reserve n,i,k,m for Nat;

theorem
  1 <= k-m & k-m <= n implies k-m in Seg n & k-m is Element of NAT
proof
  assume that
A1: 1<=k-m and
A2: k-m<=n;
  k-m is Element of NAT by A1,INT_1:3;
  hence thesis by A1,A2,FINSEQ_1:1;
end;
