let FT be non empty RelStr ; for g being FinSequence of FT
for A being Subset of FT
for x1, x2 being Element of FT
for k being Element of NAT st g is_minimum_path_in A,x1,x2 & 1 <= k & k <= len g holds
( g | k is continuous & rng (g | k) c= A & (g | k) . 1 = x1 & (g | k) . (len (g | k)) = g /. k )
let g be FinSequence of FT; for A being Subset of FT
for x1, x2 being Element of FT
for k being Element of NAT st g is_minimum_path_in A,x1,x2 & 1 <= k & k <= len g holds
( g | k is continuous & rng (g | k) c= A & (g | k) . 1 = x1 & (g | k) . (len (g | k)) = g /. k )
let A be Subset of FT; for x1, x2 being Element of FT
for k being Element of NAT st g is_minimum_path_in A,x1,x2 & 1 <= k & k <= len g holds
( g | k is continuous & rng (g | k) c= A & (g | k) . 1 = x1 & (g | k) . (len (g | k)) = g /. k )
let x1, x2 be Element of FT; for k being Element of NAT st g is_minimum_path_in A,x1,x2 & 1 <= k & k <= len g holds
( g | k is continuous & rng (g | k) c= A & (g | k) . 1 = x1 & (g | k) . (len (g | k)) = g /. k )
let k be Element of NAT ; ( g is_minimum_path_in A,x1,x2 & 1 <= k & k <= len g implies ( g | k is continuous & rng (g | k) c= A & (g | k) . 1 = x1 & (g | k) . (len (g | k)) = g /. k ) )
assume that
A1:
g is_minimum_path_in A,x1,x2
and
A2:
1 <= k
and
A3:
k <= len g
; ( g | k is continuous & rng (g | k) c= A & (g | k) . 1 = x1 & (g | k) . (len (g | k)) = g /. k )
A4:
k in dom g
by A2, A3, FINSEQ_3:25;
g is continuous
by A1;
hence
g | k is continuous
by A2, Th46; ( rng (g | k) c= A & (g | k) . 1 = x1 & (g | k) . (len (g | k)) = g /. k )
A5:
rng (g | k) c= rng g
by FINSEQ_5:19;
rng g c= A
by A1;
hence
rng (g | k) c= A
by A5; ( (g | k) . 1 = x1 & (g | k) . (len (g | k)) = g /. k )
g . 1 = x1
by A1;
hence
(g | k) . 1 = x1
by A2, FINSEQ_3:112; (g | k) . (len (g | k)) = g /. k
len (g | k) = k
by A3, FINSEQ_1:59;
hence (g | k) . (len (g | k)) =
g . k
by FINSEQ_3:112
.=
g /. k
by A4, PARTFUN1:def 6
;
verum