Metamath Proof Explorer


Theorem func0g2

Description: The source cateogry of a functor to the empty category must be empty as well. (Contributed by Zhi Wang, 19-Oct-2025)

Ref Expression
Hypotheses func0g.a
|- A = ( Base ` C )
func0g.b
|- B = ( Base ` D )
func0g.d
|- ( ph -> B = (/) )
func0g2.f
|- ( ph -> F e. ( C Func D ) )
Assertion func0g2
|- ( ph -> A = (/) )

Proof

Step Hyp Ref Expression
1 func0g.a
 |-  A = ( Base ` C )
2 func0g.b
 |-  B = ( Base ` D )
3 func0g.d
 |-  ( ph -> B = (/) )
4 func0g2.f
 |-  ( ph -> F e. ( C Func D ) )
5 4 func1st2nd
 |-  ( ph -> ( 1st ` F ) ( C Func D ) ( 2nd ` F ) )
6 1 2 3 5 func0g
 |-  ( ph -> A = (/) )