Metamath Proof Explorer


Theorem fneq12d

Description: Equality deduction for function predicate with domain. (Contributed by NM, 26-Jun-2011)

Ref Expression
Hypotheses fneq12d.1 ⊢ ( 𝜑 → 𝐹 = 𝐺 )
fneq12d.2 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion fneq12d ( 𝜑 → ( 𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐵 ) )

Proof

Step Hyp Ref Expression
1 fneq12d.1 ⊢ ( 𝜑 → 𝐹 = 𝐺 )
2 fneq12d.2 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
3 1 fneq1d ⊢ ( 𝜑 → ( 𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐴 ) )
4 2 fneq2d ⊢ ( 𝜑 → ( 𝐺 Fn 𝐴 ↔ 𝐺 Fn 𝐵 ) )
5 3 4 bitrd ⊢ ( 𝜑 → ( 𝐹 Fn 𝐴 ↔ 𝐺 Fn 𝐵 ) )