Theorem disjorsf 33054

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 2922	. . 3 ⊢ Ⅎ𝑖𝐵
3		nfcsb1v 3871	. . 3 ⊢ Ⅎ𝑥⦋𝑖 / 𝑥⦌𝐵
4		csbeq1a 3861	. . 3 ⊢ (𝑥 = 𝑖 → 𝐵 = ⦋𝑖 / 𝑥⦌𝐵)
5	1, 2, 3, 4	cbvdisjf 33045	. 2 ⊢ (Disj 𝑥 ∈ 𝐴 𝐵 ↔ Disj 𝑖 ∈ 𝐴 ⦋𝑖 / 𝑥⦌𝐵)
6		csbeq1 3850	. . 3 ⊢ (𝑖 = 𝑗 → ⦋𝑖 / 𝑥⦌𝐵 = ⦋𝑗 / 𝑥⦌𝐵)
7	6	disjor 5085	. 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 2907  ∀wral 3076  ⦋csb 3847   ∩ cin 3898  ∅c0 4279  Disj wdisj 5070

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-10 2178  ax-11 2194  ax-12 2213  ax-ext 2732

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 2100  df-mo 2564  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ral 3077  df-rmo 3365  df-v 3452  df-sbc 3740  df-csb 3848  df-dif 3902  df-in 3906  df-nul 4280  df-disj 5071

This theorem is used by:  disjif2  33055  disjdsct  33176