Metamath Proof Explorer


Theorem f1ssf1

Description: A subset of an injective function is injective. (Contributed by AV, 20-Nov-2020)

Ref Expression
Assertion f1ssf1 Fun F Fun F -1 G F Fun G -1

Proof

Step Hyp Ref Expression
1 funssres Fun F G F F dom G = G
2 funres11 Fun F -1 Fun F dom G -1
3 cnveq G = F dom G G -1 = F dom G -1
4 3 funeqd G = F dom G Fun G -1 Fun F dom G -1
5 2 4 syl5ibr G = F dom G Fun F -1 Fun G -1
6 5 eqcoms F dom G = G Fun F -1 Fun G -1
7 1 6 syl Fun F G F Fun F -1 Fun G -1
8 7 ex Fun F G F Fun F -1 Fun G -1
9 8 com23 Fun F Fun F -1 G F Fun G -1
10 9 3imp Fun F Fun F -1 G F Fun G -1