let f be non constant standard special_circular_sequence; for i2 being Element of NAT st 1 < i2 & i2 + 1 <= len f holds
f | i2 is being_S-Seq
let i2 be Element of NAT ; ( 1 < i2 & i2 + 1 <= len f implies f | i2 is being_S-Seq )
assume that
A1:
1 < i2
and
A2:
i2 + 1 <= len f
; f | i2 is being_S-Seq
i2 <= len f
by A2, NAT_1:13;
then A3:
len (f | i2) = i2
by FINSEQ_1:80;
for n1, n2 being Element of NAT st 1 <= n1 & n1 <= len (f | i2) & 1 <= n2 & n2 <= len (f | i2) & ( (f | i2) . n1 = (f | i2) . n2 or (f | i2) /. n1 = (f | i2) /. n2 ) holds
n1 = n2
proof
let n1,
n2 be
Element of
NAT ;
( 1 <= n1 & n1 <= len (f | i2) & 1 <= n2 & n2 <= len (f | i2) & ( (f | i2) . n1 = (f | i2) . n2 or (f | i2) /. n1 = (f | i2) /. n2 ) implies n1 = n2 )
assume that A4:
1
<= n1
and A5:
n1 <= len (f | i2)
and A6:
1
<= n2
and A7:
n2 <= len (f | i2)
and A8:
(
(f | i2) . n1 = (f | i2) . n2 or
(f | i2) /. n1 = (f | i2) /. n2 )
;
n1 = n2
A9:
n2 in dom (f | i2)
by A6, A7, FINSEQ_3:27;
A10:
n1 in dom (f | i2)
by A4, A5, FINSEQ_3:27;
now per cases
( n1 < n2 or n1 = n2 or n2 < n1 )
by XXREAL_0:1;
case A11:
n1 < n2
;
contradiction
n2 + 1
<= i2 + 1
by A3, A7, XREAL_1:8;
then
n2 + 1
<= len f
by A2, XXREAL_0:2;
then A12:
n2 < len f
by NAT_1:13;
len f > 4
by GOBOARD7:36;
then A13:
f /. n1 <> f /. n2
by A4, A11, A12, GOBOARD7:37;
A14:
(f | i2) /. n1 = f /. n1
by A10, FINSEQ_4:85;
A15:
(f | i2) /. n2 = f /. n2
by A9, FINSEQ_4:85;
(f | i2) /. n1 = (f | i2) . n1
by A4, A5, FINSEQ_4:24;
hence
contradiction
by A6, A7, A8, A13, A14, A15, FINSEQ_4:24;
verum end; case A16:
n2 < n1
;
contradiction
n1 + 1
<= i2 + 1
by A3, A5, XREAL_1:8;
then
n1 + 1
<= len f
by A2, XXREAL_0:2;
then A17:
n1 < len f
by NAT_1:13;
len f > 4
by GOBOARD7:36;
then A18:
f /. n2 <> f /. n1
by A6, A16, A17, GOBOARD7:37;
A19:
(f | i2) /. n2 = f /. n2
by A9, FINSEQ_4:85;
A20:
(f | i2) /. n1 = f /. n1
by A10, FINSEQ_4:85;
(f | i2) /. n2 = (f | i2) . n2
by A6, A7, FINSEQ_4:24;
hence
contradiction
by A4, A5, A8, A18, A19, A20, FINSEQ_4:24;
verum end; end; end;
hence
n1 = n2
;
verum
end;
then A21:
f | i2 is one-to-one
by Th48;
for i, j being Nat st i + 1 < j holds
LSeg (f | i2),i misses LSeg (f | i2),j
proof
let i,
j be
Nat;
( i + 1 < j implies LSeg (f | i2),i misses LSeg (f | i2),j )
assume A22:
i + 1
< j
;
LSeg (f | i2),i misses LSeg (f | i2),j
A23:
j in NAT
by ORDINAL1:def 13;
A24:
i in NAT
by ORDINAL1:def 13;
now per cases
( ( 1 <= i & j + 1 <= len (f | i2) ) or 1 > i or j + 1 > len (f | i2) )
;
case A25:
( 1
<= i &
j + 1
<= len (f | i2) )
;
LSeg (f | i2),i misses LSeg (f | i2),j
i2 <= len f
by A2, NAT_1:13;
then
len (f | i2) = i2
by FINSEQ_1:80;
then
j + 1
< i2 + 1
by A25, NAT_1:13;
then
j + 1
< len f
by A2, XXREAL_0:2;
then A26:
LSeg f,
i misses LSeg f,
j
by A22, A24, A23, GOBOARD5:def 4;
A27:
j <= len (f | i2)
by A25, NAT_1:13;
LSeg (f | i2),
j = LSeg f,
j
by A25, SPPOL_2:3;
then
(LSeg (f | i2),i) /\ (LSeg (f | i2),j) = (LSeg f,i) /\ (LSeg f,j)
by A22, A27, SPPOL_2:3, XXREAL_0:2;
then
(LSeg (f | i2),i) /\ (LSeg (f | i2),j) = {}
by A26, XBOOLE_0:def 7;
hence
LSeg (f | i2),
i misses LSeg (f | i2),
j
by XBOOLE_0:def 7;
verum end; end; end;
hence
LSeg (f | i2),
i misses LSeg (f | i2),
j
;
verum
end;
then A29:
f | i2 is s.n.c.
by TOPREAL1:def 9;
1 + 1 <= len (f | i2)
by A1, A3, NAT_1:13;
hence
f | i2 is being_S-Seq
by A29, A21, TOPREAL1:def 10; verum