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

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