Theorem xpsndisj 6162

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 4680	. 2 ⊢ (𝐵 ≠ 𝐷 → ({𝐵} ∩ {𝐷}) = ∅)
2		xpdisj2 6161	. 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 1570   ≠ wne 2960   ∩ cin 3905  ∅c0 4286  {csn 4591   × cxp 5661

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737  ax-sep 5259  ax-pr 5406

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ne 2961  df-ral 3082  df-rex 3092  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-opab 5176  df-xp 5669  df-rel 5670

This theorem is used by:  xp01disj  8482  unxpdom2  9227  sucxpdom  9228