Metamath Proof Explorer


Theorem rinvf1o

Description: Sufficient conditions for the restriction of an involution to be a bijection. (Contributed by Thierry Arnoux, 7-Dec-2016)

Ref Expression
Hypotheses rinvbij.1 ⊢ Fun ⁡ F
rinvbij.2 ⊢ F -1 = F
rinvbij.3a ⊢ F A ⊆ B
rinvbij.3b ⊢ F B ⊆ A
rinvbij.4a ⊢ A ⊆ dom ⁡ F
rinvbij.4b ⊢ B ⊆ dom ⁡ F
Assertion rinvf1o ⊢ F ↾ A : A ⟶ 1-1 onto B

Proof

Step Hyp Ref Expression
1 rinvbij.1 ⊢ Fun ⁡ F
2 rinvbij.2 ⊢ F -1 = F
3 rinvbij.3a ⊢ F A ⊆ B
4 rinvbij.3b ⊢ F B ⊆ A
5 rinvbij.4a ⊢ A ⊆ dom ⁡ F
6 rinvbij.4b ⊢ B ⊆ dom ⁡ F
7 fdmrn ⊢ Fun ⁡ F ↔ F : dom ⁡ F ⟶ ran ⁡ F
8 1 7 mpbi ⊢ F : dom ⁡ F ⟶ ran ⁡ F
9 2 funeqi ⊢ Fun ⁡ F -1 ↔ Fun ⁡ F
10 1 9 mpbir ⊢ Fun ⁡ F -1
11 df-f1 ⊢ F : dom ⁡ F ⟶ 1-1 ran ⁡ F ↔ F : dom ⁡ F ⟶ ran ⁡ F ∧ Fun ⁡ F -1
12 8 10 11 mpbir2an ⊢ F : dom ⁡ F ⟶ 1-1 ran ⁡ F
13 f1ores ⊢ F : dom ⁡ F ⟶ 1-1 ran ⁡ F ∧ A ⊆ dom ⁡ F → F ↾ A : A ⟶ 1-1 onto F A
14 12 5 13 mp2an ⊢ F ↾ A : A ⟶ 1-1 onto F A
15 funimass3 ⊢ Fun ⁡ F ∧ B ⊆ dom ⁡ F → F B ⊆ A ↔ B ⊆ F -1 A
16 1 6 15 mp2an ⊢ F B ⊆ A ↔ B ⊆ F -1 A
17 4 16 mpbi ⊢ B ⊆ F -1 A
18 2 imaeq1i ⊢ F -1 A = F A
19 17 18 sseqtri ⊢ B ⊆ F A
20 3 19 eqssi ⊢ F A = B
21 f1oeq3 ⊢ F A = B → F ↾ A : A ⟶ 1-1 onto F A ↔ F ↾ A : A ⟶ 1-1 onto B
22 20 21 ax-mp ⊢ F ↾ A : A ⟶ 1-1 onto F A ↔ F ↾ A : A ⟶ 1-1 onto B
23 14 22 mpbi ⊢ F ↾ A : A ⟶ 1-1 onto B