Metamath Proof Explorer


Theorem eqfnfvd

Description: Deduction for equality of functions. (Contributed by Mario Carneiro, 24-Jul-2014)

Ref Expression
Hypotheses eqfnfvd.1 ( 𝜑𝐹 Fn 𝐴 )
eqfnfvd.2 ( 𝜑𝐺 Fn 𝐴 )
eqfnfvd.3 ( ( 𝜑𝑥𝐴 ) → ( 𝐹𝑥 ) = ( 𝐺𝑥 ) )
Assertion eqfnfvd ( 𝜑𝐹 = 𝐺 )

Proof

Step Hyp Ref Expression
1 eqfnfvd.1 ( 𝜑𝐹 Fn 𝐴 )
2 eqfnfvd.2 ( 𝜑𝐺 Fn 𝐴 )
3 eqfnfvd.3 ( ( 𝜑𝑥𝐴 ) → ( 𝐹𝑥 ) = ( 𝐺𝑥 ) )
4 3 ralrimiva ( 𝜑 → ∀ 𝑥𝐴 ( 𝐹𝑥 ) = ( 𝐺𝑥 ) )
5 eqfnfv ( ( 𝐹 Fn 𝐴𝐺 Fn 𝐴 ) → ( 𝐹 = 𝐺 ↔ ∀ 𝑥𝐴 ( 𝐹𝑥 ) = ( 𝐺𝑥 ) ) )
6 1 2 5 syl2anc ( 𝜑 → ( 𝐹 = 𝐺 ↔ ∀ 𝑥𝐴 ( 𝐹𝑥 ) = ( 𝐺𝑥 ) ) )
7 4 6 mpbird ( 𝜑𝐹 = 𝐺 )