Theorem djudisj 6125

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 5794	. 2 ⊢ ∪ 𝑥 ∈ 𝐴 ({𝑥} × 𝐶) ⊆ (𝐴 × V)
2		incom 4161	. . 3 ⊢ ((𝐴 × V) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = (∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V))
3		djussxp 5794	. . . 4 ⊢ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V)
4		incom 4161	. . . . 5 ⊢ ((𝐵 × V) ∩ (𝐴 × V)) = ((𝐴 × V) ∩ (𝐵 × V))
5		xpdisj1 6119	. . . . 5 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × V) ∩ (𝐵 × V)) = ∅)
6	4, 5	eqtrid 2783	. . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐵 × V) ∩ (𝐴 × V)) = ∅)
7		ssdisj 4412	. . . 4 ⊢ ((∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V) ∧ ((𝐵 × V) ∩ (𝐴 × V)) = ∅) → (∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
8	3, 6, 7	sylancr 587	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
9	2, 8	eqtrid 2783	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × V) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = ∅)
10		ssdisj 4412	. 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 587	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (∪ 𝑥 ∈ 𝐴 ({𝑥} × 𝐶) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = ∅)

Syntax hints:   → wi 4   = wceq 1541  Vcvv 3440   ∩ cin 3900   ⊆ wss 3901  ∅c0 4285  {csn 4580  ∪ ciun 4946   × cxp 5622

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 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-11 2162  ax-ext 2708  ax-sep 5241  ax-nul 5251  ax-pr 5377

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2715  df-cleq 2728  df-clel 2811  df-ral 3052  df-rex 3061  df-rab 3400  df-v 3442  df-dif 3904  df-un 3906  df-in 3908  df-ss 3918  df-nul 4286  df-if 4480  df-sn 4581  df-pr 4583  df-op 4587  df-iun 4948  df-opab 5161  df-xp 5630  df-rel 5631

This theorem is referenced by:  ackbij1lem9  10139