theorem Th19: p => q in tau X implies p in tau X & q in tau X
  proof
    set t = {tau1.r where r is Element of l: r in X};
    assume p => q in tau X;
    then consider B such that
A1: B in X and
A2: p => q in tau1.B by Def5;
A3: tau1.B in t by A1;
    p in tau1.B by Th7,A2;
    then A4: p in union t by A3,TARSKI:def 4;
    q in tau1.B by Th7,A2;
    then q in union t by TARSKI:def 4,A3;
    hence thesis by A4,Th15;
  end;
