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

theorem Th4:
  X is commutative BCK-algebra iff for x,y being Element of X holds
  x\(x\y) = y\(y\(x\(x\y)))
proof
  thus X is commutative BCK-algebra implies for x,y being Element of X holds x
  \(x\y) = y\(y\(x\(x\y)))
  proof
    assume
A1: X is commutative BCK-algebra;
    let x,y be Element of X;
    y\(y\x) = y\(y\(x\(x\y))) by A1,Th3;
    hence thesis by A1,Def1;
  end;
  assume
A2: for x,y being Element of X holds x\(x\y) = y\(y\(x\(x\y)));
  for x,y being Element of X holds x\(x\y) <= y\(y\x)
  proof
    let x,y;
    x\(x\y) <= x by Th2;
    then
A3: y\x <= y\(x\(x\y)) by BCIALG_1:5;
    x\(x\y) = y\(y\(x\(x\y))) by A2;
    hence thesis by A3,BCIALG_1:5;
  end;
  hence thesis by Th1;
end;
