Theorem disjorsf 33272

Description: Two ways to say that a collection 𝐵(𝑖) for 𝑖 ∈ 𝐴 is disjoint. (Contributed by Thierry Arnoux, 8-Mar-2017.)

Hypothesis

Ref	Expression
disjorsf.1	⊢ Ⅎ𝑥𝐴

Assertion

Ref	Expression
disjorsf	⊢ (Disj 𝑥 ∈ 𝐴 𝐵 ↔ ∀𝑖 ∈ 𝐴 ∀𝑗 ∈ 𝐴 (𝑖 = 𝑗 ∨ (⦋𝑖 / 𝑥⦌𝐵 ∩ ⦋𝑗 / 𝑥⦌𝐵) = ∅))

Distinct variable groups:   𝑖,𝑗,𝑥   𝐴,𝑖,𝑗   𝐵,𝑖,𝑗

Allowed substitution hints:   𝐴(𝑥)   𝐵(𝑥)

Proof of Theorem disjorsf

Step	Hyp	Ref	Expression
1		disjorsf.1	. . 3 ⊢ Ⅎ𝑥𝐴
2		nfcv 2926	. . 3 ⊢ Ⅎ𝑖𝐵
3		nfcsb1v 3875	. . 3 ⊢ Ⅎ𝑥⦋𝑖 / 𝑥⦌𝐵
4		csbeq1a 3865	. . 3 ⊢ (𝑥 = 𝑖 → 𝐵 = ⦋𝑖 / 𝑥⦌𝐵)
5	1, 2, 3, 4	cbvdisjf 33263	. 2 ⊢ (Disj 𝑥 ∈ 𝐴 𝐵 ↔ Disj 𝑖 ∈ 𝐴 ⦋𝑖 / 𝑥⦌𝐵)
6		csbeq1 3854	. . 3 ⊢ (𝑖 = 𝑗 → ⦋𝑖 / 𝑥⦌𝐵 = ⦋𝑗 / 𝑥⦌𝐵)
7	6	disjor 5090	. 2 ⊢ (Disj 𝑖 ∈ 𝐴 ⦋𝑖 / 𝑥⦌𝐵 ↔ ∀𝑖 ∈ 𝐴 ∀𝑗 ∈ 𝐴 (𝑖 = 𝑗 ∨ (⦋𝑖 / 𝑥⦌𝐵 ∩ ⦋𝑗 / 𝑥⦌𝐵) = ∅))
8	5, 7	bitri 278	1 ⊢ (Disj 𝑥 ∈ 𝐴 𝐵 ↔ ∀𝑖 ∈ 𝐴 ∀𝑗 ∈ 𝐴 (𝑖 = 𝑗 ∨ (⦋𝑖 / 𝑥⦌𝐵 ∩ ⦋𝑗 / 𝑥⦌𝐵) = ∅))

This proof depends on syntax axioms:   ↔ wb 209   ∨ wo 861   = wceq 1570  Ⅎwnfc 2911  ∀wral 3080  ⦋csb 3851   ∩ cin 4092  ∅c0 4281  Disj wdisj 5075

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2102  df-mo 2567  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-ral 3081  df-rmo 3369  df-v 3456  df-sbc 3743  df-csb 3852  df-dif 4050  df-in 4094  df-nul 4282  df-disj 5076

This theorem is used by:  disjif2  33273  disjdsct  33393