theorem Th26:
  p => q <> prop n
proof
A1: now
    assume 1 = 1+2+n;
    then 1+0 = 1+(2+n);
    hence contradiction;
  end;
  p => q = <*1*>^(p^q) by FINSEQ_1:32;
  then (p => q).1 = 1 by FINSEQ_1:41;
  hence thesis by A1;
end;
