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 Th34:
  for f being FinSequence st
  f is one-to-one holds rng(f|n) misses rng(f/^n)
proof
  let f be FinSequence;
  assume
A1: f is one-to-one;
A2: dom(f|n) c= dom f by Th18;
  assume rng(f|n) meets rng(f/^n);
  then consider x being object such that
A3: x in rng(f|n) and
A4: x in rng(f/^n) by XBOOLE_0:3;
  consider i being object such that
A5: i in dom(f|n) and
A6: (f|n).i = x by A3,FUNCT_1:def 3;
  consider j being object such that
A7: j in dom(f/^n) and
A8: (f/^n).j = x by A4,FUNCT_1:def 3;
  reconsider ii = i, jj = j as Nat by A7,A5;
A9: jj+n in dom f by A7,Th26;
A10: now
    ii <= len(f|n) & len(f|n) <= n by A5,Th17,FINSEQ_3:25;
    then
A11: ii <= n by XXREAL_0:2;
    assume
A12: ii = jj+n;
    then n <= ii by NAT_1:11;
    then i = n by A11,XXREAL_0:1;
    then j = 0 by A12;
    hence contradiction by A7,FINSEQ_3:25;
  end;
  f.(jj+n) = (f/^n).j by A7,Th27
    .= f.i by A5,FUNCT_1:47,A6,A8;
  hence contradiction by A1,A5,A2,A9,A10;
end;
