Metamath Proof Explorer


Theorem inviso1

Description: If G is an inverse to F , then F 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 inviso1 ( 𝜑𝐹 ∈ ( 𝑋 𝐼 𝑌 ) )

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 invfun ( 𝜑 → Fun ( 𝑋 𝑁 𝑌 ) )
9 funrel ( Fun ( 𝑋 𝑁 𝑌 ) → Rel ( 𝑋 𝑁 𝑌 ) )
10 8 9 syl ( 𝜑 → Rel ( 𝑋 𝑁 𝑌 ) )
11 releldm ( ( Rel ( 𝑋 𝑁 𝑌 ) ∧ 𝐹 ( 𝑋 𝑁 𝑌 ) 𝐺 ) → 𝐹 ∈ dom ( 𝑋 𝑁 𝑌 ) )
12 10 7 11 syl2anc ( 𝜑𝐹 ∈ dom ( 𝑋 𝑁 𝑌 ) )
13 1 2 3 4 5 6 isoval ( 𝜑 → ( 𝑋 𝐼 𝑌 ) = dom ( 𝑋 𝑁 𝑌 ) )
14 12 13 eleqtrrd ( 𝜑𝐹 ∈ ( 𝑋 𝐼 𝑌 ) )