let n be Nat; for D being non empty set
for f being FinSequence of D st f is one-to-one holds
rng (f | n) misses rng (f /^ n)
let D be non empty set ; for f being FinSequence of D st f is one-to-one holds
rng (f | n) misses rng (f /^ n)
let f be FinSequence of D; ( f is one-to-one implies rng (f | n) misses rng (f /^ n) )
assume A1:
f is one-to-one
; rng (f | n) misses rng (f /^ n)
A2:
dom (f | n) c= dom f
by Th20;
assume
rng (f | n) meets rng (f /^ n)
; contradiction
then consider x being set such that
A3:
x in rng (f | n)
and
A4:
x in rng (f /^ n)
by XBOOLE_0:3;
consider i being Element of NAT such that
A5:
i in dom (f | n)
and
A6:
(f | n) /. i = x
by A3, PARTFUN2:4;
consider j being Element of NAT such that
A7:
j in dom (f /^ n)
and
A8:
(f /^ n) /. j = x
by A4, PARTFUN2:4;
A9:
j + n in dom f
by A7, Th29;
f /. (j + n) =
(f /^ n) /. j
by A7, Th30
.=
f /. i
by A5, A6, A8, FINSEQ_4:85
;
hence
contradiction
by A1, A5, A2, A9, A10, PARTFUN2:17; verum