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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 ∧ 𝐴 ∈ 𝑆 ) → ( ℂ × { 𝐴 } ) ∈ ( Poly ‘ 𝑆 ) )

Proof

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