Theorem djudisj 6161

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 5827	. 2 ⊢ ∪ 𝑥 ∈ 𝐴 ({𝑥} × 𝐶) ⊆ (𝐴 × V)
2		incom 4155	. . 3 ⊢ ((𝐴 × V) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = (∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V))
3		djussxp 5827	. . . 4 ⊢ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V)
4		incom 4155	. . . . 5 ⊢ ((𝐵 × V) ∩ (𝐴 × V)) = ((𝐴 × V) ∩ (𝐵 × V))
5		xpdisj1 6155	. . . . 5 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × V) ∩ (𝐵 × V)) = ∅)
6	4, 5	eqtrid 2807	. . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐵 × V) ∩ (𝐴 × V)) = ∅)
7		ssdisj 4413	. . . 4 ⊢ ((∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V) ∧ ((𝐵 × V) ∩ (𝐴 × V)) = ∅) → (∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
8	3, 6, 7	sylancr 599	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
9	2, 8	eqtrid 2807	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × V) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = ∅)
10		ssdisj 4413	. 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 599	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (∪ 𝑥 ∈ 𝐴 ({𝑥} × 𝐶) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = ∅)

This proof depends on syntax axioms:   → wi 4   = wceq 1570  Vcvv 3450   ∩ cin 3898   ⊆ wss 3899  ∅c0 4279  {csn 4584  ∪ ciun 4951   × cxp 5653

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-11 2194  ax-ext 2732  ax-sep 5251  ax-pr 5398

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-iun 4953  df-opab 5168  df-xp 5661  df-rel 5662

This theorem is used by:  ackbij1lem9  10254