Theorem feq1d

Description: Equality deduction for functions. (Contributed by NM, 19-Feb-2008)

	Ref	Expression
Hypothesis	feq1d.1	⊢ ( 𝜑 → 𝐹 = 𝐺 )
Assertion	feq1d	⊢ ( 𝜑 → ( 𝐹 : 𝐴 ⟶ 𝐵 ↔ 𝐺 : 𝐴 ⟶ 𝐵 ) )

Proof

Step	Hyp	Ref	Expression
1		feq1d.1	⊢ ( 𝜑 → 𝐹 = 𝐺 )
2		feq1	⊢ ( 𝐹 = 𝐺 → ( 𝐹 : 𝐴 ⟶ 𝐵 ↔ 𝐺 : 𝐴 ⟶ 𝐵 ) )
3	1 2	syl	⊢ ( 𝜑 → ( 𝐹 : 𝐴 ⟶ 𝐵 ↔ 𝐺 : 𝐴 ⟶ 𝐵 ) )