Metamath Proof Explorer


Theorem 0funcg

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

Ref Expression
Hypotheses 0funcg.c ( 𝜑𝐶𝑉 )
0funcg.b ( 𝜑 → ∅ = ( Base ‘ 𝐶 ) )
0funcg.d ( 𝜑𝐷 ∈ Cat )
Assertion 0funcg ( 𝜑 → ( 𝐶 Func 𝐷 ) = { ⟨ ∅ , ∅ ⟩ } )

Proof

Step Hyp Ref Expression
1 0funcg.c ( 𝜑𝐶𝑉 )
2 0funcg.b ( 𝜑 → ∅ = ( Base ‘ 𝐶 ) )
3 0funcg.d ( 𝜑𝐷 ∈ Cat )
4 relfunc Rel ( 𝐶 Func 𝐷 )
5 0ex ∅ ∈ V
6 5 5 relsnop Rel { ⟨ ∅ , ∅ ⟩ }
7 1 2 3 0funcg2 ( 𝜑 → ( 𝑓 ( 𝐶 Func 𝐷 ) 𝑔 ↔ ( 𝑓 = ∅ ∧ 𝑔 = ∅ ) ) )
8 brsnop ( ( ∅ ∈ V ∧ ∅ ∈ V ) → ( 𝑓 { ⟨ ∅ , ∅ ⟩ } 𝑔 ↔ ( 𝑓 = ∅ ∧ 𝑔 = ∅ ) ) )
9 5 5 8 mp2an ( 𝑓 { ⟨ ∅ , ∅ ⟩ } 𝑔 ↔ ( 𝑓 = ∅ ∧ 𝑔 = ∅ ) )
10 7 9 bitr4di ( 𝜑 → ( 𝑓 ( 𝐶 Func 𝐷 ) 𝑔𝑓 { ⟨ ∅ , ∅ ⟩ } 𝑔 ) )
11 4 6 10 eqbrrdiv ( 𝜑 → ( 𝐶 Func 𝐷 ) = { ⟨ ∅ , ∅ ⟩ } )