reserve A,B,p,q,r,s for Element of LTLB_WFF,
  n for Element of NAT,
  X for Subset of LTLB_WFF,
  g for Function of LTLB_WFF,BOOLEAN,
  x,y for set;

theorem X is without_implication implies tau X = X
  proof
    assume
A1: X is without_implication;
A2: tau X c= X
    proof
      let x be object;
      assume x in tau X;
      then consider p such that
A3:   p in X and
A4:   x in tau1.p by Def5;
      x in {p} by A3,A1,Th5,A4;
      hence thesis by A3,TARSKI:def 1;
    end;
    X c= tau X by Th16;
    hence thesis by A2;
  end;
