reserve i,j,k,n for Nat;
reserve D for non empty set,
  p for Element of D,
  f,g for FinSequence of D;

theorem Th50:
  p in rng f implies len (f:-p) = len f - p..f + 1
proof
  assume
A1: p in rng f;
  then
A2: p..f <= len f by FINSEQ_4:21;
  consider i being Element of NAT such that
A3: i+1 = p..f and
A4: f:-p = f/^i by A1,Th49;
  i <= p..f by A3,NAT_1:11;
  then i <= len f by A2,XXREAL_0:2;
  hence len(f:-p) = len f - i by A4,RFINSEQ:def 1
    .= len f - p..f + 1 by A3;
end;
