Metamath Proof Explorer
Description: A graph without vertices is connected. (Contributed by Alexander van
der Vekens, 2-Dec-2017) (Revised by AV, 15-Feb-2021)
|
|
Ref |
Expression |
|
Assertion |
0conngr |
⊢ ∅ ∈ ConnGraph |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
ral0 |
⊢ ∀ 𝑘 ∈ ∅ ∀ 𝑛 ∈ ∅ ∃ 𝑓 ∃ 𝑝 𝑓 ( 𝑘 ( PathsOn ‘ ∅ ) 𝑛 ) 𝑝 |
2 |
|
0ex |
⊢ ∅ ∈ V |
3 |
|
vtxval0 |
⊢ ( Vtx ‘ ∅ ) = ∅ |
4 |
3
|
eqcomi |
⊢ ∅ = ( Vtx ‘ ∅ ) |
5 |
4
|
isconngr |
⊢ ( ∅ ∈ V → ( ∅ ∈ ConnGraph ↔ ∀ 𝑘 ∈ ∅ ∀ 𝑛 ∈ ∅ ∃ 𝑓 ∃ 𝑝 𝑓 ( 𝑘 ( PathsOn ‘ ∅ ) 𝑛 ) 𝑝 ) ) |
6 |
2 5
|
ax-mp |
⊢ ( ∅ ∈ ConnGraph ↔ ∀ 𝑘 ∈ ∅ ∀ 𝑛 ∈ ∅ ∃ 𝑓 ∃ 𝑝 𝑓 ( 𝑘 ( PathsOn ‘ ∅ ) 𝑛 ) 𝑝 ) |
7 |
1 6
|
mpbir |
⊢ ∅ ∈ ConnGraph |