theorem Th17:
  s1 == s2 & not emp s1 implies pop s1 == pop s2
  proof
    assume
A1: s1 == s2 & not emp s1; then
A2: |.s1.| = |.s2.| & not emp s2 by Th14;
    thus |.pop s1.| = Del(|.s1.|,1) by A1,Th7 .= |.pop s2.| by A2,Th7;
  end;
