Metamath Proof Explorer


Theorem feq3d

Description: Equality deduction for functions. (Contributed by AV, 1-Jan-2020)

Ref Expression
Hypothesis feq2d.1 ( 𝜑𝐴 = 𝐵 )
Assertion feq3d ( 𝜑 → ( 𝐹 : 𝑋𝐴𝐹 : 𝑋𝐵 ) )

Proof

Step Hyp Ref Expression
1 feq2d.1 ( 𝜑𝐴 = 𝐵 )
2 feq3 ( 𝐴 = 𝐵 → ( 𝐹 : 𝑋𝐴𝐹 : 𝑋𝐵 ) )
3 1 2 syl ( 𝜑 → ( 𝐹 : 𝑋𝐴𝐹 : 𝑋𝐵 ) )