Theorem xpdisj2 6156

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

Assertion

Ref	Expression
xpdisj2	⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐶 × 𝐴) ∩ (𝐷 × 𝐵)) = ∅)

Proof of Theorem xpdisj2

Step	Hyp	Ref	Expression
1		xpeq2 5680	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐶 ∩ 𝐷) × (𝐴 ∩ 𝐵)) = ((𝐶 ∩ 𝐷) × ∅))
2		inxp 5816	. 2 ⊢ ((𝐶 × 𝐴) ∩ (𝐷 × 𝐵)) = ((𝐶 ∩ 𝐷) × (𝐴 ∩ 𝐵))
3		xp0 6152	. . 3 ⊢ ((𝐶 ∩ 𝐷) × ∅) = ∅
4	3	eqcomi 2745	. 2 ⊢ ∅ = ((𝐶 ∩ 𝐷) × ∅)
5	1, 2, 4	3eqtr4g 2796	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐶 × 𝐴) ∩ (𝐷 × 𝐵)) = ∅)

Syntax hints:   → wi 4   = wceq 1540   ∩ cin 3930  ∅c0 4313   × cxp 5657

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-11 2158  ax-12 2178  ax-ext 2708  ax-sep 5271  ax-nul 5281  ax-pr 5407

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2715  df-cleq 2728  df-clel 2810  df-ral 3053  df-rex 3062  df-rab 3421  df-v 3466  df-dif 3934  df-un 3936  df-in 3938  df-ss 3948  df-nul 4314  df-if 4506  df-sn 4607  df-pr 4609  df-op 4613  df-br 5125  df-opab 5187  df-xp 5665  df-rel 5666  df-cnv 5667

This theorem is referenced by:  xpsndisj  6157