reserve X for BCK-algebra;
reserve x,y for Element of X;
reserve IT for non empty Subset of X;

theorem
  X is BCK-implicative BCK-algebra iff for x,y,z being Element of X
  holds x\(x\(y\z)) = (y\z)\((y\z)\(x\z))
proof
  thus X is BCK-implicative BCK-algebra implies for x,y,z being Element of X
  holds x\(x\(y\z)) = (y\z)\((y\z)\(x\z))
  proof
    assume
A1: X is BCK-implicative BCK-algebra;
    then
A2: X is BCK-positive-implicative BCK-algebra by Th34;
    for x,y,z being Element of X holds x\(x\(y\z)) = (y\z)\((y\z)\(x\z))
    proof
      let x,y,z be Element of X;
      (x\0.X)\(x\(y\z)) = ((y\z)\0.X)\(((y\z)\x)\0.X) by A1,Th40;
      then x\(x\(y\z)) = ((y\z)\0.X)\(((y\z)\x)\0.X) by BCIALG_1:2
        .= (y\z)\(((y\z)\x)\0.X) by BCIALG_1:2;
      then x\(x\(y\z)) = (y\z)\((y\z)\x) by BCIALG_1:2
        .= (y\z)\((y\x)\z) by BCIALG_1:7;
      hence thesis by A2,Def11;
    end;
    hence thesis;
  end;
  assume
A3: for x,y,z being Element of X holds x\(x\(y\z)) = (y\z)\((y\z)\(x\z));
  for x,y being Element of X holds x\(y\x)=x
  proof
    let x,y be Element of X;
    x\(x\(y\x)) = (y\x)\((y\x)\(x\x)) by A3;
    then x\(x\(y\x)) = (y\x)\((y\x)\0.X) by BCIALG_1:def 5;
    then x\(x\(y\x)) = (y\x)\(y\x) by BCIALG_1:2;
    then
A4: x\(x\(y\x)) = 0.X by BCIALG_1:def 5;
    (x\(y\x))\x = (x\x)\(y\x) by BCIALG_1:7
      .= (y\x)` by BCIALG_1:def 5
      .= 0.X by BCIALG_1:def 8;
    hence thesis by A4,BCIALG_1:def 7;
  end;
  hence thesis by Def12;
end;
