set W2 = W .remove m,n;
A1:
len (W .remove m,n) <= len W
by Lm26;
now per cases
( ( not m is even & not n is even & m <= n & n <= len W & W . m = W . n ) or m is even or n is even or not m <= n or not n <= len W or not W . m = W . n )
;
suppose A2:
( not
m is
even & not
n is
even &
m <= n &
n <= len W &
W . m = W . n )
;
W .remove m,n is Trail-like then reconsider m9 =
m,
n9 =
n as
odd Element of
NAT ;
now given a,
b being
even Element of
NAT such that A3:
1
<= a
and A4:
a < b
and A5:
b <= len (W .remove m,n)
and A6:
(W .remove m,n) . a = (W .remove m,n) . b
;
contradiction
1
<= b
by A3, A4, XXREAL_0:2;
then A7:
b in dom (W .remove m,n)
by A5, FINSEQ_3:27;
a <= len (W .remove m,n)
by A4, A5, XXREAL_0:2;
then A8:
a in dom (W .remove m,n)
by A3, FINSEQ_3:27;
now per cases
( a in Seg m or ( m <= a & a <= len (W .remove m,n) ) )
by A2, A8, Lm34;
suppose
a in Seg m
;
contradictionthen A9:
(W .remove m,n) . a = W . a
by A2, Lm29;
now per cases
( b in Seg m or ( m <= b & b <= len (W .remove m,n) ) )
by A2, A7, Lm34;
suppose A10:
b in Seg m
;
contradictionA11:
b <= len W
by A1, A5, XXREAL_0:2;
(W .remove m,n) . b = W . b
by A2, A10, Lm29;
hence
contradiction
by A3, A4, A6, A9, A11, Lm57;
verum end; end; end; hence
contradiction
;
verum end; end; end; hence
contradiction
;
verum end; hence
W .remove m,
n is
Trail-like
by Lm57;
verum end; end; end;
hence
W .remove m,n is Trail-like
; verum