let X be set ; for g being SimpleGraph of X
for v, e being set st e in the SEdges of g & degree g,v = 0 holds
not v in e
let g be SimpleGraph of X; for v, e being set st e in the SEdges of g & degree g,v = 0 holds
not v in e
let v, e be set ; ( e in the SEdges of g & degree g,v = 0 implies not v in e )
assume that
A1:
e in the SEdges of g
and
A2:
degree g,v = 0
; not v in e
consider Y being finite set such that
A3:
for z being set holds
( z in Y iff ( z in the SEdges of g & v in z ) )
and
A4:
degree g,v = card Y
by Def11;
assume
v in e
; contradiction
then
not Y is empty
by A1, A3;
hence
contradiction
by A2, A4; verum