Metamath Proof Explorer


Theorem feq2dd

Description: Equality deduction for functions. (Contributed by Thierry Arnoux, 27-May-2025)

Ref Expression
Hypotheses feq2dd.eq ( 𝜑𝐴 = 𝐵 )
feq2dd.f ( 𝜑𝐹 : 𝐴𝐶 )
Assertion feq2dd ( 𝜑𝐹 : 𝐵𝐶 )

Proof

Step Hyp Ref Expression
1 feq2dd.eq ( 𝜑𝐴 = 𝐵 )
2 feq2dd.f ( 𝜑𝐹 : 𝐴𝐶 )
3 1 feq2d ( 𝜑 → ( 𝐹 : 𝐴𝐶𝐹 : 𝐵𝐶 ) )
4 2 3 mpbid ( 𝜑𝐹 : 𝐵𝐶 )