theorem Th5:
  ASeq.n c= BSeq.n implies (P * ASeq).n <= (P * BSeq).n
proof
A1: n in NAT by ORDINAL1:def 12;
  assume ASeq.n c= BSeq.n;
  then P.(ASeq.n) <= P.(BSeq.n) by PROB_1:34;
  then (P * ASeq).n <= P.(BSeq.n) by A1,FUNCT_2:15;
  hence thesis by A1,FUNCT_2:15;
end;
