Metamath Proof Explorer


Theorem vtxnbuvtx

Description: A universal vertex has all other vertices as neighbors. (Contributed by Alexander van der Vekens, 14-Oct-2017) (Revised by AV, 30-Oct-2020) (Proof shortened by AV, 14-Feb-2022)

Ref Expression
Hypothesis uvtxel.v 𝑉 = ( Vtx ‘ 𝐺 )
Assertion vtxnbuvtx ( 𝑁 ∈ ( UnivVtx ‘ 𝐺 ) → ∀ 𝑛 ∈ ( 𝑉 ∖ { 𝑁 } ) 𝑛 ∈ ( 𝐺 NeighbVtx 𝑁 ) )

Proof

Step Hyp Ref Expression
1 uvtxel.v 𝑉 = ( Vtx ‘ 𝐺 )
2 1 uvtxel ( 𝑁 ∈ ( UnivVtx ‘ 𝐺 ) ↔ ( 𝑁𝑉 ∧ ∀ 𝑛 ∈ ( 𝑉 ∖ { 𝑁 } ) 𝑛 ∈ ( 𝐺 NeighbVtx 𝑁 ) ) )
3 2 simprbi ( 𝑁 ∈ ( UnivVtx ‘ 𝐺 ) → ∀ 𝑛 ∈ ( 𝑉 ∖ { 𝑁 } ) 𝑛 ∈ ( 𝐺 NeighbVtx 𝑁 ) )