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Theorem difxp1 6154
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 6153 . 2 ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶)) = (((𝐴𝐵) × 𝐶) ∪ (𝐴 × (𝐶𝐶)))
2 difid 4362 . . . . 5 (𝐶𝐶) = ∅
32xpeq2i 5693 . . . 4 (𝐴 × (𝐶𝐶)) = (𝐴 × ∅)
4 xp0 6147 . . . 4 (𝐴 × ∅) = ∅
53, 4eqtri 2752 . . 3 (𝐴 × (𝐶𝐶)) = ∅
65uneq2i 4152 . 2 (((𝐴𝐵) × 𝐶) ∪ (𝐴 × (𝐶𝐶))) = (((𝐴𝐵) × 𝐶) ∪ ∅)
7 un0 4382 . 2 (((𝐴𝐵) × 𝐶) ∪ ∅) = ((𝐴𝐵) × 𝐶)
81, 6, 73eqtrri 2757 1 ((𝐴𝐵) × 𝐶) = ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1533  cdif 3937  cun 3938  c0 4314   × cxp 5664
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 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-10 2129  ax-11 2146  ax-12 2163  ax-ext 2695  ax-sep 5289  ax-nul 5296  ax-pr 5417
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-3an 1086  df-tru 1536  df-fal 1546  df-ex 1774  df-nf 1778  df-sb 2060  df-clab 2702  df-cleq 2716  df-clel 2802  df-ral 3054  df-rex 3063  df-rab 3425  df-v 3468  df-dif 3943  df-un 3945  df-in 3947  df-ss 3957  df-nul 4315  df-if 4521  df-sn 4621  df-pr 4623  df-op 4627  df-br 5139  df-opab 5201  df-xp 5672  df-rel 5673  df-cnv 5674
This theorem is referenced by:  resdifdi  6225  difxp1ss  32229  sxbrsigalem2  33774
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