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 holds
  x\(y\(y\x))=x\y & (x\y)\((x\y)\x)=x\y
proof
  assume
A1: X is commutative BCK-algebra;
  let x,y be Element of X;
A2: (x\y)\((x\y)\x) = x\(x\(x\y)) by A1,Def1
    .= x\y by BCIALG_1:8;
  x\(y\(y\x)) = x\(x\(x\y)) by A1,Def1
    .= x\y by BCIALG_1:8;
  hence thesis by A2;
end;
