Metamath Proof Explorer


Theorem usgrvd00

Description: If every vertex in a simple graph has degree 0, there is no edge in the graph. (Contributed by Alexander van der Vekens, 12-Jul-2018) (Revised by AV, 17-Dec-2020) (Proof shortened by AV, 23-Dec-2020)

Ref Expression
Hypotheses vtxdusgradjvtx.v
|- V = ( Vtx ` G )
vtxdusgradjvtx.e
|- E = ( Edg ` G )
Assertion usgrvd00
|- ( G e. USGraph -> ( A. v e. V ( ( VtxDeg ` G ) ` v ) = 0 -> E = (/) ) )

Proof

Step Hyp Ref Expression
1 vtxdusgradjvtx.v
 |-  V = ( Vtx ` G )
2 vtxdusgradjvtx.e
 |-  E = ( Edg ` G )
3 usgruhgr
 |-  ( G e. USGraph -> G e. UHGraph )
4 1 2 uhgrvd00
 |-  ( G e. UHGraph -> ( A. v e. V ( ( VtxDeg ` G ) ` v ) = 0 -> E = (/) ) )
5 3 4 syl
 |-  ( G e. USGraph -> ( A. v e. V ( ( VtxDeg ` G ) ` v ) = 0 -> E = (/) ) )