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Theorem xpdisj2 5475
 Description: Cartesian products with disjoint sets are disjoint. (Contributed by NM, 13-Sep-2004.)
Assertion
Ref Expression
xpdisj2 ((𝐴𝐵) = ∅ → ((𝐶 × 𝐴) ∩ (𝐷 × 𝐵)) = ∅)

Proof of Theorem xpdisj2
StepHypRef Expression
1 xpeq2 5053 . 2 ((𝐴𝐵) = ∅ → ((𝐶𝐷) × (𝐴𝐵)) = ((𝐶𝐷) × ∅))
2 inxp 5176 . 2 ((𝐶 × 𝐴) ∩ (𝐷 × 𝐵)) = ((𝐶𝐷) × (𝐴𝐵))
3 xp0 5471 . . 3 ((𝐶𝐷) × ∅) = ∅
43eqcomi 2619 . 2 ∅ = ((𝐶𝐷) × ∅)
51, 2, 43eqtr4g 2669 1 ((𝐴𝐵) = ∅ → ((𝐶 × 𝐴) ∩ (𝐷 × 𝐵)) = ∅)
 Colors of variables: wff setvar class Syntax hints:   → wi 4   = wceq 1475   ∩ cin 3539  ∅c0 3874   × cxp 5036 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-9 1986  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590  ax-sep 4709  ax-nul 4717  ax-pr 4833 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-3an 1033  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-eu 2462  df-mo 2463  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-rab 2905  df-v 3175  df-dif 3543  df-un 3545  df-in 3547  df-ss 3554  df-nul 3875  df-if 4037  df-sn 4126  df-pr 4128  df-op 4132  df-br 4584  df-opab 4644  df-xp 5044  df-rel 5045  df-cnv 5046 This theorem is referenced by:  xpsndisj  5476
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