Theorem xpdisj1 6137

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 5655	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷)) = (∅ × (𝐶 ∩ 𝐷)))
2		inxp 5798	. 2 ⊢ ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷))
3		0xp 5740	. . 3 ⊢ (∅ × (𝐶 ∩ 𝐷)) = ∅
4	3	eqcomi 2739	. 2 ⊢ ∅ = (∅ × (𝐶 ∩ 𝐷))
5	1, 2, 4	3eqtr4g 2790	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ∅)

Syntax hints:   → wi 4   = wceq 1540   ∩ cin 3916  ∅c0 4299   × cxp 5639

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 2008  ax-8 2111  ax-9 2119  ax-ext 2702  ax-sep 5254  ax-nul 5264  ax-pr 5390

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 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-ral 3046  df-rex 3055  df-rab 3409  df-v 3452  df-dif 3920  df-un 3922  df-in 3924  df-ss 3934  df-nul 4300  df-if 4492  df-sn 4593  df-pr 4595  df-op 4599  df-opab 5173  df-xp 5647  df-rel 5648

This theorem is referenced by:  djudisj  6143  xp01disjl  8459  djuin  9878  nosupbnd2lem1  27634  noetasuplem3  27654  noetasuplem4  27655  xpdisjres  32534  esum2dlem  34089  disjxp1  45070