let G1, G2 be _Graph; for v being set st G1 == G2 & not v in the_Vertices_of G1 holds
G2 is removeVertex of G1,v
let v be set ; ( G1 == G2 & not v in the_Vertices_of G1 implies G2 is removeVertex of G1,v )
assume A1:
( G1 == G2 & not v in the_Vertices_of G1 )
; G2 is removeVertex of G1,v
then A2:
G2 is Subgraph of G1
by GLIB_000:87;
set V = (the_Vertices_of G1) \ {v};
A3:
(the_Vertices_of G1) \ {v} = the_Vertices_of G1
by A1, ZFMISC_1:57;
A4: G1 .edgesBetween ((the_Vertices_of G1) \ {v}) =
the_Edges_of G1
by A3, GLIB_000:34
.=
the_Edges_of G2
by A1, GLIB_000:def 34
;
(the_Vertices_of G1) \ {v} = the_Vertices_of G2
by A1, A3, GLIB_000:def 34;
hence
G2 is removeVertex of G1,v
by A2, A4, GLIB_000:def 37; verum