Theorem feqmptd

Description: Deduction form of dffn5 . (Contributed by Mario Carneiro, 8-Jan-2015)

	Ref	Expression
Hypothesis	feqmptd.1	⊢ ( 𝜑 → 𝐹 : 𝐴 ⟶ 𝐵 )
Assertion	feqmptd	⊢ ( 𝜑 → 𝐹 = ( 𝑥 ∈ 𝐴 ↦ ( 𝐹 ‘ 𝑥 ) ) )

Proof

Step	Hyp	Ref	Expression
1		feqmptd.1	⊢ ( 𝜑 → 𝐹 : 𝐴 ⟶ 𝐵 )
2	1	ffnd	⊢ ( 𝜑 → 𝐹 Fn 𝐴 )
3		dffn5	⊢ ( 𝐹 Fn 𝐴 ↔ 𝐹 = ( 𝑥 ∈ 𝐴 ↦ ( 𝐹 ‘ 𝑥 ) ) )
4	2 3	sylib	⊢ ( 𝜑 → 𝐹 = ( 𝑥 ∈ 𝐴 ↦ ( 𝐹 ‘ 𝑥 ) ) )