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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 ) )