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Theorem djudisj 5719
 Description: Disjoint unions with disjoint index sets are disjoint. (Contributed by Stefan O'Rear, 21-Nov-2014.)
Assertion
Ref Expression
djudisj ((𝐴𝐵) = ∅ → ( 𝑥𝐴 ({𝑥} × 𝐶) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
Distinct variable groups:   𝑥,𝐴   𝑦,𝐵
Allowed substitution hints:   𝐴(𝑦)   𝐵(𝑥)   𝐶(𝑥,𝑦)   𝐷(𝑥,𝑦)

Proof of Theorem djudisj
StepHypRef Expression
1 djussxp 5423 . 2 𝑥𝐴 ({𝑥} × 𝐶) ⊆ (𝐴 × V)
2 incom 3948 . . 3 ((𝐴 × V) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ( 𝑦𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V))
3 djussxp 5423 . . . 4 𝑦𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V)
4 incom 3948 . . . . 5 ((𝐵 × V) ∩ (𝐴 × V)) = ((𝐴 × V) ∩ (𝐵 × V))
5 xpdisj1 5713 . . . . 5 ((𝐴𝐵) = ∅ → ((𝐴 × V) ∩ (𝐵 × V)) = ∅)
64, 5syl5eq 2806 . . . 4 ((𝐴𝐵) = ∅ → ((𝐵 × V) ∩ (𝐴 × V)) = ∅)
7 ssdisj 4170 . . . 4 (( 𝑦𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V) ∧ ((𝐵 × V) ∩ (𝐴 × V)) = ∅) → ( 𝑦𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
83, 6, 7sylancr 698 . . 3 ((𝐴𝐵) = ∅ → ( 𝑦𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
92, 8syl5eq 2806 . 2 ((𝐴𝐵) = ∅ → ((𝐴 × V) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
10 ssdisj 4170 . 2 (( 𝑥𝐴 ({𝑥} × 𝐶) ⊆ (𝐴 × V) ∧ ((𝐴 × V) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅) → ( 𝑥𝐴 ({𝑥} × 𝐶) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
111, 9, 10sylancr 698 1 ((𝐴𝐵) = ∅ → ( 𝑥𝐴 ({𝑥} × 𝐶) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
 Colors of variables: wff setvar class Syntax hints:   → wi 4   = wceq 1632  Vcvv 3340   ∩ cin 3714   ⊆ wss 3715  ∅c0 4058  {csn 4321  ∪ ciun 4672   × cxp 5264 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886  ax-5 1988  ax-6 2054  ax-7 2090  ax-9 2148  ax-10 2168  ax-11 2183  ax-12 2196  ax-13 2391  ax-ext 2740  ax-sep 4933  ax-nul 4941  ax-pr 5055 This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-3an 1074  df-tru 1635  df-ex 1854  df-nf 1859  df-sb 2047  df-clab 2747  df-cleq 2753  df-clel 2756  df-nfc 2891  df-ral 3055  df-rex 3056  df-rab 3059  df-v 3342  df-dif 3718  df-un 3720  df-in 3722  df-ss 3729  df-nul 4059  df-if 4231  df-sn 4322  df-pr 4324  df-op 4328  df-iun 4674  df-opab 4865  df-xp 5272  df-rel 5273 This theorem is referenced by:  ackbij1lem9  9262
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