Metamath Proof Explorer


Theorem elfunsg

Description: Closed form of elfuns . (Contributed by Scott Fenton, 2-May-2014)

Ref Expression
Assertion elfunsg ( 𝐹 ∈ 𝑉 → ( 𝐹 ∈ Funs ↔ Fun 𝐹 ) )

Proof

Step Hyp Ref Expression
1 eleq1 ⊢ ( 𝑓 = 𝐹 → ( 𝑓 ∈ Funs ↔ 𝐹 ∈ Funs ) )
2 funeq ⊢ ( 𝑓 = 𝐹 → ( Fun 𝑓 ↔ Fun 𝐹 ) )
3 vex ⊢ 𝑓 ∈ V
4 3 elfuns ⊢ ( 𝑓 ∈ Funs ↔ Fun 𝑓 )
5 1 2 4 vtoclbg ⊢ ( 𝐹 ∈ 𝑉 → ( 𝐹 ∈ Funs ↔ Fun 𝐹 ) )