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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