Metamath Proof Explorer


Theorem dvcl

Description: The derivative function takes values in the complex numbers. (Contributed by Mario Carneiro, 7-Aug-2014) (Revised by Mario Carneiro, 9-Feb-2015)

Ref Expression
Hypotheses dvcl.s ( 𝜑𝑆 ⊆ ℂ )
dvcl.f ( 𝜑𝐹 : 𝐴 ⟶ ℂ )
dvcl.a ( 𝜑𝐴𝑆 )
Assertion dvcl ( ( 𝜑𝐵 ( 𝑆 D 𝐹 ) 𝐶 ) → 𝐶 ∈ ℂ )

Proof

Step Hyp Ref Expression
1 dvcl.s ( 𝜑𝑆 ⊆ ℂ )
2 dvcl.f ( 𝜑𝐹 : 𝐴 ⟶ ℂ )
3 dvcl.a ( 𝜑𝐴𝑆 )
4 limccl ( ( 𝑧 ∈ ( 𝐴 ∖ { 𝐵 } ) ↦ ( ( ( 𝐹𝑧 ) − ( 𝐹𝐵 ) ) / ( 𝑧𝐵 ) ) ) lim 𝐵 ) ⊆ ℂ
5 eqid ( ( TopOpen ‘ ℂfld ) ↾t 𝑆 ) = ( ( TopOpen ‘ ℂfld ) ↾t 𝑆 )
6 eqid ( TopOpen ‘ ℂfld ) = ( TopOpen ‘ ℂfld )
7 eqid ( 𝑧 ∈ ( 𝐴 ∖ { 𝐵 } ) ↦ ( ( ( 𝐹𝑧 ) − ( 𝐹𝐵 ) ) / ( 𝑧𝐵 ) ) ) = ( 𝑧 ∈ ( 𝐴 ∖ { 𝐵 } ) ↦ ( ( ( 𝐹𝑧 ) − ( 𝐹𝐵 ) ) / ( 𝑧𝐵 ) ) )
8 5 6 7 1 2 3 eldv ( 𝜑 → ( 𝐵 ( 𝑆 D 𝐹 ) 𝐶 ↔ ( 𝐵 ∈ ( ( int ‘ ( ( TopOpen ‘ ℂfld ) ↾t 𝑆 ) ) ‘ 𝐴 ) ∧ 𝐶 ∈ ( ( 𝑧 ∈ ( 𝐴 ∖ { 𝐵 } ) ↦ ( ( ( 𝐹𝑧 ) − ( 𝐹𝐵 ) ) / ( 𝑧𝐵 ) ) ) lim 𝐵 ) ) ) )
9 8 simplbda ( ( 𝜑𝐵 ( 𝑆 D 𝐹 ) 𝐶 ) → 𝐶 ∈ ( ( 𝑧 ∈ ( 𝐴 ∖ { 𝐵 } ) ↦ ( ( ( 𝐹𝑧 ) − ( 𝐹𝐵 ) ) / ( 𝑧𝐵 ) ) ) lim 𝐵 ) )
10 4 9 sseldi ( ( 𝜑𝐵 ( 𝑆 D 𝐹 ) 𝐶 ) → 𝐶 ∈ ℂ )