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

theorem Th5:
  X is commutative BCK-algebra iff for x,y being Element of X st x
  <= y holds x= y\(y\x)
proof
  thus X is commutative BCK-algebra implies for x,y being Element of X st x<=
  y holds x= y\(y\x)
  proof
    assume
A1: X is commutative BCK-algebra;
    let x,y be Element of X;
    assume x<= y;
    then x\y = 0.X;
    then y\(y\x) = x\0.X by A1,Def1
      .=x by BCIALG_1:2;
    hence thesis;
  end;
  assume
A2: for x,y being Element of X st x<= y holds x= y\(y\x);
  for x,y being Element of X holds x\(x\y) <= y\(y\x)
  proof
    let x,y be Element of X;
    (x\(x\y))\x = (x\x)\(x\y) by BCIALG_1:7
      .= (x\y)` by BCIALG_1:def 5
      .= 0.X by BCIALG_1:def 8;
    then (x\(x\y)) <= x;
    then
A3: y\x <= y\(x\(x\y)) by BCIALG_1:5;
    (x\(x\y))\y = (x\y)\(x\y) by BCIALG_1:7
      .= 0.X by BCIALG_1:def 5;
    then (x\(x\y)) <= y;
    then (x\(x\y))= y\(y\(x\(x\y))) by A2;
    hence thesis by A3,BCIALG_1:5;
  end;
  hence thesis by Th1;
end;
