 2 (((𝐴𝐵) × 𝐶) ∪ (𝐴 × (𝐶𝐶))) = (((𝐴𝐵) × 𝐶) ∪ ∅)
7 un0 4354 . 2 (((𝐴𝐵) × 𝐶) ∪ ∅) = ((𝐴𝐵) × 𝐶)
81, 6, 73eqtrri 2766 1 ((𝐴𝐵) × 𝐶) = ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1542  cdif 3911  cun 3912  c0 4286   × cxp 5635
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-10 2138  ax-11 2155  ax-12 2172  ax-ext 2704  ax-sep 5260  ax-nul 5267  ax-pr 5388
This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-3an 1090  df-tru 1545  df-fal 1555  df-ex 1783  df-nf 1787  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-ral 3062  df-rex 3071  df-rab 3407  df-v 3449  df-dif 3917  df-un 3919  df-in 3921  df-ss 3931  df-nul 4287  df-if 4491  df-sn 4591  df-pr 4593  df-op 4597  df-br 5110  df-opab 5172  df-xp 5643  df-rel 5644  df-cnv 5645
This theorem is referenced by:  resdifdi  6192  difxp1ss  31499  sxbrsigalem2  32950
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