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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 ⊢ ( 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 inviso1 ( 𝜑 → 𝐹 ∈ ( 𝑋 𝐼 𝑌 ) )

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 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 ⊢ ( 𝜑 → 𝐹 ∈ dom ( 𝑋 𝑁 𝑌 ) )
13 1 2 3 4 5 6 isoval ⊢ ( 𝜑 → ( 𝑋 𝐼 𝑌 ) = dom ( 𝑋 𝑁 𝑌 ) )
14 12 13 eleqtrrd ⊢ ( 𝜑 → 𝐹 ∈ ( 𝑋 𝐼 𝑌 ) )