Metamath Proof Explorer


Theorem idfu1stalem

Description: Lemma for idfu1sta . (Contributed by Zhi Wang, 10-Nov-2025)

Ref Expression
Hypotheses idfu2nda.i 𝐼 = ( idfunc𝐶 )
idfu2nda.d ( 𝜑𝐼 ∈ ( 𝐷 Func 𝐸 ) )
idfu2nda.b ( 𝜑𝐵 = ( Base ‘ 𝐷 ) )
Assertion idfu1stalem ( 𝜑𝐵 = ( Base ‘ 𝐶 ) )

Proof

Step Hyp Ref Expression
1 idfu2nda.i 𝐼 = ( idfunc𝐶 )
2 idfu2nda.d ( 𝜑𝐼 ∈ ( 𝐷 Func 𝐸 ) )
3 idfu2nda.b ( 𝜑𝐵 = ( Base ‘ 𝐷 ) )
4 1 2 eqeltrrid ( 𝜑 → ( idfunc𝐶 ) ∈ ( 𝐷 Func 𝐸 ) )
5 idfurcl ( ( idfunc𝐶 ) ∈ ( 𝐷 Func 𝐸 ) → 𝐶 ∈ Cat )
6 1 idfucl ( 𝐶 ∈ Cat → 𝐼 ∈ ( 𝐶 Func 𝐶 ) )
7 4 5 6 3syl ( 𝜑𝐼 ∈ ( 𝐶 Func 𝐶 ) )
8 7 func1st2nd ( 𝜑 → ( 1st𝐼 ) ( 𝐶 Func 𝐶 ) ( 2nd𝐼 ) )
9 2 func1st2nd ( 𝜑 → ( 1st𝐼 ) ( 𝐷 Func 𝐸 ) ( 2nd𝐼 ) )
10 8 9 funchomf ( 𝜑 → ( Homf𝐶 ) = ( Homf𝐷 ) )
11 10 homfeqbas ( 𝜑 → ( Base ‘ 𝐶 ) = ( Base ‘ 𝐷 ) )
12 3 11 eqtr4d ( 𝜑𝐵 = ( Base ‘ 𝐶 ) )