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 being Element of X holds
  (x\(x\y))\(x\y) = y\(y\(x\(x\y)))
proof
  thus X is BCK-implicative BCK-algebra implies for x,y being Element of X
  holds (x\(x\y))\(x\y) = y\(y\(x\(x\y)))
  proof
    assume
A1: X is BCK-implicative BCK-algebra;
    let x,y be Element of X;
    X is commutative BCK-algebra by A1,Th34;
    then y\(y\(x\(x\y))) = y\(y\(y\(y\x))) by Def1
      .= y\(y\x) by BCIALG_1:8;
    hence thesis by A1,Th35;
  end;
  assume
A2: for x,y being Element of X holds (x\(x\y))\(x\y) = y\(y\(x\(x\y)));
A3: for x,y being Element of X holds (x\y)\y = x\y
  proof
    let x,y be Element of X;
A4: (x\y)\((x\y)\(x\(x\(x\y)))) = (x\y)\((x\y)\(x\y)) by BCIALG_1:8
      .= (x\y)\0.X by BCIALG_1:def 5
      .= x\y by BCIALG_1:2;
    (x\(x\(x\y)))\(x\(x\y)) = (x\y)\(x\(x\y)) by BCIALG_1:8
      .= (x\(x\(x\y)))\y by BCIALG_1:7
      .= (x\y)\y by BCIALG_1:8;
    hence thesis by A2,A4;
  end;
  for x,y being Element of X holds x\(x\y) = y\(y\(x\(x\y)))
  proof
    let x,y be Element of X;
    x\(x\y) = (x\(x\y))\(x\y) by A3;
    hence thesis by A2;
  end;
  then
A5: X is commutative BCK-algebra by Th4;
  for x,y being Element of X holds (x\(x\y))\(x\y) = (y\(y\x))
  proof
    let x,y be Element of X;
    (x\(x\y))\(x\y) = y\(y\(x\(x\y))) by A2
      .= y\(y\(y\(y\x))) by A5,Def1;
    hence thesis by BCIALG_1:8;
  end;
  hence thesis by Th35;
end;
