Metamath Proof Explorer


Theorem iundisj2cnt

Description: A countable disjoint union is disjoint. Cf. iundisj2 . (Contributed by Thierry Arnoux, 16-Feb-2017)

Ref Expression
Hypotheses iundisj2cnt.0 𝑛 𝐵
iundisj2cnt.1 ( 𝑛 = 𝑘𝐴 = 𝐵 )
iundisj2cnt.2 ( 𝜑 → ( 𝑁 = ℕ ∨ 𝑁 = ( 1 ..^ 𝑀 ) ) )
Assertion iundisj2cnt ( 𝜑Disj 𝑛𝑁 ( 𝐴 𝑘 ∈ ( 1 ..^ 𝑛 ) 𝐵 ) )

Proof

Step Hyp Ref Expression
1 iundisj2cnt.0 𝑛 𝐵
2 iundisj2cnt.1 ( 𝑛 = 𝑘𝐴 = 𝐵 )
3 iundisj2cnt.2 ( 𝜑 → ( 𝑁 = ℕ ∨ 𝑁 = ( 1 ..^ 𝑀 ) ) )
4 nfcv 𝑘 𝐴
5 4 1 2 iundisj2f Disj 𝑛 ∈ ℕ ( 𝐴 𝑘 ∈ ( 1 ..^ 𝑛 ) 𝐵 )
6 disjeq1 ( 𝑁 = ℕ → ( Disj 𝑛𝑁 ( 𝐴 𝑘 ∈ ( 1 ..^ 𝑛 ) 𝐵 ) ↔ Disj 𝑛 ∈ ℕ ( 𝐴 𝑘 ∈ ( 1 ..^ 𝑛 ) 𝐵 ) ) )
7 5 6 mpbiri ( 𝑁 = ℕ → Disj 𝑛𝑁 ( 𝐴 𝑘 ∈ ( 1 ..^ 𝑛 ) 𝐵 ) )
8 1 2 iundisj2fi Disj 𝑛 ∈ ( 1 ..^ 𝑀 ) ( 𝐴 𝑘 ∈ ( 1 ..^ 𝑛 ) 𝐵 )
9 disjeq1 ( 𝑁 = ( 1 ..^ 𝑀 ) → ( Disj 𝑛𝑁 ( 𝐴 𝑘 ∈ ( 1 ..^ 𝑛 ) 𝐵 ) ↔ Disj 𝑛 ∈ ( 1 ..^ 𝑀 ) ( 𝐴 𝑘 ∈ ( 1 ..^ 𝑛 ) 𝐵 ) ) )
10 8 9 mpbiri ( 𝑁 = ( 1 ..^ 𝑀 ) → Disj 𝑛𝑁 ( 𝐴 𝑘 ∈ ( 1 ..^ 𝑛 ) 𝐵 ) )
11 7 10 jaoi ( ( 𝑁 = ℕ ∨ 𝑁 = ( 1 ..^ 𝑀 ) ) → Disj 𝑛𝑁 ( 𝐴 𝑘 ∈ ( 1 ..^ 𝑛 ) 𝐵 ) )
12 3 11 syl ( 𝜑Disj 𝑛𝑁 ( 𝐴 𝑘 ∈ ( 1 ..^ 𝑛 ) 𝐵 ) )