Theorem xpdisj1 6110

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 5633	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷)) = (∅ × (𝐶 ∩ 𝐷)))
2		inxp 5774	. 2 ⊢ ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ((𝐴 ∩ 𝐵) × (𝐶 ∩ 𝐷))
3		0xp 5718	. . 3 ⊢ (∅ × (𝐶 ∩ 𝐷)) = ∅
4	3	eqcomi 2738	. 2 ⊢ ∅ = (∅ × (𝐶 ∩ 𝐷))
5	1, 2, 4	3eqtr4g 2789	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × 𝐶) ∩ (𝐵 × 𝐷)) = ∅)

Syntax hints:   → wi 4   = wceq 1540   ∩ cin 3902  ∅c0 4284   × cxp 5617

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 2701  ax-sep 5235  ax-nul 5245  ax-pr 5371

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 2708  df-cleq 2721  df-clel 2803  df-ral 3045  df-rex 3054  df-rab 3395  df-v 3438  df-dif 3906  df-un 3908  df-in 3910  df-ss 3920  df-nul 4285  df-if 4477  df-sn 4578  df-pr 4580  df-op 4584  df-opab 5155  df-xp 5625  df-rel 5626

This theorem is referenced by:  djudisj  6116  xp01disjl  8410  djuin  9814  nosupbnd2lem1  27625  noetasuplem3  27645  noetasuplem4  27646  xpdisjres  32542  esum2dlem  34059  disjxp1  45051