Theorem xpdisj1 6150

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 5668	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷)) = (∅ × (𝐶 ∩ 𝐷)))
2		inxp 5811	. 2 ⊢ ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷))
3		0xp 5753	. . 3 ⊢ (∅ × (𝐶 ∩ 𝐷)) = ∅
4	3	eqcomi 2744	. 2 ⊢ ∅ = (∅ × (𝐶 ∩ 𝐷))
5	1, 2, 4	3eqtr4g 2795	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ∅)

Syntax hints:   → wi 4   = wceq 1540   ∩ cin 3925  ∅c0 4308   × cxp 5652

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2707  ax-sep 5266  ax-nul 5276  ax-pr 5402

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2065  df-clab 2714  df-cleq 2727  df-clel 2809  df-ral 3052  df-rex 3061  df-rab 3416  df-v 3461  df-dif 3929  df-un 3931  df-in 3933  df-ss 3943  df-nul 4309  df-if 4501  df-sn 4602  df-pr 4604  df-op 4608  df-opab 5182  df-xp 5660  df-rel 5661

This theorem is referenced by:  djudisj  6156  xp01disjl  8504  djuin  9932  nosupbnd2lem1  27679  noetasuplem3  27699  noetasuplem4  27700  xpdisjres  32579  esum2dlem  34123  disjxp1  45093