Metamath Proof Explorer


Theorem inviso2

Description: If G is an inverse to F , then G is an isomorphism. (Contributed by Mario Carneiro, 3-Jan-2017)

Ref Expression
Hypotheses invfval.b ⊢ B = Base C
invfval.n ⊢ N = Inv ⁡ C
invfval.c ⊢ φ → C ∈ Cat
invss.x ⊢ φ → X ∈ B
invss.y ⊢ φ → Y ∈ B
isoval.n ⊢ I = Iso ⁡ C
inviso1.1 ⊢ φ → F X N Y G
Assertion inviso2 ⊢ φ → G ∈ Y I X

Proof

Step Hyp Ref Expression
1 invfval.b ⊢ B = Base C
2 invfval.n ⊢ N = Inv ⁡ C
3 invfval.c ⊢ φ → C ∈ Cat
4 invss.x ⊢ φ → X ∈ B
5 invss.y ⊢ φ → Y ∈ B
6 isoval.n ⊢ I = Iso ⁡ C
7 inviso1.1 ⊢ φ → F X N Y G
8 1 2 3 4 5 invsym ⊢ φ → F X N Y G ↔ G Y N X F
9 7 8 mpbid ⊢ φ → G Y N X F
10 1 2 3 5 4 6 9 inviso1 ⊢ φ → G ∈ Y I X