reserve x for set,
  D for non empty set,
  k, n for Nat,
  z for Nat;
reserve
  N for with_zero set,
  S for IC-Ins-separated non empty
          with_non-empty_values AMI-Struct over N,
  i for Element of the InstructionsF of S,
  l, l1, l2, l3 for Nat,
  s for State of S;
reserve ss for Element of product the_Values_of S;

theorem Th1:
  for i being Element of the InstructionsF of S
   st for l being Nat holds NIC(i,l)={l}
  holds JUMP i is empty
proof
  let i be Element of the InstructionsF of S;
  set p = 0, q = 1;
  set X = the set of all  NIC(i, l) where l is Nat;
  reconsider p,q as Nat;
  assume
A1: for l being Nat holds NIC(i,l)={l};
  assume not thesis;
  then consider x being object such that
A2: x in meet X;
  NIC(i,p) = {p} by A1;
  then {p} in X;
  then
A3: x in {p} by A2,SETFAM_1:def 1;
  NIC(i,q) = {q} by A1;
  then {q} in X;
  then x in {q} by A2,SETFAM_1:def 1;
  hence contradiction by A3,TARSKI:def 1;
end;
