reserve x for set,
  t,t1,t2 for DecoratedTree;
reserve C for set;
reserve X,Y for non empty constituted-DTrees set;

theorem Th29:
  for p being FinTree-yielding FinSequence, n being empty Element
  of tree p holds card succ n = len p
proof
  let p be FinTree-yielding FinSequence, n be empty Element of tree p;
  assume
A1: not thesis;
  per cases by A1,XXREAL_0:1;
  suppose
A2: card succ n < len p;
    then (card succ n)+1 in dom p by Lm3;
    then reconsider t = p.((card succ n)+1) as finite Tree by TREES_3:23;
A3: n^<*card succ n*> = <*card succ n*> by FINSEQ_1:34;
    n in t & <*card succ n*>^n = <*card succ n*> by FINSEQ_1:34,TREES_1:22;
    then n^<*card succ n*> in tree(p) by A2,A3,TREES_3:def 15;
    then n^<*card succ n*> in succ n;
    hence contradiction by Th7;
  end;
  suppose
    card succ n > len p;
    then n^<*len p*> in succ n by Th7;
    then n^<*len p*> in tree(p);
    then <*len p*> in tree(p) by FINSEQ_1:34;
    then consider i being Nat, q being FinSequence such that
A4: i < len p and
    q in p.(i+1) and
A5: <*len p*> = <*i*>^q by TREES_3:def 15;
    len p = <*len p*>.1
      .= i by A5,FINSEQ_1:41;
    hence contradiction by A4;
  end;
end;
