Metamath Proof Explorer
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 ) = { 〈 ∅ , ∅ 〉 } ) |