Theorem xpdisj1 6125

Description: Cartesian 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		xpeq1 5645	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷)) = (∅ × (𝐶 ∩ 𝐷)))
2		inxp 5787	. 2 ⊢ ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷))
3		0xp 5730	. . 3 ⊢ (∅ × (𝐶 ∩ 𝐷)) = ∅
4	3	eqcomi 2745	. 2 ⊢ ∅ = (∅ × (𝐶 ∩ 𝐷))
5	1, 2, 4	3eqtr4g 2796	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ∅)

Syntax hints:   → wi 4   = wceq 1542   ∩ cin 3888  ∅c0 4273   × cxp 5629

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2708  ax-sep 5231  ax-pr 5375

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2715  df-cleq 2728  df-clel 2811  df-ral 3052  df-rex 3062  df-rab 3390  df-v 3431  df-dif 3892  df-un 3894  df-in 3896  df-ss 3906  df-nul 4274  df-if 4467  df-sn 4568  df-pr 4570  df-op 4574  df-opab 5148  df-xp 5637  df-rel 5638

This theorem is referenced by:  djudisj  6131  xp01disjl  8427  djuin  9842  nosupbnd2lem1  27679  noetasuplem3  27699  noetasuplem4  27700  xpdisjres  32668  esum2dlem  34236  disjxp1  45500