Theorem xpsndisj 6155

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 6154	. 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 2955   ∩ cin 3898  ∅c0 4279  {csn 4584   × cxp 5653

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 2147  ax-9 2155  ax-ext 2732  ax-sep 5251  ax-pr 5398

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 2739  df-cleq 2752  df-clel 2835  df-ne 2956  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-opab 5168  df-xp 5661  df-rel 5662

This theorem is used by:  xp01disj  8481  unxpdom2  9233  sucxpdom  9234