Theorem plyval

Description: Value of the polynomial set function. (Contributed by Mario Carneiro, 17-Jul-2014)

	Ref	Expression
Assertion	plyval	⊢ ( 𝑆 ⊆ ℂ → ( Poly ‘ 𝑆 ) = { 𝑓 ∣ ∃ 𝑛 ∈ ℕ0 ∃ 𝑎 ∈ ( ( 𝑆 ∪ { 0 } ) ↑m ℕ0 ) 𝑓 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎 ‘ 𝑘 ) · ( 𝑧 ↑ 𝑘 ) ) ) } )

Proof

Step	Hyp	Ref	Expression
1		cnex	⊢ ℂ ∈ V
2	1	elpw2	⊢ ( 𝑆 ∈ 𝒫 ℂ ↔ 𝑆 ⊆ ℂ )
3		uneq1	⊢ ( 𝑥 = 𝑆 → ( 𝑥 ∪ { 0 } ) = ( 𝑆 ∪ { 0 } ) )
4	3	oveq1d	⊢ ( 𝑥 = 𝑆 → ( ( 𝑥 ∪ { 0 } ) ↑m ℕ0 ) = ( ( 𝑆 ∪ { 0 } ) ↑m ℕ0 ) )
5	4	rexeqdv	⊢ ( 𝑥 = 𝑆 → ( ∃ 𝑎 ∈ ( ( 𝑥 ∪ { 0 } ) ↑m ℕ0 ) 𝑓 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎 ‘ 𝑘 ) · ( 𝑧 ↑ 𝑘 ) ) ) ↔ ∃ 𝑎 ∈ ( ( 𝑆 ∪ { 0 } ) ↑m ℕ0 ) 𝑓 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎 ‘ 𝑘 ) · ( 𝑧 ↑ 𝑘 ) ) ) ) )
6	5	rexbidv	⊢ ( 𝑥 = 𝑆 → ( ∃ 𝑛 ∈ ℕ0 ∃ 𝑎 ∈ ( ( 𝑥 ∪ { 0 } ) ↑m ℕ0 ) 𝑓 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎 ‘ 𝑘 ) · ( 𝑧 ↑ 𝑘 ) ) ) ↔ ∃ 𝑛 ∈ ℕ0 ∃ 𝑎 ∈ ( ( 𝑆 ∪ { 0 } ) ↑m ℕ0 ) 𝑓 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎 ‘ 𝑘 ) · ( 𝑧 ↑ 𝑘 ) ) ) ) )
7	6	abbidv	⊢ ( 𝑥 = 𝑆 → { 𝑓 ∣ ∃ 𝑛 ∈ ℕ0 ∃ 𝑎 ∈ ( ( 𝑥 ∪ { 0 } ) ↑m ℕ0 ) 𝑓 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎 ‘ 𝑘 ) · ( 𝑧 ↑ 𝑘 ) ) ) } = { 𝑓 ∣ ∃ 𝑛 ∈ ℕ0 ∃ 𝑎 ∈ ( ( 𝑆 ∪ { 0 } ) ↑m ℕ0 ) 𝑓 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎 ‘ 𝑘 ) · ( 𝑧 ↑ 𝑘 ) ) ) } )
8		df-ply	⊢ Poly = ( 𝑥 ∈ 𝒫 ℂ ↦ { 𝑓 ∣ ∃ 𝑛 ∈ ℕ0 ∃ 𝑎 ∈ ( ( 𝑥 ∪ { 0 } ) ↑m ℕ0 ) 𝑓 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎 ‘ 𝑘 ) · ( 𝑧 ↑ 𝑘 ) ) ) } )
9		nn0ex	⊢ ℕ0 ∈ V
10		ovex	⊢ ( ( 𝑆 ∪ { 0 } ) ↑m ℕ0 ) ∈ V
11	9 10	ab2rexex	⊢ { 𝑓 ∣ ∃ 𝑛 ∈ ℕ0 ∃ 𝑎 ∈ ( ( 𝑆 ∪ { 0 } ) ↑m ℕ0 ) 𝑓 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎 ‘ 𝑘 ) · ( 𝑧 ↑ 𝑘 ) ) ) } ∈ V
12	7 8 11	fvmpt	⊢ ( 𝑆 ∈ 𝒫 ℂ → ( Poly ‘ 𝑆 ) = { 𝑓 ∣ ∃ 𝑛 ∈ ℕ0 ∃ 𝑎 ∈ ( ( 𝑆 ∪ { 0 } ) ↑m ℕ0 ) 𝑓 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎 ‘ 𝑘 ) · ( 𝑧 ↑ 𝑘 ) ) ) } )
13	2 12	sylbir	⊢ ( 𝑆 ⊆ ℂ → ( Poly ‘ 𝑆 ) = { 𝑓 ∣ ∃ 𝑛 ∈ ℕ0 ∃ 𝑎 ∈ ( ( 𝑆 ∪ { 0 } ) ↑m ℕ0 ) 𝑓 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎 ‘ 𝑘 ) · ( 𝑧 ↑ 𝑘 ) ) ) } )