Metamath Proof Explorer

Theorem 0func

Description: The functor from the empty category. (Contributed by Zhi Wang, 7-Oct-2025) (Proof shortened by Zhi Wang, 17-Oct-2025)

Ref Expression
Hypothesis 0func.c 
|- ( ph -> C e. Cat )
Assertion 0func 
|- ( ph -> ( (/) Func C ) = { <. (/) , (/) >. } )

Proof

Step Hyp Ref Expression
1 0func.c 
 |-  ( ph -> C e. Cat )
2 0ex 
 |-  (/) e. _V
3 2 a1i 
 |-  ( ph -> (/) e. _V )
4 base0 
 |-  (/) = ( Base ` (/) )
5 4 a1i 
 |-  ( ph -> (/) = ( Base ` (/) ) )
6 3 5 1 0funcg 
 |-  ( ph -> ( (/) Func C ) = { <. (/) , (/) >. } )