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