let G be _Graph; for P being Path of G
for n, m being odd Element of NAT st n < m & m <= len P & ( n <> 1 or m <> len P ) holds
P .cut (n,m) is open
let P be Path of G; for n, m being odd Element of NAT st n < m & m <= len P & ( n <> 1 or m <> len P ) holds
P .cut (n,m) is open
let n, m be odd Element of NAT ; ( n < m & m <= len P & ( n <> 1 or m <> len P ) implies P .cut (n,m) is open )
assume A1:
( n < m & m <= len P & ( n <> 1 or m <> len P ) )
; P .cut (n,m) is open
then A2:
( (P .cut (n,m)) .first() = P . n & (P .cut (n,m)) .last() = P . m )
by GLIB_001:37;
assume
P .cut (n,m) is closed
; contradiction
then
P . n = P . m
by A2, GLIB_001:def 24;
hence
contradiction
by A1, GLIB_001:def 28; verum