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Theorem djudisj 5561
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 5267 . 2 𝑥𝐴 ({𝑥} × 𝐶) ⊆ (𝐴 × V)
2 incom 3805 . . 3 ((𝐴 × V) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ( 𝑦𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V))
3 djussxp 5267 . . . 4 𝑦𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V)
4 incom 3805 . . . . 5 ((𝐵 × V) ∩ (𝐴 × V)) = ((𝐴 × V) ∩ (𝐵 × V))
5 xpdisj1 5555 . . . . 5 ((𝐴𝐵) = ∅ → ((𝐴 × V) ∩ (𝐵 × V)) = ∅)
64, 5syl5eq 2668 . . . 4 ((𝐴𝐵) = ∅ → ((𝐵 × V) ∩ (𝐴 × V)) = ∅)
7 ssdisj 4026 . . . 4 (( 𝑦𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V) ∧ ((𝐵 × V) ∩ (𝐴 × V)) = ∅) → ( 𝑦𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
83, 6, 7sylancr 695 . . 3 ((𝐴𝐵) = ∅ → ( 𝑦𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
92, 8syl5eq 2668 . 2 ((𝐴𝐵) = ∅ → ((𝐴 × V) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
10 ssdisj 4026 . 2 (( 𝑥𝐴 ({𝑥} × 𝐶) ⊆ (𝐴 × V) ∧ ((𝐴 × V) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅) → ( 𝑥𝐴 ({𝑥} × 𝐶) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
111, 9, 10sylancr 695 1 ((𝐴𝐵) = ∅ → ( 𝑥𝐴 ({𝑥} × 𝐶) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1483  Vcvv 3200  cin 3573  wss 3574  c0 3915  {csn 4177   ciun 4520   × cxp 5112
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602  ax-sep 4781  ax-nul 4789  ax-pr 4906
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ral 2917  df-rex 2918  df-rab 2921  df-v 3202  df-dif 3577  df-un 3579  df-in 3581  df-ss 3588  df-nul 3916  df-if 4087  df-sn 4178  df-pr 4180  df-op 4184  df-iun 4522  df-opab 4713  df-xp 5120  df-rel 5121
This theorem is referenced by:  ackbij1lem9  9050
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