theorem Th24:
  p '&' q <> prop n
proof
A1: now
    assume 2 = 2+1+n;
    then 2+0 = 2+(1+n);
    hence contradiction;
  end;
  p '&' q = <*2*>^(p^q) by FINSEQ_1:32;
  then (p '&' q).1 = 2 by FINSEQ_1:41;
  hence thesis by A1;
end;
