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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 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 bitrdi ⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴 ) → ( 𝐹 = 𝐺 ↔ ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) ) )