theorem Th11:
  0 < m implies len (Sgm (Seg n \/ {n+m})) = n+1
proof
A1: m <= n+m by NAT_1:11;
  assume
A2: 0 < m;
  then card (Seg n \/ {n+m}) = card (Seg n) + 1 by CARD_2:41,FINSEQ_3:10;
  then
A3: card (Seg n \/ {n+m}) = n + 1 by FINSEQ_1:57;
  0+1 <= m by A2,NAT_1:13;
  then 1 <= m+n by A1,XXREAL_0:2;
  then n+m in Seg (n+m) by FINSEQ_1:1;
  then {n+m} c= Seg (n+m) by ZFMISC_1:31;
  then {n+m} is included_in_Seg by FINSEQ_1:def 13;
  hence thesis by A3,FINSEQ_3:39;
end;
