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Theorem djudisj 5465
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 5176 . 2 𝑥𝐴 ({𝑥} × 𝐶) ⊆ (𝐴 × V)
2 incom 3766 . . 3 ((𝐴 × V) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ( 𝑦𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V))
3 djussxp 5176 . . . 4 𝑦𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V)
4 incom 3766 . . . . 5 ((𝐵 × V) ∩ (𝐴 × V)) = ((𝐴 × V) ∩ (𝐵 × V))
5 xpdisj1 5459 . . . . 5 ((𝐴𝐵) = ∅ → ((𝐴 × V) ∩ (𝐵 × V)) = ∅)
64, 5syl5eq 2655 . . . 4 ((𝐴𝐵) = ∅ → ((𝐵 × V) ∩ (𝐴 × V)) = ∅)
7 ssdisj 3977 . . . 4 (( 𝑦𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V) ∧ ((𝐵 × V) ∩ (𝐴 × V)) = ∅) → ( 𝑦𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
83, 6, 7sylancr 693 . . 3 ((𝐴𝐵) = ∅ → ( 𝑦𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
92, 8syl5eq 2655 . 2 ((𝐴𝐵) = ∅ → ((𝐴 × V) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
10 ssdisj 3977 . 2 (( 𝑥𝐴 ({𝑥} × 𝐶) ⊆ (𝐴 × V) ∧ ((𝐴 × V) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅) → ( 𝑥𝐴 ({𝑥} × 𝐶) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
111, 9, 10sylancr 693 1 ((𝐴𝐵) = ∅ → ( 𝑥𝐴 ({𝑥} × 𝐶) ∩ 𝑦𝐵 ({𝑦} × 𝐷)) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1474  Vcvv 3172  cin 3538  wss 3539  c0 3873  {csn 4124   ciun 4449   × cxp 5025
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1712  ax-4 1727  ax-5 1826  ax-6 1874  ax-7 1921  ax-9 1985  ax-10 2005  ax-11 2020  ax-12 2033  ax-13 2233  ax-ext 2589  ax-sep 4703  ax-nul 4711  ax-pr 4827
This theorem depends on definitions:  df-bi 195  df-or 383  df-an 384  df-3an 1032  df-tru 1477  df-ex 1695  df-nf 1700  df-sb 1867  df-clab 2596  df-cleq 2602  df-clel 2605  df-nfc 2739  df-ral 2900  df-rex 2901  df-rab 2904  df-v 3174  df-dif 3542  df-un 3544  df-in 3546  df-ss 3553  df-nul 3874  df-if 4036  df-sn 4125  df-pr 4127  df-op 4131  df-iun 4451  df-opab 4638  df-xp 5033  df-rel 5034
This theorem is referenced by:  ackbij1lem9  8910
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