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 ( 𝜑𝐵 = ∅ )
func0g2.f ( 𝜑𝐹 ∈ ( 𝐶 Func 𝐷 ) )
Assertion func0g2 ( 𝜑𝐴 = ∅ )

Proof

Step Hyp Ref Expression
1 func0g.a 𝐴 = ( Base ‘ 𝐶 )
2 func0g.b 𝐵 = ( Base ‘ 𝐷 )
3 func0g.d ( 𝜑𝐵 = ∅ )
4 func0g2.f ( 𝜑𝐹 ∈ ( 𝐶 Func 𝐷 ) )
5 4 func1st2nd ( 𝜑 → ( 1st𝐹 ) ( 𝐶 Func 𝐷 ) ( 2nd𝐹 ) )
6 1 2 3 5 func0g ( 𝜑𝐴 = ∅ )