theorem Th3:
  X is top-push pop-push & push(e1,s1) = push(e2,s2) implies e1 = e2 & s1 = s2
  proof assume X is top-push; then
A1: e1 = top push(e1,s1) & e2 = top push(e2,s2);
    assume X is pop-push; then
    s1 = pop push(e1,s1) & s2 = pop push(e2,s2);
    hence thesis by A1;
  end;
