Metamath Proof Explorer


Theorem func0g

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 = (/) )
func0g.f
|- ( ph -> F ( C Func D ) G )
Assertion func0g
|- ( 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 func0g.f
 |-  ( ph -> F ( C Func D ) G )
5 1 2 4 funcf1
 |-  ( ph -> F : A --> B )
6 5 f002
 |-  ( ph -> ( B = (/) -> A = (/) ) )
7 3 6 mpd
 |-  ( ph -> A = (/) )