reserve x,y for set;
reserve i, j, k for Nat;
reserve I,J,K for Element of Segm 9,
  a,a1 for Nat,
  b,b1,c for Element of Data-Locations SCM;
reserve a, b for Data-Location,
  loc for Nat;
reserve I,J,K for Element of Segm 9,
  a,a1 for Nat,
  b,b1,c for Element of Data-Locations SCM,
  da,db for Data-Location;

theorem
  for ins being Instruction of SCM st InsCode ins = 5 holds ex da,
  db st ins = Divide(da,db)
proof
  let ins be Instruction of SCM such that
A1: InsCode ins = 5;
A2: now
    assume ins in { [J,<*a*>,{}] : J = 6 };
    then ex J,a st ins = [J,<*a*>,{}] & J = 6;
    hence contradiction by A1;
  end;
A3: now
    assume ins in { [K,<*a1*>,<*b1*>] : K in { 7,8 } };
    then consider K,a1,b1 such that
A4: ins = [K,<*a1*>,<*b1*>] and
A5: K in { 7,8 };
    InsCode ins = K by A4;
    hence contradiction by A1,A5,TARSKI:def 2;
  end;
  InsCode halt SCM = 0 by COMPOS_1:70;
  then not ins in { [SCM-Halt,{},{}] } by A1,AMI_3:26,TARSKI:def 1;
  then not ins in { [SCM-Halt,{},{}] } \/ { [J,<*a*>,{}] : J = 6 } by A2,
XBOOLE_0:def 3;
  then
  not ins in { [SCM-Halt,{},{}] } \/ { [J,<*a*>,{}] : J = 6 }
   \/ { [K,<*a1*>,<*b1*>] : K in { 7,8 } } by A3,XBOOLE_0:def 3;
  then ins in { [I,{},<*b,c*>] : I in { 1,2,3,4,5} }
   by AMI_3:27,XBOOLE_0:def 3;
  then consider I,b,c such that
A6: ins = [I,{},<*b,c*>] and
  I in { 1,2,3,4,5};
  reconsider da = b ,db = c as Data-Location by Lm1;
  take da,db;
  thus thesis by A1,A6;
end;
