Metamath Proof Explorer


Theorem vdgn0frgrv2

Description: A vertex in a friendship graph with more than one vertex cannot have degree 0. (Contributed by Alexander van der Vekens, 9-Dec-2017) (Revised by AV, 4-Apr-2021)

Ref Expression
Hypothesis vdn1frgrv2.v
|- V = ( Vtx ` G )
Assertion vdgn0frgrv2
|- ( ( G e. FriendGraph /\ N e. V ) -> ( 1 < ( # ` V ) -> ( ( VtxDeg ` G ) ` N ) =/= 0 ) )

Proof

Step Hyp Ref Expression
1 vdn1frgrv2.v
 |-  V = ( Vtx ` G )
2 frgrconngr
 |-  ( G e. FriendGraph -> G e. ConnGraph )
3 frgrusgr
 |-  ( G e. FriendGraph -> G e. USGraph )
4 usgrumgr
 |-  ( G e. USGraph -> G e. UMGraph )
5 3 4 syl
 |-  ( G e. FriendGraph -> G e. UMGraph )
6 1 vdn0conngrumgrv2
 |-  ( ( ( G e. ConnGraph /\ G e. UMGraph ) /\ ( N e. V /\ 1 < ( # ` V ) ) ) -> ( ( VtxDeg ` G ) ` N ) =/= 0 )
7 6 ex
 |-  ( ( G e. ConnGraph /\ G e. UMGraph ) -> ( ( N e. V /\ 1 < ( # ` V ) ) -> ( ( VtxDeg ` G ) ` N ) =/= 0 ) )
8 2 5 7 syl2anc
 |-  ( G e. FriendGraph -> ( ( N e. V /\ 1 < ( # ` V ) ) -> ( ( VtxDeg ` G ) ` N ) =/= 0 ) )
9 8 expdimp
 |-  ( ( G e. FriendGraph /\ N e. V ) -> ( 1 < ( # ` V ) -> ( ( VtxDeg ` G ) ` N ) =/= 0 ) )