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Theorem djudisj 6059
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 5743 . 2 𝑥𝐴 ({𝑥} × 𝐶) ⊆ (𝐴 × V)
2 incom 4131 . . 3 ((𝐴 × V) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ( 𝑦𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V))
3 djussxp 5743 . . . 4 𝑦𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V)
4 incom 4131 . . . . 5 ((𝐵 × V) ∩ (𝐴 × V)) = ((𝐴 × V) ∩ (𝐵 × V))
5 xpdisj1 6053 . . . . 5 ((𝐴𝐵) = ∅ → ((𝐴 × V) ∩ (𝐵 × V)) = ∅)
64, 5eqtrid 2790 . . . 4 ((𝐴𝐵) = ∅ → ((𝐵 × V) ∩ (𝐴 × V)) = ∅)
7 ssdisj 4390 . . . 4 (( 𝑦𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V) ∧ ((𝐵 × V) ∩ (𝐴 × V)) = ∅) → ( 𝑦𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
83, 6, 7sylancr 586 . . 3 ((𝐴𝐵) = ∅ → ( 𝑦𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
92, 8eqtrid 2790 . 2 ((𝐴𝐵) = ∅ → ((𝐴 × V) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
10 ssdisj 4390 . 2 (( 𝑥𝐴 ({𝑥} × 𝐶) ⊆ (𝐴 × V) ∧ ((𝐴 × V) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅) → ( 𝑥𝐴 ({𝑥} × 𝐶) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
111, 9, 10sylancr 586 1 ((𝐴𝐵) = ∅ → ( 𝑥𝐴 ({𝑥} × 𝐶) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1539  Vcvv 3422  cin 3882  wss 3883  c0 4253  {csn 4558   ciun 4921   × cxp 5578
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2156  ax-12 2173  ax-ext 2709  ax-sep 5218  ax-nul 5225  ax-pr 5347
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1784  df-nf 1788  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-nfc 2888  df-ral 3068  df-rex 3069  df-rab 3072  df-v 3424  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4254  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-iun 4923  df-opab 5133  df-xp 5586  df-rel 5587
This theorem is referenced by:  ackbij1lem9  9915
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