Theorem djudisj 6133

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 5802	. 2 ⊢ ∪ 𝑥 ∈ 𝐴 ({𝑥} × 𝐶) ⊆ (𝐴 × V)
2		incom 4163	. . 3 ⊢ ((𝐴 × V) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = (∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V))
3		djussxp 5802	. . . 4 ⊢ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V)
4		incom 4163	. . . . 5 ⊢ ((𝐵 × V) ∩ (𝐴 × V)) = ((𝐴 × V) ∩ (𝐵 × V))
5		xpdisj1 6127	. . . . 5 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × V) ∩ (𝐵 × V)) = ∅)
6	4, 5	eqtrid 2784	. . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐵 × V) ∩ (𝐴 × V)) = ∅)
7		ssdisj 4414	. . . 4 ⊢ ((∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V) ∧ ((𝐵 × V) ∩ (𝐴 × V)) = ∅) → (∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
8	3, 6, 7	sylancr 588	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
9	2, 8	eqtrid 2784	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × V) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = ∅)
10		ssdisj 4414	. 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 588	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (∪ 𝑥 ∈ 𝐴 ({𝑥} × 𝐶) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = ∅)

Syntax hints:   → wi 4   = wceq 1542  Vcvv 3442   ∩ cin 3902   ⊆ wss 3903  ∅c0 4287  {csn 4582  ∪ ciun 4948   × cxp 5630

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-11 2163  ax-ext 2709  ax-sep 5243  ax-pr 5379

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ral 3053  df-rex 3063  df-rab 3402  df-v 3444  df-dif 3906  df-un 3908  df-in 3910  df-ss 3920  df-nul 4288  df-if 4482  df-sn 4583  df-pr 4585  df-op 4589  df-iun 4950  df-opab 5163  df-xp 5638  df-rel 5639

This theorem is referenced by:  ackbij1lem9  10149