let k be natural number ; :: thesis: il. SCM ,k = k
deffunc H₁( Element of NAT ) -> Element of NAT = $1;
A1: for k being Element of NAT holds H₁(k) is Element of NAT ;
consider f being Function of NAT ,NAT such that
A2: for k being Element of NAT holds f . k = H₁(k) from FUNCT_2:sch 9(A1);
reconsider f = f as Function of NAT ,NAT ;
X: k in NAT by ORDINAL1:def 13;
A3: ( f is bijective & ( for k being Element of NAT holds
( f . (k + 1) in SUCC (f . k),SCM & ( for j being Element of NAT st f . j in SUCC (f . k),SCM holds
k <= j ) ) ) ) by A2, Th53;
ex f being Function of NAT ,NAT st
( f is bijective & ( for m, n being Element of NAT holds
( m <= n iff f . m <= f . n, SCM ) ) & k = f . k )

proof

take f ; :: thesis: ( f is bijective & ( for m, n being Element of NAT holds
( m <= n iff f . m <= f . n, SCM ) ) & k = f . k )
thus f is bijective by A2, Th53; :: thesis: ( ( for m, n being Element of NAT holds
( m <= n iff f . m <= f . n, SCM ) ) & k = f . k )
thus for m, n being Element of NAT holds
( m <= n iff f . m <= f . n, SCM ) by A3, AMISTD_1:18; :: thesis: k = f . k
k in NAT by ORDINAL1:def 13;
hence k = f . k by A2; :: thesis: verum

end;

hence il. SCM ,k = k by X, AMISTD_1:def 12; :: thesis: verum