Theorem xpdisj1 5161

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 4864	. 2 ⊢ ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷))
2		xpeq1 4739	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷)) = (∅ × (𝐶 ∩ 𝐷)))
3		0xp 4806	. . 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 1397   ∩ cin 3199  ∅c0 3494   × cxp 4723

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 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-14 2205  ax-ext 2213  ax-sep 4207  ax-pow 4264  ax-pr 4299

This theorem depends on definitions:  df-bi 117  df-3an 1006  df-tru 1400  df-fal 1403  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2363  df-ral 2515  df-rex 2516  df-v 2804  df-dif 3202  df-un 3204  df-in 3206  df-ss 3213  df-nul 3495  df-pw 3654  df-sn 3675  df-pr 3676  df-op 3678  df-opab 4151  df-xp 4731  df-rel 4732

This theorem is referenced by:  djudisj  5164  xp01disjl  6601  xpfi  7123  djuinr  7261