Metamath Proof Explorer


Theorem reprfi2

Description: Corollary of reprinfz1 . (Contributed by Thierry Arnoux, 15-Dec-2021)

Ref Expression
Hypotheses reprinfz1.n ( 𝜑𝑁 ∈ ℕ0 )
reprinfz1.s ( 𝜑𝑆 ∈ ℕ0 )
reprinfz1.a ( 𝜑𝐴 ⊆ ℕ )
Assertion reprfi2 ( 𝜑 → ( 𝐴 ( repr ‘ 𝑆 ) 𝑁 ) ∈ Fin )

Proof

Step Hyp Ref Expression
1 reprinfz1.n ( 𝜑𝑁 ∈ ℕ0 )
2 reprinfz1.s ( 𝜑𝑆 ∈ ℕ0 )
3 reprinfz1.a ( 𝜑𝐴 ⊆ ℕ )
4 1 2 3 reprinfz1 ( 𝜑 → ( 𝐴 ( repr ‘ 𝑆 ) 𝑁 ) = ( ( 𝐴 ∩ ( 1 ... 𝑁 ) ) ( repr ‘ 𝑆 ) 𝑁 ) )
5 inss2 ( 𝐴 ∩ ( 1 ... 𝑁 ) ) ⊆ ( 1 ... 𝑁 )
6 fz1ssnn ( 1 ... 𝑁 ) ⊆ ℕ
7 5 6 sstri ( 𝐴 ∩ ( 1 ... 𝑁 ) ) ⊆ ℕ
8 7 a1i ( 𝜑 → ( 𝐴 ∩ ( 1 ... 𝑁 ) ) ⊆ ℕ )
9 1 nn0zd ( 𝜑𝑁 ∈ ℤ )
10 fzfi ( 1 ... 𝑁 ) ∈ Fin
11 10 a1i ( 𝜑 → ( 1 ... 𝑁 ) ∈ Fin )
12 5 a1i ( 𝜑 → ( 𝐴 ∩ ( 1 ... 𝑁 ) ) ⊆ ( 1 ... 𝑁 ) )
13 11 12 ssfid ( 𝜑 → ( 𝐴 ∩ ( 1 ... 𝑁 ) ) ∈ Fin )
14 8 9 2 13 reprfi ( 𝜑 → ( ( 𝐴 ∩ ( 1 ... 𝑁 ) ) ( repr ‘ 𝑆 ) 𝑁 ) ∈ Fin )
15 4 14 eqeltrd ( 𝜑 → ( 𝐴 ( repr ‘ 𝑆 ) 𝑁 ) ∈ Fin )