Metamath Proof Explorer


Theorem sseqfres

Description: The first elements in the strong recursive sequence are the sequence initializer. (Contributed by Thierry Arnoux, 23-Apr-2019)

Ref Expression
Hypotheses sseqval.1 ( 𝜑𝑆 ∈ V )
sseqval.2 ( 𝜑𝑀 ∈ Word 𝑆 )
sseqval.3 𝑊 = ( Word 𝑆 ∩ ( ♯ “ ( ℤ ‘ ( ♯ ‘ 𝑀 ) ) ) )
sseqval.4 ( 𝜑𝐹 : 𝑊𝑆 )
Assertion sseqfres ( 𝜑 → ( ( 𝑀 seqstr 𝐹 ) ↾ ( 0 ..^ ( ♯ ‘ 𝑀 ) ) ) = 𝑀 )

Proof

Step Hyp Ref Expression
1 sseqval.1 ( 𝜑𝑆 ∈ V )
2 sseqval.2 ( 𝜑𝑀 ∈ Word 𝑆 )
3 sseqval.3 𝑊 = ( Word 𝑆 ∩ ( ♯ “ ( ℤ ‘ ( ♯ ‘ 𝑀 ) ) ) )
4 sseqval.4 ( 𝜑𝐹 : 𝑊𝑆 )
5 1 adantr ( ( 𝜑𝑖 ∈ ( 0 ..^ ( ♯ ‘ 𝑀 ) ) ) → 𝑆 ∈ V )
6 2 adantr ( ( 𝜑𝑖 ∈ ( 0 ..^ ( ♯ ‘ 𝑀 ) ) ) → 𝑀 ∈ Word 𝑆 )
7 4 adantr ( ( 𝜑𝑖 ∈ ( 0 ..^ ( ♯ ‘ 𝑀 ) ) ) → 𝐹 : 𝑊𝑆 )
8 simpr ( ( 𝜑𝑖 ∈ ( 0 ..^ ( ♯ ‘ 𝑀 ) ) ) → 𝑖 ∈ ( 0 ..^ ( ♯ ‘ 𝑀 ) ) )
9 5 6 3 7 8 sseqfv1 ( ( 𝜑𝑖 ∈ ( 0 ..^ ( ♯ ‘ 𝑀 ) ) ) → ( ( 𝑀 seqstr 𝐹 ) ‘ 𝑖 ) = ( 𝑀𝑖 ) )
10 9 ralrimiva ( 𝜑 → ∀ 𝑖 ∈ ( 0 ..^ ( ♯ ‘ 𝑀 ) ) ( ( 𝑀 seqstr 𝐹 ) ‘ 𝑖 ) = ( 𝑀𝑖 ) )
11 1 2 3 4 sseqfn ( 𝜑 → ( 𝑀 seqstr 𝐹 ) Fn ℕ0 )
12 wrdfn ( 𝑀 ∈ Word 𝑆𝑀 Fn ( 0 ..^ ( ♯ ‘ 𝑀 ) ) )
13 2 12 syl ( 𝜑𝑀 Fn ( 0 ..^ ( ♯ ‘ 𝑀 ) ) )
14 fzo0ssnn0 ( 0 ..^ ( ♯ ‘ 𝑀 ) ) ⊆ ℕ0
15 14 a1i ( 𝜑 → ( 0 ..^ ( ♯ ‘ 𝑀 ) ) ⊆ ℕ0 )
16 fvreseq1 ( ( ( ( 𝑀 seqstr 𝐹 ) Fn ℕ0𝑀 Fn ( 0 ..^ ( ♯ ‘ 𝑀 ) ) ) ∧ ( 0 ..^ ( ♯ ‘ 𝑀 ) ) ⊆ ℕ0 ) → ( ( ( 𝑀 seqstr 𝐹 ) ↾ ( 0 ..^ ( ♯ ‘ 𝑀 ) ) ) = 𝑀 ↔ ∀ 𝑖 ∈ ( 0 ..^ ( ♯ ‘ 𝑀 ) ) ( ( 𝑀 seqstr 𝐹 ) ‘ 𝑖 ) = ( 𝑀𝑖 ) ) )
17 11 13 15 16 syl21anc ( 𝜑 → ( ( ( 𝑀 seqstr 𝐹 ) ↾ ( 0 ..^ ( ♯ ‘ 𝑀 ) ) ) = 𝑀 ↔ ∀ 𝑖 ∈ ( 0 ..^ ( ♯ ‘ 𝑀 ) ) ( ( 𝑀 seqstr 𝐹 ) ‘ 𝑖 ) = ( 𝑀𝑖 ) ) )
18 10 17 mpbird ( 𝜑 → ( ( 𝑀 seqstr 𝐹 ) ↾ ( 0 ..^ ( ♯ ‘ 𝑀 ) ) ) = 𝑀 )