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 Th84:
  X is weakly-positive-implicative iff for x,y being Element of X
  holds x\y=((x\y)\y)\y`
proof
  thus X is weakly-positive-implicative implies for x,y being Element of X
  holds x\y=((x\y)\y)\y`
  proof
    assume
A1: X is weakly-positive-implicative;
    let x,y be Element of X;
    (x\0.X)\y=((x\y)\y)\y` by A1;
    hence thesis by Th2;
  end;
  assume
A2: for x,y being Element of X holds x\y=((x\y)\y)\y`;
  for x,y,z being Element of X holds (x\y)\z=((x\z)\z)\(y\z)
  proof
    let x,y,z be Element of X;
    ((x\z)\(y\z))\(x\y)=0.X by Def3;
    then (((x\z)\(y\z))\z)\((x\y)\z)=0.X by Th4;
    then
A3: (((x\z)\z)\(y\z))\((x\y)\z)=0.X by Th7;
    (((x\z)\z)\(((x\z)\z)\(y\z)))\(y\z)=0.X by Th1;
    then ((((x\z)\z)\(((x\z)\z)\(y\z)))\z`)\((y\z)\z`)=0.X by Th4;
    then (((((x\z)\z)\(((x\z)\z)\(y\z)))\z`)\y)\(((y\z)\z`)\y) =0.X by Th4;
    then (((((x\z)\z)\(((x\z)\z)\(y\z)))\z`)\y)\(((y\z)\z`)\(y\0.X))=0.X by Th2
;
    then
A4: (((((x\z)\z)\(((x\z)\z)\(y\z)))\z`)\y)\0.X=0.X by Def3;
    ((x\y)\z)\(((x\z)\z)\(y\z)) =((x\z)\y)\(((x\z)\z)\(y\z)) by Th7
      .=((((x\z)\z)\z`)\y)\(((x\z)\z)\(y\z)) by A2
      .=((((x\z)\z)\z`)\(((x\z)\z)\(y\z)))\y by Th7
      .=(((x\z)\z)\(((x\z)\z)\(y\z))\z`)\y by Th7;
    then ((x\y)\z)\(((x\z)\z)\(y\z))=0.X by A4,Th2;
    hence thesis by A3,Def7;
  end;
  hence thesis;
end;
