Metamath Proof Explorer


Theorem issect2

Description: Property of being a section. (Contributed by Mario Carneiro, 2-Jan-2017)

Ref Expression
Hypotheses issect.b 𝐵 = ( Base ‘ 𝐶 )
issect.h 𝐻 = ( Hom ‘ 𝐶 )
issect.o · = ( comp ‘ 𝐶 )
issect.i 1 = ( Id ‘ 𝐶 )
issect.s 𝑆 = ( Sect ‘ 𝐶 )
issect.c ( 𝜑𝐶 ∈ Cat )
issect.x ( 𝜑𝑋𝐵 )
issect.y ( 𝜑𝑌𝐵 )
issect.f ( 𝜑𝐹 ∈ ( 𝑋 𝐻 𝑌 ) )
issect.g ( 𝜑𝐺 ∈ ( 𝑌 𝐻 𝑋 ) )
Assertion issect2 ( 𝜑 → ( 𝐹 ( 𝑋 𝑆 𝑌 ) 𝐺 ↔ ( 𝐺 ( ⟨ 𝑋 , 𝑌· 𝑋 ) 𝐹 ) = ( 1𝑋 ) ) )

Proof

Step Hyp Ref Expression
1 issect.b 𝐵 = ( Base ‘ 𝐶 )
2 issect.h 𝐻 = ( Hom ‘ 𝐶 )
3 issect.o · = ( comp ‘ 𝐶 )
4 issect.i 1 = ( Id ‘ 𝐶 )
5 issect.s 𝑆 = ( Sect ‘ 𝐶 )
6 issect.c ( 𝜑𝐶 ∈ Cat )
7 issect.x ( 𝜑𝑋𝐵 )
8 issect.y ( 𝜑𝑌𝐵 )
9 issect.f ( 𝜑𝐹 ∈ ( 𝑋 𝐻 𝑌 ) )
10 issect.g ( 𝜑𝐺 ∈ ( 𝑌 𝐻 𝑋 ) )
11 9 10 jca ( 𝜑 → ( 𝐹 ∈ ( 𝑋 𝐻 𝑌 ) ∧ 𝐺 ∈ ( 𝑌 𝐻 𝑋 ) ) )
12 1 2 3 4 5 6 7 8 issect ( 𝜑 → ( 𝐹 ( 𝑋 𝑆 𝑌 ) 𝐺 ↔ ( 𝐹 ∈ ( 𝑋 𝐻 𝑌 ) ∧ 𝐺 ∈ ( 𝑌 𝐻 𝑋 ) ∧ ( 𝐺 ( ⟨ 𝑋 , 𝑌· 𝑋 ) 𝐹 ) = ( 1𝑋 ) ) ) )
13 df-3an ( ( 𝐹 ∈ ( 𝑋 𝐻 𝑌 ) ∧ 𝐺 ∈ ( 𝑌 𝐻 𝑋 ) ∧ ( 𝐺 ( ⟨ 𝑋 , 𝑌· 𝑋 ) 𝐹 ) = ( 1𝑋 ) ) ↔ ( ( 𝐹 ∈ ( 𝑋 𝐻 𝑌 ) ∧ 𝐺 ∈ ( 𝑌 𝐻 𝑋 ) ) ∧ ( 𝐺 ( ⟨ 𝑋 , 𝑌· 𝑋 ) 𝐹 ) = ( 1𝑋 ) ) )
14 12 13 bitrdi ( 𝜑 → ( 𝐹 ( 𝑋 𝑆 𝑌 ) 𝐺 ↔ ( ( 𝐹 ∈ ( 𝑋 𝐻 𝑌 ) ∧ 𝐺 ∈ ( 𝑌 𝐻 𝑋 ) ) ∧ ( 𝐺 ( ⟨ 𝑋 , 𝑌· 𝑋 ) 𝐹 ) = ( 1𝑋 ) ) ) )
15 11 14 mpbirand ( 𝜑 → ( 𝐹 ( 𝑋 𝑆 𝑌 ) 𝐺 ↔ ( 𝐺 ( ⟨ 𝑋 , 𝑌· 𝑋 ) 𝐹 ) = ( 1𝑋 ) ) )