Theorem xpdisj2 5991

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 5545	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐶 ∩ 𝐷) × (𝐴 ∩ 𝐵)) = ((𝐶 ∩ 𝐷) × ∅))
2		inxp 5672	. 2 ⊢ ((𝐶 × 𝐴) ∩ (𝐷 × 𝐵)) = ((𝐶 ∩ 𝐷) × (𝐴 ∩ 𝐵))
3		xp0 5987	. . 3 ⊢ ((𝐶 ∩ 𝐷) × ∅) = ∅
4	3	eqcomi 2767	. 2 ⊢ ∅ = ((𝐶 ∩ 𝐷) × ∅)
5	1, 2, 4	3eqtr4g 2818	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐶 × 𝐴) ∩ (𝐷 × 𝐵)) = ∅)

Syntax hints:   → wi 4   = wceq 1538   ∩ cin 3857  ∅c0 4225   × cxp 5522

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-10 2142  ax-11 2158  ax-12 2175  ax-ext 2729  ax-sep 5169  ax-nul 5176  ax-pr 5298

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-3an 1086  df-tru 1541  df-fal 1551  df-ex 1782  df-nf 1786  df-sb 2070  df-clab 2736  df-cleq 2750  df-clel 2830  df-rab 3079  df-v 3411  df-dif 3861  df-un 3863  df-in 3865  df-ss 3875  df-nul 4226  df-if 4421  df-sn 4523  df-pr 4525  df-op 4529  df-br 5033  df-opab 5095  df-xp 5530  df-rel 5531  df-cnv 5532

This theorem is referenced by:  xpsndisj  5992