Theorem djudisj 6164

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 5831	. 2 ⊢ ∪ 𝑥 ∈ 𝐴 ({𝑥} × 𝐶) ⊆ (𝐴 × V)
2		incom 4162	. . 3 ⊢ ((𝐴 × V) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = (∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V))
3		djussxp 5831	. . . 4 ⊢ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V)
4		incom 4162	. . . . 5 ⊢ ((𝐵 × V) ∩ (𝐴 × V)) = ((𝐴 × V) ∩ (𝐵 × V))
5		xpdisj1 6158	. . . . 5 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × V) ∩ (𝐵 × V)) = ∅)
6	4, 5	eqtrid 2810	. . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐵 × V) ∩ (𝐴 × V)) = ∅)
7		ssdisj 4420	. . . 4 ⊢ ((∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ⊆ (𝐵 × V) ∧ ((𝐵 × V) ∩ (𝐴 × V)) = ∅) → (∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
8	3, 6, 7	sylancr 598	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷) ∩ (𝐴 × V)) = ∅)
9	2, 8	eqtrid 2810	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 × V) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = ∅)
10		ssdisj 4420	. 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 598	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (∪ 𝑥 ∈ 𝐴 ({𝑥} × 𝐶) ∩ ∪ 𝑦 ∈ 𝐵 ({𝑦} × 𝐷)) = ∅)

Syntax hints:   → wi 4   = wceq 1570  Vcvv 3455   ∩ cin 3904   ⊆ wss 3905  ∅c0 4286  {csn 4589  ∪ ciun 4956   × cxp 5659

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-11 2192  ax-ext 2735  ax-sep 5257  ax-pr 5404

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-ral 3080  df-rex 3090  df-rab 3417  df-v 3457  df-dif 3908  df-un 3910  df-in 3912  df-ss 3922  df-nul 4287  df-if 4488  df-sn 4590  df-pr 4592  df-op 4596  df-iun 4958  df-opab 5174  df-xp 5667  df-rel 5668

This theorem is referenced by:  ackbij1lem9  10206