Metamath Proof Explorer


Theorem 0conngr

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 ∈ ∅ ∃ 𝑓𝑝 𝑓 ( 𝑘 ( PathsOn ‘ ∅ ) 𝑛 ) 𝑝
2 0ex ∅ ∈ V
3 vtxval0 ( Vtx ‘ ∅ ) = ∅
4 3 eqcomi ∅ = ( Vtx ‘ ∅ )
5 4 isconngr ( ∅ ∈ V → ( ∅ ∈ ConnGraph ↔ ∀ 𝑘 ∈ ∅ ∀ 𝑛 ∈ ∅ ∃ 𝑓𝑝 𝑓 ( 𝑘 ( PathsOn ‘ ∅ ) 𝑛 ) 𝑝 ) )
6 2 5 ax-mp ( ∅ ∈ ConnGraph ↔ ∀ 𝑘 ∈ ∅ ∀ 𝑛 ∈ ∅ ∃ 𝑓𝑝 𝑓 ( 𝑘 ( PathsOn ‘ ∅ ) 𝑛 ) 𝑝 )
7 1 6 mpbir ∅ ∈ ConnGraph