Metamath Proof Explorer


Theorem invsym

Description: The inverse relation is symmetric. (Contributed by Mario Carneiro, 2-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 )
Assertion invsym ( 𝜑 → ( 𝐹 ( 𝑋 𝑁 𝑌 ) 𝐺 ↔ 𝐺 ( 𝑌 𝑁 𝑋 ) 𝐹 ) )

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 eqid ⊢ ( Sect ‘ 𝐶 ) = ( Sect ‘ 𝐶 )
7 1 2 3 4 5 6 isinv ⊢ ( 𝜑 → ( 𝐹 ( 𝑋 𝑁 𝑌 ) 𝐺 ↔ ( 𝐹 ( 𝑋 ( Sect ‘ 𝐶 ) 𝑌 ) 𝐺 ∧ 𝐺 ( 𝑌 ( Sect ‘ 𝐶 ) 𝑋 ) 𝐹 ) ) )
8 7 biancomd ⊢ ( 𝜑 → ( 𝐹 ( 𝑋 𝑁 𝑌 ) 𝐺 ↔ ( 𝐺 ( 𝑌 ( Sect ‘ 𝐶 ) 𝑋 ) 𝐹 ∧ 𝐹 ( 𝑋 ( Sect ‘ 𝐶 ) 𝑌 ) 𝐺 ) ) )
9 1 2 3 5 4 6 isinv ⊢ ( 𝜑 → ( 𝐺 ( 𝑌 𝑁 𝑋 ) 𝐹 ↔ ( 𝐺 ( 𝑌 ( Sect ‘ 𝐶 ) 𝑋 ) 𝐹 ∧ 𝐹 ( 𝑋 ( Sect ‘ 𝐶 ) 𝑌 ) 𝐺 ) ) )
10 8 9 bitr4d ⊢ ( 𝜑 → ( 𝐹 ( 𝑋 𝑁 𝑌 ) 𝐺 ↔ 𝐺 ( 𝑌 𝑁 𝑋 ) 𝐹 ) )