Metamath Proof Explorer


Theorem feqmptd

Description: Deduction form of dffn5 . (Contributed by Mario Carneiro, 8-Jan-2015)

Ref Expression
Hypothesis feqmptd.1 ⊢ φ → F : A ⟶ B
Assertion feqmptd ⊢ φ → F = x ∈ A ⟼ F ⁡ x

Proof

Step Hyp Ref Expression
1 feqmptd.1 ⊢ φ → F : A ⟶ B
2 1 ffnd ⊢ φ → F Fn A
3 dffn5 ⊢ F Fn A ↔ F = x ∈ A ⟼ F ⁡ x
4 2 3 sylib ⊢ φ → F = x ∈ A ⟼ F ⁡ x