Theorem xpdisj2 6113

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 5639	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐶 ∩ 𝐷) × (𝐴 ∩ 𝐵)) = ((𝐶 ∩ 𝐷) × ∅))
2		inxp 5774	. 2 ⊢ ((𝐶 × 𝐴) ∩ (𝐷 × 𝐵)) = ((𝐶 ∩ 𝐷) × (𝐴 ∩ 𝐵))
3		xp0 5718	. . 3 ⊢ ((𝐶 ∩ 𝐷) × ∅) = ∅
4	3	eqcomi 2748	. 2 ⊢ ∅ = ((𝐶 ∩ 𝐷) × ∅)
5	1, 2, 4	3eqtr4g 2799	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐶 × 𝐴) ∩ (𝐷 × 𝐵)) = ∅)

Syntax hints:   → wi 4   = wceq 1547   ∩ cin 3882  ∅c0 4261   × cxp 5616

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2711  ax-sep 5218  ax-pr 5362

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-3an 1094  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-ral 3054  df-rex 3064  df-rab 3392  df-v 3433  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4262  df-if 4455  df-sn 4556  df-pr 4558  df-op 4562  df-opab 5135  df-xp 5624  df-rel 5625

This theorem is referenced by:  xpsndisj  6114