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Metamath Proof Explorer

Theorem plypow

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

Ref Expression
Assertion plypow ( ( 𝑆 ⊆ ℂ ∧ 1 ∈ 𝑆 ∧ 𝑁 ∈ ℕ0 ) → ( 𝑧 ∈ ℂ ↦ ( 𝑧 ↑ 𝑁 ) ) ∈ ( Poly ‘ 𝑆 ) )

Proof

Step Hyp Ref Expression
1 id ⊢ ( 𝑧 ∈ ℂ → 𝑧 ∈ ℂ )
2 simp3 ⊢ ( ( 𝑆 ⊆ ℂ ∧ 1 ∈ 𝑆 ∧ 𝑁 ∈ ℕ0 ) → 𝑁 ∈ ℕ0 )
3 expcl ⊢ ( ( 𝑧 ∈ ℂ ∧ 𝑁 ∈ ℕ0 ) → ( 𝑧 ↑ 𝑁 ) ∈ ℂ )
4 1 2 3 syl2anr ⊢ ( ( ( 𝑆 ⊆ ℂ ∧ 1 ∈ 𝑆 ∧ 𝑁 ∈ ℕ0 ) ∧ 𝑧 ∈ ℂ ) → ( 𝑧 ↑ 𝑁 ) ∈ ℂ )
5 4 mullidd ⊢ ( ( ( 𝑆 ⊆ ℂ ∧ 1 ∈ 𝑆 ∧ 𝑁 ∈ ℕ0 ) ∧ 𝑧 ∈ ℂ ) → ( 1 · ( 𝑧 ↑ 𝑁 ) ) = ( 𝑧 ↑ 𝑁 ) )
6 5 mpteq2dva ⊢ ( ( 𝑆 ⊆ ℂ ∧ 1 ∈ 𝑆 ∧ 𝑁 ∈ ℕ0 ) → ( 𝑧 ∈ ℂ ↦ ( 1 · ( 𝑧 ↑ 𝑁 ) ) ) = ( 𝑧 ∈ ℂ ↦ ( 𝑧 ↑ 𝑁 ) ) )
7 eqid ⊢ ( 𝑧 ∈ ℂ ↦ ( 1 · ( 𝑧 ↑ 𝑁 ) ) ) = ( 𝑧 ∈ ℂ ↦ ( 1 · ( 𝑧 ↑ 𝑁 ) ) )
8 7 ply1term ⊢ ( ( 𝑆 ⊆ ℂ ∧ 1 ∈ 𝑆 ∧ 𝑁 ∈ ℕ0 ) → ( 𝑧 ∈ ℂ ↦ ( 1 · ( 𝑧 ↑ 𝑁 ) ) ) ∈ ( Poly ‘ 𝑆 ) )
9 6 8 eqeltrrd ⊢ ( ( 𝑆 ⊆ ℂ ∧ 1 ∈ 𝑆 ∧ 𝑁 ∈ ℕ0 ) → ( 𝑧 ∈ ℂ ↦ ( 𝑧 ↑ 𝑁 ) ) ∈ ( Poly ‘ 𝑆 ) )