Metamath Proof Explorer


Theorem cofid1a

Description: Express the object part of ( G o.func F ) = I explicitly. (Contributed by Zhi Wang, 15-Nov-2025)

Ref Expression
Hypotheses cofid1a.i 𝐼 = ( idfunc𝐷 )
cofid1a.b 𝐵 = ( Base ‘ 𝐷 )
cofid1a.x ( 𝜑𝑋𝐵 )
cofid1a.f ( 𝜑𝐹 ∈ ( 𝐷 Func 𝐸 ) )
cofid1a.g ( 𝜑𝐺 ∈ ( 𝐸 Func 𝐷 ) )
cofid1a.o ( 𝜑 → ( 𝐺func 𝐹 ) = 𝐼 )
Assertion cofid1a ( 𝜑 → ( ( 1st𝐺 ) ‘ ( ( 1st𝐹 ) ‘ 𝑋 ) ) = 𝑋 )

Proof

Step Hyp Ref Expression
1 cofid1a.i 𝐼 = ( idfunc𝐷 )
2 cofid1a.b 𝐵 = ( Base ‘ 𝐷 )
3 cofid1a.x ( 𝜑𝑋𝐵 )
4 cofid1a.f ( 𝜑𝐹 ∈ ( 𝐷 Func 𝐸 ) )
5 cofid1a.g ( 𝜑𝐺 ∈ ( 𝐸 Func 𝐷 ) )
6 cofid1a.o ( 𝜑 → ( 𝐺func 𝐹 ) = 𝐼 )
7 6 fveq2d ( 𝜑 → ( 1st ‘ ( 𝐺func 𝐹 ) ) = ( 1st𝐼 ) )
8 7 fveq1d ( 𝜑 → ( ( 1st ‘ ( 𝐺func 𝐹 ) ) ‘ 𝑋 ) = ( ( 1st𝐼 ) ‘ 𝑋 ) )
9 2 4 5 3 cofu1 ( 𝜑 → ( ( 1st ‘ ( 𝐺func 𝐹 ) ) ‘ 𝑋 ) = ( ( 1st𝐺 ) ‘ ( ( 1st𝐹 ) ‘ 𝑋 ) ) )
10 4 func1st2nd ( 𝜑 → ( 1st𝐹 ) ( 𝐷 Func 𝐸 ) ( 2nd𝐹 ) )
11 10 funcrcl2 ( 𝜑𝐷 ∈ Cat )
12 1 2 11 3 idfu1 ( 𝜑 → ( ( 1st𝐼 ) ‘ 𝑋 ) = 𝑋 )
13 8 9 12 3eqtr3d ( 𝜑 → ( ( 1st𝐺 ) ‘ ( ( 1st𝐹 ) ‘ 𝑋 ) ) = 𝑋 )