let z be natural number ; for N being non empty with_non-empty_elements set holds il. (STC N),z = z
let N be non empty with_non-empty_elements set ; il. (STC N),z = z
set M = STC N;
reconsider f2 = id NAT as Function of NAT ,NAT ;
consider f being Function of NAT ,NAT such that
A1:
( f is bijective & ( for m, n being Element of NAT holds
( m <= n iff f . m <= f . n, STC N ) ) )
and
A2:
il. (STC N),z = f . z
by Def12;
now let k be
Element of
NAT ;
( f2 . (k + 1) in SUCC (f2 . k),(STC N) & ( for j being Element of NAT st f2 . j in SUCC (f2 . k),(STC N) holds
k <= j ) )reconsider fk =
f2 . k as
Element of
NAT ;
A3:
SUCC fk,
(STC N) = {k,(k + 1)}
by Th24, FUNCT_1:35;
f2 . (k + 1) = k + 1
by FUNCT_1:35;
hence
f2 . (k + 1) in SUCC (f2 . k),
(STC N)
by A3, TARSKI:def 2;
for j being Element of NAT st f2 . j in SUCC (f2 . k),(STC N) holds
k <= jlet j be
Element of
NAT ;
( f2 . j in SUCC (f2 . k),(STC N) implies k <= j )A4:
j = f2 . j
by FUNCT_1:35;
assume
f2 . j in SUCC (f2 . k),
(STC N)
;
k <= jthen
(
j = k or
j = k + 1 )
by A3, A4, TARSKI:def 2;
hence
k <= j
by NAT_1:11;
verum end;
then
for m, n being Element of NAT holds
( m <= n iff f2 . m <= f2 . n, STC N )
by Th18;
then
( z is Element of NAT & f = f2 )
by A1, Th17, ORDINAL1:def 13;
hence
il. (STC N),z = z
by A2, FUNCT_1:35; verum