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
  p in rng f & f is one-to-one implies f:-p is one-to-one
proof
  assume p in rng f;
  then ex i being Element of NAT st i+1 = p..f & f:-p = f/^i by Th49;
  hence thesis;
end;
