reserve X for BCI-algebra;
reserve x,y,z,u,a,b for Element of X;
reserve IT for non empty Subset of X;

theorem
  for X being BCI-algebra holds X is being_K iff X is BCK-algebra
proof
  let X be BCI-algebra;
  thus X is being_K implies X is BCK-algebra
  proof
    assume
A1: X is being_K;
    now
      let s be Element of X;
      s`\0.X = 0.X by A1;
      hence s` = 0.X by Th2;
    end;
    hence thesis by Def8;
  end;
  assume
A2: X is BCK-algebra;
  let x,y be Element of X;
  y` = 0.X by A2,Def8;
  then (x\y)\(x\0.X) = 0.X by Th4;
  hence thesis by Th2;
end;
