Theorem xpdisj1 5090

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 4796	. 2 ⊢ ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷))
2		xpeq1 4673	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷)) = (∅ × (𝐶 ∩ 𝐷)))
3		0xp 4739	. . 3 ⊢ (∅ × (𝐶 ∩ 𝐷)) = ∅
4	2, 3	eqtrdi 2242	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷)) = ∅)
5	1, 4	eqtrid 2238	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ∅)

Syntax hints:   → wi 4   = wceq 1364   ∩ cin 3152  ∅c0 3446   × cxp 4657

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 615  ax-in2 616  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-14 2167  ax-ext 2175  ax-sep 4147  ax-pow 4203  ax-pr 4238

This theorem depends on definitions:  df-bi 117  df-3an 982  df-tru 1367  df-fal 1370  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-ral 2477  df-rex 2478  df-v 2762  df-dif 3155  df-un 3157  df-in 3159  df-ss 3166  df-nul 3447  df-pw 3603  df-sn 3624  df-pr 3625  df-op 3627  df-opab 4091  df-xp 4665  df-rel 4666

This theorem is referenced by:  djudisj  5093  xp01disjl  6487  xpfi  6986  djuinr  7122