Theorem djudisj 6118

Description: Disjoint unions with disjoint index sets are disjoint. (Contributed by Stefan O'Rear, 21-Nov-2014.)

Assertion

Ref	Expression
djudisj	⊢ ((𝐴 ∩ 𝐵) = ∅ → (∪ 𝑥 ∈ 𝐴 ({𝑥} × 𝐶) ∩ ∪ 𝑦 ∈ 𝐵 ({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

Step	Hyp	Ref	Expression
1		djussxp 5787	. 2 ⊢ ∪ 𝑥 ∈ 𝐴 ({𝑥} × 𝐶) ⊆ (𝐴 × V)
2		incom 4138	. . 3 ⊢ ((𝐴 × V) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = (∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V))
3		djussxp 5787	. . . 4 ⊢ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V)
4		incom 4138	. . . . 5 ⊢ ((𝐵 × V) ∩ (𝐴 × V)) = ((𝐴 × V) ∩ (𝐵 × V))
5		xpdisj1 6112	. . . . 5 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × V) ∩ (𝐵 × V)) = ∅)
6	4, 5	eqtrid 2786	. . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐵 × V) ∩ (𝐴 × V)) = ∅)
7		ssdisj 4388	. . . 4 ⊢ ((∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V) ∧ ((𝐵 × V) ∩ (𝐴 × V)) = ∅) → (∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
8	3, 6, 7	sylancr 593	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
9	2, 8	eqtrid 2786	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × V) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = ∅)
10		ssdisj 4388	. 2 ⊢ ((∪ 𝑥 ∈ 𝐴 ({𝑥} × 𝐶) ⊆ (𝐴 × V) ∧ ((𝐴 × V) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = ∅) → (∪ 𝑥 ∈ 𝐴 ({𝑥} × 𝐶) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = ∅)
11	1, 9, 10	sylancr 593	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (∪ 𝑥 ∈ 𝐴 ({𝑥} × 𝐶) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = ∅)

Syntax hints:   → wi 4   = wceq 1547  Vcvv 3431   ∩ cin 3882   ⊆ wss 3883  ∅c0 4261  {csn 4555  ∪ ciun 4921   × cxp 5616

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-11 2168  ax-ext 2711  ax-sep 5218  ax-pr 5362

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-3an 1094  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-ral 3054  df-rex 3064  df-rab 3392  df-v 3433  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4262  df-if 4455  df-sn 4556  df-pr 4558  df-op 4562  df-iun 4923  df-opab 5135  df-xp 5624  df-rel 5625

This theorem is referenced by:  ackbij1lem9  10140