reserve k, l, m, n, i, j for Nat,
  K, N for non empty Subset of NAT,
  Ke, Ne, Me for Subset of NAT,
  X,Y for set;

theorem
  N c= n implies n-1 is Element of NAT
proof
A1: min* N in N by NAT_1:def 1;
  assume N c= n;
  then min*N in n by A1;
  then min* N in {l where l is Nat:l<n} by AXIOMS:4;
  then ex l be Nat st min* N = l & l<n;
  hence thesis by NAT_1:20;
end;
