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

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