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 Th10:
  k in Segm n implies k<=n-1 & n-1 is Element of NAT
proof
  assume k in Segm n;
  then k in {l where l is Nat:l<n} by AXIOMS:4;
  then
A1: ex l be Nat st k=l & l<n;
  then
A2: n-1 is Element of NAT by NAT_1:20;
  k<(n-1)+1 by A1;
  hence thesis by A2,NAT_1:13;
end;
