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 ∈ ∅ ∃ 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