theorem Th13: tau1.q c= tau1.(p '&&' q)
  proof
    set pq = p '&&' q,np = 'not' p,nq = 'not' q;
A1: tau1.(p => nq) c= tau1.('not' (p => nq)) by Th12;
    tau1.(p => nq) = {p => nq} \/ tau1.p \/ tau1.nq by Def4;
    then A2: tau1.nq c= tau1.(p => nq) by XBOOLE_1:7;
    tau1.q c= tau1.nq by Th12;
    then tau1.q c= tau1.(p => nq) by A2;
    hence thesis by A1;
  end;
