let i, m, n be Nat; ( i in Seg m implies (Sgm ((Seg (n + m)) \ (Seg n))) . i = n + i )
assume A1:
i in Seg m
; (Sgm ((Seg (n + m)) \ (Seg n))) . i = n + i
reconsider N = n as Element of NAT by ORDINAL1:def 13;
set I = idseq m;
A2:
dom (idseq m) = Seg (len (idseq m))
by FINSEQ_1:def 3;
A3:
len (idseq m) = m
by FINSEQ_1:def 18;
A4:
dom (N Shift (idseq m)) = { (N + k) where k is Element of NAT : k in dom (idseq m) }
by PNPROC_1:def 15;
A5:
(Seg (n + m)) \ (Seg n) c= dom (N Shift (idseq m))
proof
let x be
set ;
TARSKI:def 3 ( not x in (Seg (n + m)) \ (Seg n) or x in dom (N Shift (idseq m)) )
assume A6:
x in (Seg (n + m)) \ (Seg n)
;
x in dom (N Shift (idseq m))
reconsider i =
x as
Element of
NAT by A6;
A7:
i in Seg (n + m)
by A6, XBOOLE_0:def 5;
not
i in Seg n
by A6, XBOOLE_0:def 5;
then A8:
(
i < 1 or
i > n )
;
then reconsider IN =
i - n as
Element of
NAT by A7, FINSEQ_1:3, NAT_1:21;
A9:
n + IN = i
;
i <= n + m
by A7, FINSEQ_1:3;
then A10:
IN <= m
by A9, XREAL_1:10;
IN >= 1
by A7, A8, A9, FINSEQ_1:3, NAT_1:19;
then
IN in dom (idseq m)
by A2, A3, A10;
then
n + IN in dom (N Shift (idseq m))
by A4;
hence
x in dom (N Shift (idseq m))
;
verum
end;
dom (N Shift (idseq m)) c= (Seg (n + m)) \ (Seg n)
then
(Seg (n + m)) \ (Seg n) = dom (N Shift (idseq m))
by A5, XBOOLE_0:def 10;
hence
(Sgm ((Seg (n + m)) \ (Seg n))) . i = n + i
by A1, A2, A3, PNPROC_1:59; verum