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Theorem difxp2 5460
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 5458 . 2 ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) = (((𝐴𝐴) × 𝐵) ∪ (𝐴 × (𝐵𝐶)))
2 difid 3896 . . . . 5 (𝐴𝐴) = ∅
32xpeq1i 5044 . . . 4 ((𝐴𝐴) × 𝐵) = (∅ × 𝐵)
4 0xp 5107 . . . 4 (∅ × 𝐵) = ∅
53, 4eqtri 2626 . . 3 ((𝐴𝐴) × 𝐵) = ∅
65uneq1i 3719 . 2 (((𝐴𝐴) × 𝐵) ∪ (𝐴 × (𝐵𝐶))) = (∅ ∪ (𝐴 × (𝐵𝐶)))
7 uncom 3713 . . 3 (∅ ∪ (𝐴 × (𝐵𝐶))) = ((𝐴 × (𝐵𝐶)) ∪ ∅)
8 un0 3913 . . 3 ((𝐴 × (𝐵𝐶)) ∪ ∅) = (𝐴 × (𝐵𝐶))
97, 8eqtri 2626 . 2 (∅ ∪ (𝐴 × (𝐵𝐶))) = (𝐴 × (𝐵𝐶))
101, 6, 93eqtrri 2631 1 (𝐴 × (𝐵𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶))
Colors of variables: wff setvar class
Syntax hints:   = wceq 1474  cdif 3531  cun 3532  c0 3868   × cxp 5021
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1711  ax-4 1726  ax-5 1825  ax-6 1873  ax-7 1920  ax-9 1984  ax-10 2004  ax-11 2019  ax-12 2031  ax-13 2227  ax-ext 2584  ax-sep 4698  ax-nul 4707  ax-pr 4823
This theorem depends on definitions:  df-bi 195  df-or 383  df-an 384  df-3an 1032  df-tru 1477  df-ex 1695  df-nf 1700  df-sb 1866  df-clab 2591  df-cleq 2597  df-clel 2600  df-nfc 2734  df-ral 2895  df-rex 2896  df-rab 2899  df-v 3169  df-dif 3537  df-un 3539  df-in 3541  df-ss 3548  df-nul 3869  df-if 4031  df-sn 4120  df-pr 4122  df-op 4126  df-opab 4633  df-xp 5029  df-rel 5030
This theorem is referenced by:  imadifxp  28597  sxbrsigalem2  29476
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