Metamath Proof Explorer


Theorem f1ocnvfvrneq

Description: If the values of a one-to-one function for two arguments from the range of the function are equal, the arguments themselves must be equal. (Contributed by Alexander van der Vekens, 12-Nov-2017)

Ref Expression
Assertion f1ocnvfvrneq ⊢ F : A ⟶ 1-1 B ∧ C ∈ ran ⁡ F ∧ D ∈ ran ⁡ F → F -1 ⁡ C = F -1 ⁡ D → C = D

Proof

Step Hyp Ref Expression
1 f1f1orn ⊢ F : A ⟶ 1-1 B → F : A ⟶ 1-1 onto ran ⁡ F
2 f1ocnv ⊢ F : A ⟶ 1-1 onto ran ⁡ F → F -1 : ran ⁡ F ⟶ 1-1 onto A
3 f1of1 ⊢ F -1 : ran ⁡ F ⟶ 1-1 onto A → F -1 : ran ⁡ F ⟶ 1-1 A
4 f1veqaeq ⊢ F -1 : ran ⁡ F ⟶ 1-1 A ∧ C ∈ ran ⁡ F ∧ D ∈ ran ⁡ F → F -1 ⁡ C = F -1 ⁡ D → C = D
5 4 ex ⊢ F -1 : ran ⁡ F ⟶ 1-1 A → C ∈ ran ⁡ F ∧ D ∈ ran ⁡ F → F -1 ⁡ C = F -1 ⁡ D → C = D
6 1 2 3 5 4syl ⊢ F : A ⟶ 1-1 B → C ∈ ran ⁡ F ∧ D ∈ ran ⁡ F → F -1 ⁡ C = F -1 ⁡ D → C = D
7 6 imp ⊢ F : A ⟶ 1-1 B ∧ C ∈ ran ⁡ F ∧ D ∈ ran ⁡ F → F -1 ⁡ C = F -1 ⁡ D → C = D