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Theorem djudisj 6026
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 5718 . 2 𝑥𝐴 ({𝑥} × 𝐶) ⊆ (𝐴 × V)
2 incom 4180 . . 3 ((𝐴 × V) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ( 𝑦𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V))
3 djussxp 5718 . . . 4 𝑦𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V)
4 incom 4180 . . . . 5 ((𝐵 × V) ∩ (𝐴 × V)) = ((𝐴 × V) ∩ (𝐵 × V))
5 xpdisj1 6020 . . . . 5 ((𝐴𝐵) = ∅ → ((𝐴 × V) ∩ (𝐵 × V)) = ∅)
64, 5syl5eq 2870 . . . 4 ((𝐴𝐵) = ∅ → ((𝐵 × V) ∩ (𝐴 × V)) = ∅)
7 ssdisj 4411 . . . 4 (( 𝑦𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V) ∧ ((𝐵 × V) ∩ (𝐴 × V)) = ∅) → ( 𝑦𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
83, 6, 7sylancr 589 . . 3 ((𝐴𝐵) = ∅ → ( 𝑦𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
92, 8syl5eq 2870 . 2 ((𝐴𝐵) = ∅ → ((𝐴 × V) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
10 ssdisj 4411 . 2 (( 𝑥𝐴 ({𝑥} × 𝐶) ⊆ (𝐴 × V) ∧ ((𝐴 × V) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅) → ( 𝑥𝐴 ({𝑥} × 𝐶) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
111, 9, 10sylancr 589 1 ((𝐴𝐵) = ∅ → ( 𝑥𝐴 ({𝑥} × 𝐶) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1537  Vcvv 3496  cin 3937  wss 3938  c0 4293  {csn 4569   ciun 4921   × cxp 5555
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2795  ax-sep 5205  ax-nul 5212  ax-pr 5332
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2802  df-cleq 2816  df-clel 2895  df-nfc 2965  df-ral 3145  df-rex 3146  df-rab 3149  df-v 3498  df-dif 3941  df-un 3943  df-in 3945  df-ss 3954  df-nul 4294  df-if 4470  df-sn 4570  df-pr 4572  df-op 4576  df-iun 4923  df-opab 5131  df-xp 5563  df-rel 5564
This theorem is referenced by:  ackbij1lem9  9652
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