reserve x,y,z for object, X,Y for set,
  i,k,n for Nat,
  p,q,r,s for FinSequence,
  w for FinSequence of NAT,
  f for Function;

theorem
  for p,q being Tree-yielding FinSequence st tree(p) = tree(q) holds p = q
proof
  let p,q be Tree-yielding FinSequence such that
A1: tree(p) = tree(q);
A2: tree(p)-level 1 = {<*n*>: n < len p} by Th49;
A3: tree(q)-level 1 = {<*k*>: k < len q} by Th49;
A4: now
    let n1,n2 be Element of NAT;
    assume {<*i*>: i < n1} = {<*k*>: k < n2} & n1 < n2;
    then <*n1*> in {<*i*>: i < n1};
    then
A5: ex i st ( <*n1*> = <*i*>)&( i < n1);
    <*n1*>.1 = n1;
    hence contradiction by A5,FINSEQ_1:40;
  end;
  then
A6: not len p < len q by A1,A2,A3;
A7: not len p > len q by A1,A2,A3,A4;
  then
A8: len p = len q by A6,XXREAL_0:1;
  now
    let i be Nat;
    assume that
A9: i >= 1 and
A10: i <= len p;
    consider k be Nat such that
A11: i = 1+k by A9,NAT_1:10;
    reconsider k as Element of NAT by ORDINAL1:def 12;
A12: k < len p by A10,A11,NAT_1:13;
    then p.i = (tree(p))|<*k*> by A11,Th49;
    hence p.i = q.i by A1,A8,A11,A12,Th49;
  end;
  hence thesis by A6,A7,FINSEQ_1:14,XXREAL_0:1;
end;
