Theorem xpdisj1 5168

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 4870	. 2 ⊢ ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷))
2		xpeq1 4745	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷)) = (∅ × (𝐶 ∩ 𝐷)))
3		0xp 4812	. . 3 ⊢ (∅ × (𝐶 ∩ 𝐷)) = ∅
4	2, 3	eqtrdi 2280	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷)) = ∅)
5	1, 4	eqtrid 2276	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ∅)

Syntax hints:   → wi 4   = wceq 1398   ∩ cin 3200  ∅c0 3496   × cxp 4729

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 2205  ax-ext 2213  ax-sep 4212  ax-pow 4270  ax-pr 4305

This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-fal 1404  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2364  df-ral 2516  df-rex 2517  df-v 2805  df-dif 3203  df-un 3205  df-in 3207  df-ss 3214  df-nul 3497  df-pw 3658  df-sn 3679  df-pr 3680  df-op 3682  df-opab 4156  df-xp 4737  df-rel 4738

This theorem is referenced by:  djudisj  5171  xp01disjl  6645  xpfi  7167  djuinr  7305