reserve i,j,k,n for Nat;

theorem Th2:
  for i,n being Nat holds i in Seg n implies n - i + 1 in Seg n
proof
  let i,n be Nat;
  assume
A1: i in Seg n;
  then i <= n by FINSEQ_1:1;
  then reconsider ni = n - i as Element of NAT by INT_1:5;
  reconsider j = ni + 1 as Element of NAT;
A2: now
    assume j = n + 1;
    then -0 = -i;
    hence contradiction by A1,FINSEQ_1:1;
  end;
  j <= n + 1 by XREAL_1:6,XREAL_1:43;
  then j < n + 1 by A2,XXREAL_0:1;
  then 0+1 <= j & j <= n by NAT_1:13;
  hence thesis;
end;
