Metamath Proof Explorer


Theorem chnsubseqword

Description: A subsequence of a chain is a word. (Contributed by Ender Ting, 22-Jan-2026)

Ref Expression
Hypotheses chnsubseq.1 ( 𝜑𝑊 ∈ ( < Chain 𝐴 ) )
chnsubseq.2 ( 𝜑𝐼 ∈ ( < Chain ( 0 ..^ ( ♯ ‘ 𝑊 ) ) ) )
Assertion chnsubseqword ( 𝜑 → ( 𝑊𝐼 ) ∈ Word 𝐴 )

Proof

Step Hyp Ref Expression
1 chnsubseq.1 ( 𝜑𝑊 ∈ ( < Chain 𝐴 ) )
2 chnsubseq.2 ( 𝜑𝐼 ∈ ( < Chain ( 0 ..^ ( ♯ ‘ 𝑊 ) ) ) )
3 1 adantr ( ( 𝜑𝑥 = ( ♯ ‘ 𝐼 ) ) → 𝑊 ∈ ( < Chain 𝐴 ) )
4 3 chnwrd ( ( 𝜑𝑥 = ( ♯ ‘ 𝐼 ) ) → 𝑊 ∈ Word 𝐴 )
5 wrdf ( 𝑊 ∈ Word 𝐴𝑊 : ( 0 ..^ ( ♯ ‘ 𝑊 ) ) ⟶ 𝐴 )
6 4 5 syl ( ( 𝜑𝑥 = ( ♯ ‘ 𝐼 ) ) → 𝑊 : ( 0 ..^ ( ♯ ‘ 𝑊 ) ) ⟶ 𝐴 )
7 2 chnwrd ( 𝜑𝐼 ∈ Word ( 0 ..^ ( ♯ ‘ 𝑊 ) ) )
8 7 adantr ( ( 𝜑𝑥 = ( ♯ ‘ 𝐼 ) ) → 𝐼 ∈ Word ( 0 ..^ ( ♯ ‘ 𝑊 ) ) )
9 wrdf ( 𝐼 ∈ Word ( 0 ..^ ( ♯ ‘ 𝑊 ) ) → 𝐼 : ( 0 ..^ ( ♯ ‘ 𝐼 ) ) ⟶ ( 0 ..^ ( ♯ ‘ 𝑊 ) ) )
10 8 9 syl ( ( 𝜑𝑥 = ( ♯ ‘ 𝐼 ) ) → 𝐼 : ( 0 ..^ ( ♯ ‘ 𝐼 ) ) ⟶ ( 0 ..^ ( ♯ ‘ 𝑊 ) ) )
11 6 10 fcod ( ( 𝜑𝑥 = ( ♯ ‘ 𝐼 ) ) → ( 𝑊𝐼 ) : ( 0 ..^ ( ♯ ‘ 𝐼 ) ) ⟶ 𝐴 )
12 simpr ( ( 𝜑𝑥 = ( ♯ ‘ 𝐼 ) ) → 𝑥 = ( ♯ ‘ 𝐼 ) )
13 12 oveq2d ( ( 𝜑𝑥 = ( ♯ ‘ 𝐼 ) ) → ( 0 ..^ 𝑥 ) = ( 0 ..^ ( ♯ ‘ 𝐼 ) ) )
14 13 feq2d ( ( 𝜑𝑥 = ( ♯ ‘ 𝐼 ) ) → ( ( 𝑊𝐼 ) : ( 0 ..^ 𝑥 ) ⟶ 𝐴 ↔ ( 𝑊𝐼 ) : ( 0 ..^ ( ♯ ‘ 𝐼 ) ) ⟶ 𝐴 ) )
15 11 14 mpbird ( ( 𝜑𝑥 = ( ♯ ‘ 𝐼 ) ) → ( 𝑊𝐼 ) : ( 0 ..^ 𝑥 ) ⟶ 𝐴 )
16 lencl ( 𝐼 ∈ Word ( 0 ..^ ( ♯ ‘ 𝑊 ) ) → ( ♯ ‘ 𝐼 ) ∈ ℕ0 )
17 7 16 syl ( 𝜑 → ( ♯ ‘ 𝐼 ) ∈ ℕ0 )
18 15 17 rspcime ( 𝜑 → ∃ 𝑥 ∈ ℕ0 ( 𝑊𝐼 ) : ( 0 ..^ 𝑥 ) ⟶ 𝐴 )
19 iswrd ( ( 𝑊𝐼 ) ∈ Word 𝐴 ↔ ∃ 𝑥 ∈ ℕ0 ( 𝑊𝐼 ) : ( 0 ..^ 𝑥 ) ⟶ 𝐴 )
20 18 19 sylibr ( 𝜑 → ( 𝑊𝐼 ) ∈ Word 𝐴 )