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

theorem Th3:
  X is commutative BCK-algebra iff for x,y being Element of X holds
  x\y = x\(y\(y\x))
proof
  thus X is commutative BCK-algebra implies for x,y being Element of X holds x
  \y = x\(y\(y\x))
  proof
    assume
A1: X is commutative BCK-algebra;
    let x,y be Element of X;
    x\y = x\(x\(x\y)) by BCIALG_1:8
      .= x\(y\(y\x)) by A1,Def1;
    hence thesis;
  end;
  assume
A2: for x,y being Element of X holds x\y = x\(y\(y\x));
  for x,y being Element of X holds x\(x\y) <= y\(y\x)
  proof
    let x,y;
    x\(x\(y\(y\x))) <= y\(y\x) by Th2;
    hence thesis by A2;
  end;
  hence thesis by Th1;
end;
