theorem Th20:
  p => q = r => s implies p = r & s = q
proof
  assume p => q = r => s;
  then <*1*>^(p^q) = r => s by FINSEQ_1:32
    .= <*1*>^(r^s) by FINSEQ_1:32;
  then p^q = r^s by Th2;
  hence thesis by Th18;
end;
