Theorem xpdisj2 2654

Description: Cross products with disjoint sets are disjoint.

Assertion

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

Proof of Theorem xpdisj2

Step	Hyp	Ref	Expression
1		xpeq2 2441	. . 3 ⊢ ((A ∩ B) = ∅ → ((C ∩ D) × (A ∩ B)) = ((C ∩ D) × ∅))
2		xp0 2652	. . 3 ⊢ ((C ∩ D) × ∅) = ∅
3	1, 2	syl6eq 1140	. 2 ⊢ ((A ∩ B) = ∅ → ((C ∩ D) × (A ∩ B)) = ∅)
4		inxp 2496	. 2 ⊢ ((C × A) ∩ (D × B)) = ((C ∩ D) × (A ∩ B))
5	3, 4	syl5eq 1136	1 ⊢ ((A ∩ B) = ∅ → ((C × A) ∩ (D × B)) = ∅)

Syntax hints:   → wi 2   = wceq 1091   ∩ cin 1486  ∅c0 1707   × cxp 2408

This theorem is referenced by:  xpsndisj 2655  infxpidmlem11 4943

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-13 804  ax-14 805  ax-16 922  ax-17 925  ax-ext 1074  ax-rep 1075  ax-pow 1077

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-dif 1489  df-un 1490  df-in 1491  df-ss 1492  df-nul 1708  df-pw 1799  df-sn 1811  df-pr 1812  df-op 1815  df-br 2063  df-opab 2098  df-xp 2424  df-rel 2425  df-cnv 2426

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