theorem Th30:
  for O1,O2 being Operation of X st for L holds L|O1 = L|O2 holds O1 = O2
  proof
    let O1,O2 be Operation of X;
    assume
A1: for L holds L|O1 = L|O2;
    now let x;
      per cases;
      suppose X <> {}; then
        reconsider L = {x} as Subset of X by ZFMISC_1:31;
        thus x.O1 = L|O1 .= x.O2 by A1;
      end;
      suppose X = {};
        hence x.O1 = x.O2;
      end;
    end;
    hence thesis by Th29;
  end;
