Theorem eqfnfv

Description: Equality of functions is determined by their values. Special case of Exercise 4 of TakeutiZaring p. 28 (with domain equality omitted). (Contributed by NM, 3-Aug-1994) (Proof shortened by Andrew Salmon, 22-Oct-2011) (Proof shortened by Mario Carneiro, 31-Aug-2015)

	Ref	Expression
Assertion	eqfnfv	⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴 ) → ( 𝐹 = 𝐺 ↔ ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) ) )

Proof

Step	Hyp	Ref	Expression
1		dffn5	⊢ ( 𝐹 Fn 𝐴 ↔ 𝐹 = ( 𝑥 ∈ 𝐴 ↦ ( 𝐹 ‘ 𝑥 ) ) )
2		dffn5	⊢ ( 𝐺 Fn 𝐴 ↔ 𝐺 = ( 𝑥 ∈ 𝐴 ↦ ( 𝐺 ‘ 𝑥 ) ) )
3		eqeq12	⊢ ( ( 𝐹 = ( 𝑥 ∈ 𝐴 ↦ ( 𝐹 ‘ 𝑥 ) ) ∧ 𝐺 = ( 𝑥 ∈ 𝐴 ↦ ( 𝐺 ‘ 𝑥 ) ) ) → ( 𝐹 = 𝐺 ↔ ( 𝑥 ∈ 𝐴 ↦ ( 𝐹 ‘ 𝑥 ) ) = ( 𝑥 ∈ 𝐴 ↦ ( 𝐺 ‘ 𝑥 ) ) ) )
4	1 2 3	syl2anb	⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴 ) → ( 𝐹 = 𝐺 ↔ ( 𝑥 ∈ 𝐴 ↦ ( 𝐹 ‘ 𝑥 ) ) = ( 𝑥 ∈ 𝐴 ↦ ( 𝐺 ‘ 𝑥 ) ) ) )
5		fvex	⊢ ( 𝐹 ‘ 𝑥 ) ∈ V
6	5	rgenw	⊢ ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) ∈ V
7		mpteqb	⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) ∈ V → ( ( 𝑥 ∈ 𝐴 ↦ ( 𝐹 ‘ 𝑥 ) ) = ( 𝑥 ∈ 𝐴 ↦ ( 𝐺 ‘ 𝑥 ) ) ↔ ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) ) )
8	6 7	ax-mp	⊢ ( ( 𝑥 ∈ 𝐴 ↦ ( 𝐹 ‘ 𝑥 ) ) = ( 𝑥 ∈ 𝐴 ↦ ( 𝐺 ‘ 𝑥 ) ) ↔ ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) )
9	4 8	syl6bb	⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴 ) → ( 𝐹 = 𝐺 ↔ ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) ) )