Metamath Proof Explorer


Theorem plyconst

Description: A constant function is a polynomial. (Contributed by Mario Carneiro, 17-Jul-2014)

Ref Expression
Assertion plyconst ( ( 𝑆 ⊆ ℂ ∧ 𝐴𝑆 ) → ( ℂ × { 𝐴 } ) ∈ ( Poly ‘ 𝑆 ) )

Proof

Step Hyp Ref Expression
1 exp0 ( 𝑧 ∈ ℂ → ( 𝑧 ↑ 0 ) = 1 )
2 1 adantl ( ( ( 𝑆 ⊆ ℂ ∧ 𝐴𝑆 ) ∧ 𝑧 ∈ ℂ ) → ( 𝑧 ↑ 0 ) = 1 )
3 2 oveq2d ( ( ( 𝑆 ⊆ ℂ ∧ 𝐴𝑆 ) ∧ 𝑧 ∈ ℂ ) → ( 𝐴 · ( 𝑧 ↑ 0 ) ) = ( 𝐴 · 1 ) )
4 ssel2 ( ( 𝑆 ⊆ ℂ ∧ 𝐴𝑆 ) → 𝐴 ∈ ℂ )
5 4 adantr ( ( ( 𝑆 ⊆ ℂ ∧ 𝐴𝑆 ) ∧ 𝑧 ∈ ℂ ) → 𝐴 ∈ ℂ )
6 5 mulid1d ( ( ( 𝑆 ⊆ ℂ ∧ 𝐴𝑆 ) ∧ 𝑧 ∈ ℂ ) → ( 𝐴 · 1 ) = 𝐴 )
7 3 6 eqtrd ( ( ( 𝑆 ⊆ ℂ ∧ 𝐴𝑆 ) ∧ 𝑧 ∈ ℂ ) → ( 𝐴 · ( 𝑧 ↑ 0 ) ) = 𝐴 )
8 7 mpteq2dva ( ( 𝑆 ⊆ ℂ ∧ 𝐴𝑆 ) → ( 𝑧 ∈ ℂ ↦ ( 𝐴 · ( 𝑧 ↑ 0 ) ) ) = ( 𝑧 ∈ ℂ ↦ 𝐴 ) )
9 fconstmpt ( ℂ × { 𝐴 } ) = ( 𝑧 ∈ ℂ ↦ 𝐴 )
10 8 9 eqtr4di ( ( 𝑆 ⊆ ℂ ∧ 𝐴𝑆 ) → ( 𝑧 ∈ ℂ ↦ ( 𝐴 · ( 𝑧 ↑ 0 ) ) ) = ( ℂ × { 𝐴 } ) )
11 0nn0 0 ∈ ℕ0
12 eqid ( 𝑧 ∈ ℂ ↦ ( 𝐴 · ( 𝑧 ↑ 0 ) ) ) = ( 𝑧 ∈ ℂ ↦ ( 𝐴 · ( 𝑧 ↑ 0 ) ) )
13 12 ply1term ( ( 𝑆 ⊆ ℂ ∧ 𝐴𝑆 ∧ 0 ∈ ℕ0 ) → ( 𝑧 ∈ ℂ ↦ ( 𝐴 · ( 𝑧 ↑ 0 ) ) ) ∈ ( Poly ‘ 𝑆 ) )
14 11 13 mp3an3 ( ( 𝑆 ⊆ ℂ ∧ 𝐴𝑆 ) → ( 𝑧 ∈ ℂ ↦ ( 𝐴 · ( 𝑧 ↑ 0 ) ) ) ∈ ( Poly ‘ 𝑆 ) )
15 10 14 eqeltrrd ( ( 𝑆 ⊆ ℂ ∧ 𝐴𝑆 ) → ( ℂ × { 𝐴 } ) ∈ ( Poly ‘ 𝑆 ) )