theorem Th14: tau1.q c= tau1.(p 'or' q)
  proof
    set pq = p 'or' q,np = 'not' p,nq = 'not' q,npq = np '&&' nq;
    tau1.nq c= tau1.npq & tau1.q c= tau1.nq by Th13, Th12;
    then A1: tau1.q c= tau1.npq;
    tau1.npq c= tau1.pq by Th12;
    hence thesis by A1;
  end;
