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 ( 𝜑𝐶 ∈ Cat )
invfval.x ( 𝜑𝑋𝐵 )
invfval.y ( 𝜑𝑌𝐵 )
isoval.n 𝐼 = ( Iso ‘ 𝐶 )
inviso1.1 ( 𝜑𝐹 ( 𝑋 𝑁 𝑌 ) 𝐺 )
Assertion inviso2 ( 𝜑𝐺 ∈ ( 𝑌 𝐼 𝑋 ) )

Proof

Step Hyp Ref Expression
1 invfval.b 𝐵 = ( Base ‘ 𝐶 )
2 invfval.n 𝑁 = ( Inv ‘ 𝐶 )
3 invfval.c ( 𝜑𝐶 ∈ Cat )
4 invfval.x ( 𝜑𝑋𝐵 )
5 invfval.y ( 𝜑𝑌𝐵 )
6 isoval.n 𝐼 = ( Iso ‘ 𝐶 )
7 inviso1.1 ( 𝜑𝐹 ( 𝑋 𝑁 𝑌 ) 𝐺 )
8 1 2 3 4 5 invsym ( 𝜑 → ( 𝐹 ( 𝑋 𝑁 𝑌 ) 𝐺𝐺 ( 𝑌 𝑁 𝑋 ) 𝐹 ) )
9 7 8 mpbid ( 𝜑𝐺 ( 𝑌 𝑁 𝑋 ) 𝐹 )
10 1 2 3 5 4 6 9 inviso1 ( 𝜑𝐺 ∈ ( 𝑌 𝐼 𝑋 ) )