Metamath Proof Explorer


Theorem feq123d

Description: Equality deduction for functions. (Contributed by Paul Chapman, 22-Jun-2011)

Ref Expression
Hypotheses feq12d.1 ⊢ ( 𝜑 → 𝐹 = 𝐺 )
feq12d.2 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
feq123d.3 ⊢ ( 𝜑 → 𝐶 = 𝐷 )
Assertion feq123d ( 𝜑 → ( 𝐹 : 𝐴 ⟶ 𝐶 ↔ 𝐺 : 𝐵 ⟶ 𝐷 ) )

Proof

Step Hyp Ref Expression
1 feq12d.1 ⊢ ( 𝜑 → 𝐹 = 𝐺 )
2 feq12d.2 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
3 feq123d.3 ⊢ ( 𝜑 → 𝐶 = 𝐷 )
4 1 2 feq12d ⊢ ( 𝜑 → ( 𝐹 : 𝐴 ⟶ 𝐶 ↔ 𝐺 : 𝐵 ⟶ 𝐶 ) )
5 3 feq3d ⊢ ( 𝜑 → ( 𝐺 : 𝐵 ⟶ 𝐶 ↔ 𝐺 : 𝐵 ⟶ 𝐷 ) )
6 4 5 bitrd ⊢ ( 𝜑 → ( 𝐹 : 𝐴 ⟶ 𝐶 ↔ 𝐺 : 𝐵 ⟶ 𝐷 ) )