Theorem xpsndisj 6160

Description: Cartesian products with two different singletons are disjoint. (Contributed by NM, 28-Jul-2004.)

Assertion

Ref	Expression
xpsndisj	⊢ (𝐵 ≠ 𝐷 → ((𝐴 × {𝐵}) ∩ (𝐶 × {𝐷})) = ∅)

Proof of Theorem xpsndisj

Step	Hyp	Ref	Expression
1		disjsn2 4673	. 2 ⊢ (𝐵 ≠ 𝐷 → ({𝐵} ∩ {𝐷}) = ∅)
2		xpdisj2 6159	. 2 ⊢ (({𝐵} ∩ {𝐷}) = ∅ → ((𝐴 × {𝐵}) ∩ (𝐶 × {𝐷})) = ∅)
3	1, 2	syl 18	1 ⊢ (𝐵 ≠ 𝐷 → ((𝐴 × {𝐵}) ∩ (𝐶 × {𝐷})) = ∅)

This proof depends on syntax axioms:   → wi 4   = wceq 1569   ≠ wne 2959   ∩ cin 4092  ∅c0 4281  {csn 4588   × cxp 5653

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735  ax-sep 5252  ax-pr 5395

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2839  df-ne 2960  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-opab 5171  df-xp 5661  df-rel 5662

This theorem is used by:  xp01disj  8513  unxpdom2  9265  sucxpdom  9266