theorem Th14:
  s1 == s2 & emp s1 implies emp s2
  proof
    assume
A1: |.s1.| = |.s2.| & emp s1;
    assume not emp s2; then
    |.s2.| = <*top s2*>^|.pop s2.| by Th6;
    hence thesis by A1,Th5;
  end;
