Metamath Proof Explorer


Theorem frgrconngr

Description: A friendship graph is connected, see remark 1 in MertziosUnger p. 153 (after Proposition 1): "An arbitrary friendship graph has to be connected, ... ". (Contributed by Alexander van der Vekens, 6-Dec-2017) (Revised by AV, 1-Apr-2021)

Ref Expression
Assertion frgrconngr
|- ( G e. FriendGraph -> G e. ConnGraph )

Proof

Step Hyp Ref Expression
1 eqid
 |-  ( Vtx ` G ) = ( Vtx ` G )
2 1 2pthfrgr
 |-  ( G e. FriendGraph -> A. k e. ( Vtx ` G ) A. n e. ( ( Vtx ` G ) \ { k } ) E. f E. p ( f ( k ( SPathsOn ` G ) n ) p /\ ( # ` f ) = 2 ) )
3 spthonpthon
 |-  ( f ( k ( SPathsOn ` G ) n ) p -> f ( k ( PathsOn ` G ) n ) p )
4 3 adantr
 |-  ( ( f ( k ( SPathsOn ` G ) n ) p /\ ( # ` f ) = 2 ) -> f ( k ( PathsOn ` G ) n ) p )
5 4 2eximi
 |-  ( E. f E. p ( f ( k ( SPathsOn ` G ) n ) p /\ ( # ` f ) = 2 ) -> E. f E. p f ( k ( PathsOn ` G ) n ) p )
6 5 2ralimi
 |-  ( A. k e. ( Vtx ` G ) A. n e. ( ( Vtx ` G ) \ { k } ) E. f E. p ( f ( k ( SPathsOn ` G ) n ) p /\ ( # ` f ) = 2 ) -> A. k e. ( Vtx ` G ) A. n e. ( ( Vtx ` G ) \ { k } ) E. f E. p f ( k ( PathsOn ` G ) n ) p )
7 2 6 syl
 |-  ( G e. FriendGraph -> A. k e. ( Vtx ` G ) A. n e. ( ( Vtx ` G ) \ { k } ) E. f E. p f ( k ( PathsOn ` G ) n ) p )
8 1 isconngr1
 |-  ( G e. FriendGraph -> ( G e. ConnGraph <-> A. k e. ( Vtx ` G ) A. n e. ( ( Vtx ` G ) \ { k } ) E. f E. p f ( k ( PathsOn ` G ) n ) p ) )
9 7 8 mpbird
 |-  ( G e. FriendGraph -> G e. ConnGraph )