Theorem xpdisj2 5190

Description: Cross 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		inxp 4891	. 2 ⊢ ((𝐶 × 𝐴) ∩ (𝐷 × 𝐵)) = ((𝐶 ∩ 𝐷) × (𝐴 ∩ 𝐵))
2		xpeq2 4766	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐶 ∩ 𝐷) × (𝐴 ∩ 𝐵)) = ((𝐶 ∩ 𝐷) × ∅))
3		xp0 5184	. . 3 ⊢ ((𝐶 ∩ 𝐷) × ∅) = ∅
4	2, 3	eqtrdi 2283	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐶 ∩ 𝐷) × (𝐴 ∩ 𝐵)) = ∅)
5	1, 4	eqtrid 2279	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐶 × 𝐴) ∩ (𝐷 × 𝐵)) = ∅)

Syntax hints:   → wi 4   = wceq 1398   ∩ cin 3212  ∅c0 3510   × cxp 4749

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 619  ax-in2 620  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-14 2208  ax-ext 2216  ax-sep 4230  ax-pow 4289  ax-pr 4324

This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-fal 1404  df-nf 1510  df-sb 1812  df-eu 2085  df-mo 2086  df-clab 2221  df-cleq 2227  df-clel 2230  df-nfc 2375  df-ral 2527  df-rex 2528  df-v 2817  df-dif 3215  df-un 3217  df-in 3219  df-ss 3226  df-nul 3511  df-pw 3673  df-sn 3697  df-pr 3698  df-op 3700  df-br 4112  df-opab 4174  df-xp 4757  df-rel 4758  df-cnv 4759

This theorem is referenced by:  xpsndisj  5191