Theorem xpdisj1 5152

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

Assertion

Ref	Expression
xpdisj1	⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ∅)

Proof of Theorem xpdisj1

Step	Hyp	Ref	Expression
1		inxp 4855	. 2 ⊢ ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷))
2		xpeq1 4732	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷)) = (∅ × (𝐶 ∩ 𝐷)))
3		0xp 4798	. . 3 ⊢ (∅ × (𝐶 ∩ 𝐷)) = ∅
4	2, 3	eqtrdi 2278	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷)) = ∅)
5	1, 4	eqtrid 2274	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ∅)

Syntax hints:   → wi 4   = wceq 1395   ∩ cin 3196  ∅c0 3491   × cxp 4716

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 617  ax-in2 618  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-14 2203  ax-ext 2211  ax-sep 4201  ax-pow 4257  ax-pr 4292

This theorem depends on definitions:  df-bi 117  df-3an 1004  df-tru 1398  df-fal 1401  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-ral 2513  df-rex 2514  df-v 2801  df-dif 3199  df-un 3201  df-in 3203  df-ss 3210  df-nul 3492  df-pw 3651  df-sn 3672  df-pr 3673  df-op 3675  df-opab 4145  df-xp 4724  df-rel 4725

This theorem is referenced by:  djudisj  5155  xp01disjl  6578  xpfi  7090  djuinr  7226