reserve a, b for Int_position,
  i for Instruction of SCMPDS,
  l for Element of NAT,
  k, k1, k2 for Integer;

theorem Th2:
  (for l being Element of NAT holds NIC(i,l)={l+1}) implies JUMP i is empty
proof
  set p=1, q=2;
  assume
A1: for l being Element of NAT holds NIC(i,l)={l+1};
  set X = the set of all  NIC(i,f) where f is Nat;
  reconsider p, q as Element of NAT;
  assume not thesis;
  then consider x being object such that
A2: x in meet X;
  NIC(i,p) = {p+1} by A1;
  then {p+1} in X;
  then x in {p+1} by A2,SETFAM_1:def 1;
  then
A3: x = p+1 by TARSKI:def 1;
  NIC(i,q) = {q+1} by A1;
  then {q+1} in X;
  then x in {q+1} by A2,SETFAM_1:def 1;
  hence contradiction by A3,TARSKI:def 1;
end;
