Metamath Proof Explorer


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 𝐴𝐺 Fn 𝐴 ) → ( 𝐹 = 𝐺 ↔ ∀ 𝑥𝐴 ( 𝐹𝑥 ) = ( 𝐺𝑥 ) ) )

Proof

Step Hyp Ref Expression
1 dffn5 ( 𝐹 Fn 𝐴𝐹 = ( 𝑥𝐴 ↦ ( 𝐹𝑥 ) ) )
2 dffn5 ( 𝐺 Fn 𝐴𝐺 = ( 𝑥𝐴 ↦ ( 𝐺𝑥 ) ) )
3 eqeq12 ( ( 𝐹 = ( 𝑥𝐴 ↦ ( 𝐹𝑥 ) ) ∧ 𝐺 = ( 𝑥𝐴 ↦ ( 𝐺𝑥 ) ) ) → ( 𝐹 = 𝐺 ↔ ( 𝑥𝐴 ↦ ( 𝐹𝑥 ) ) = ( 𝑥𝐴 ↦ ( 𝐺𝑥 ) ) ) )
4 1 2 3 syl2anb ( ( 𝐹 Fn 𝐴𝐺 Fn 𝐴 ) → ( 𝐹 = 𝐺 ↔ ( 𝑥𝐴 ↦ ( 𝐹𝑥 ) ) = ( 𝑥𝐴 ↦ ( 𝐺𝑥 ) ) ) )
5 fvex ( 𝐹𝑥 ) ∈ V
6 5 rgenw 𝑥𝐴 ( 𝐹𝑥 ) ∈ V
7 mpteqb ( ∀ 𝑥𝐴 ( 𝐹𝑥 ) ∈ V → ( ( 𝑥𝐴 ↦ ( 𝐹𝑥 ) ) = ( 𝑥𝐴 ↦ ( 𝐺𝑥 ) ) ↔ ∀ 𝑥𝐴 ( 𝐹𝑥 ) = ( 𝐺𝑥 ) ) )
8 6 7 ax-mp ( ( 𝑥𝐴 ↦ ( 𝐹𝑥 ) ) = ( 𝑥𝐴 ↦ ( 𝐺𝑥 ) ) ↔ ∀ 𝑥𝐴 ( 𝐹𝑥 ) = ( 𝐺𝑥 ) )
9 4 8 bitrdi ( ( 𝐹 Fn 𝐴𝐺 Fn 𝐴 ) → ( 𝐹 = 𝐺 ↔ ∀ 𝑥𝐴 ( 𝐹𝑥 ) = ( 𝐺𝑥 ) ) )