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 imbitrrid ⊢ 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