Metamath Proof Explorer
Theorem fn0
Description: A function with empty domain is empty. (Contributed by NM, 15-Apr-1998)
(Proof shortened by Andrew Salmon, 17-Sep-2011)
|
|
Ref |
Expression |
|
Assertion |
fn0 |
⊢ ( 𝐹 Fn ∅ ↔ 𝐹 = ∅ ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
fnrel |
⊢ ( 𝐹 Fn ∅ → Rel 𝐹 ) |
2 |
|
fndm |
⊢ ( 𝐹 Fn ∅ → dom 𝐹 = ∅ ) |
3 |
|
reldm0 |
⊢ ( Rel 𝐹 → ( 𝐹 = ∅ ↔ dom 𝐹 = ∅ ) ) |
4 |
3
|
biimpar |
⊢ ( ( Rel 𝐹 ∧ dom 𝐹 = ∅ ) → 𝐹 = ∅ ) |
5 |
1 2 4
|
syl2anc |
⊢ ( 𝐹 Fn ∅ → 𝐹 = ∅ ) |
6 |
|
fun0 |
⊢ Fun ∅ |
7 |
|
dm0 |
⊢ dom ∅ = ∅ |
8 |
|
df-fn |
⊢ ( ∅ Fn ∅ ↔ ( Fun ∅ ∧ dom ∅ = ∅ ) ) |
9 |
6 7 8
|
mpbir2an |
⊢ ∅ Fn ∅ |
10 |
|
fneq1 |
⊢ ( 𝐹 = ∅ → ( 𝐹 Fn ∅ ↔ ∅ Fn ∅ ) ) |
11 |
9 10
|
mpbiri |
⊢ ( 𝐹 = ∅ → 𝐹 Fn ∅ ) |
12 |
5 11
|
impbii |
⊢ ( 𝐹 Fn ∅ ↔ 𝐹 = ∅ ) |