Metamath Proof Explorer


Theorem feq1dd

Description: Equality deduction for functions. (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Hypotheses feq1dd.eq ( 𝜑𝐹 = 𝐺 )
feq1dd.f ( 𝜑𝐹 : 𝐴𝐵 )
Assertion feq1dd ( 𝜑𝐺 : 𝐴𝐵 )

Proof

Step Hyp Ref Expression
1 feq1dd.eq ( 𝜑𝐹 = 𝐺 )
2 feq1dd.f ( 𝜑𝐹 : 𝐴𝐵 )
3 1 feq1d ( 𝜑 → ( 𝐹 : 𝐴𝐵𝐺 : 𝐴𝐵 ) )
4 2 3 mpbid ( 𝜑𝐺 : 𝐴𝐵 )