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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 ⊢ ( 𝜑 → 𝐶 ∈ Cat )
Assertion 0func ( 𝜑 → ( ∅ Func 𝐶 ) = { ⟨ ∅ , ∅ ⟩ } )

Proof

Step Hyp Ref Expression
1 0func.c ⊢ ( 𝜑 → 𝐶 ∈ Cat )
2 0ex ⊢ ∅ ∈ V
3 2 a1i ⊢ ( 𝜑 → ∅ ∈ V )
4 base0 ⊢ ∅ = ( Base ‘ ∅ )
5 4 a1i ⊢ ( 𝜑 → ∅ = ( Base ‘ ∅ ) )
6 3 5 1 0funcg ⊢ ( 𝜑 → ( ∅ Func 𝐶 ) = { ⟨ ∅ , ∅ ⟩ } )