theorem Th18:
  s1 == s2 & not emp s1 implies top s1 = top s2
  proof
    assume
A1: s1 == s2 & not emp s1; then
A2: |.s1.| = |.s2.| & not emp s2 by Th14;
    thus top s1 = |.s1.|.1 by A1,Th9 .= top s2 by A2,Th9;
  end;
