Theorem xpdisj1 6183

Description: Cartesian products with disjoint sets are disjoint. (Contributed by NM, 13-Sep-2004.)

Assertion

Ref	Expression
xpdisj1	⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ∅)

Proof of Theorem xpdisj1

Step	Hyp	Ref	Expression
1		xpeq1 5703	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷)) = (∅ × (𝐶 ∩ 𝐷)))
2		inxp 5845	. 2 ⊢ ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷))
3		0xp 5787	. . 3 ⊢ (∅ × (𝐶 ∩ 𝐷)) = ∅
4	3	eqcomi 2744	. 2 ⊢ ∅ = (∅ × (𝐶 ∩ 𝐷))
5	1, 2, 4	3eqtr4g 2800	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ∅)

Syntax hints:   → wi 4   = wceq 1537   ∩ cin 3962  ∅c0 4339   × cxp 5687

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706  ax-sep 5302  ax-nul 5312  ax-pr 5438

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-fal 1550  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-ral 3060  df-rex 3069  df-rab 3434  df-v 3480  df-dif 3966  df-un 3968  df-in 3970  df-ss 3980  df-nul 4340  df-if 4532  df-sn 4632  df-pr 4634  df-op 4638  df-opab 5211  df-xp 5695  df-rel 5696

This theorem is referenced by:  djudisj  6189  xp01disjl  8529  djuin  9956  nosupbnd2lem1  27775  noetasuplem3  27795  noetasuplem4  27796  xpdisjres  32618  esum2dlem  34073  disjxp1  45009