reserve x for set,
  t,t1,t2 for DecoratedTree;
reserve C for set;
reserve X,Y for non empty constituted-DTrees set;
reserve T for DecoratedTree,
  p for FinSequence of NAT;
reserve T for finite-branching DecoratedTree,
  t for Element of dom T,
  x for FinSequence,
  n, m for Nat;
reserve x, x9 for Element of dom T,
  y9 for set;
reserve n,k1,k2,l,k,m for Nat,
  x,y for set;

theorem Th46:
  for T being finite-branching Tree, n being Nat holds
  T-level n is finite
proof
  let T be finite-branching Tree;
  defpred Q[Nat] means T-level $1 is finite;
A1: for n st Q[n] holds Q[n+1]
  proof
    let n such that
A2: T-level n is finite;
    set X = { succ w where w is Element of T : len w = n };
A3: X is finite
    proof
      defpred P[object,object] means
         ex w be Element of T st $1=w & $2=succ w;
A4:   for x being object st x in T-level n ex y being object st P[x,y]
      proof
        let x be object;
        assume x in T-level n;
        then consider w being Element of T such that
A5:     w = x;
        consider y such that
A6:     y = succ w;
        take y,w;
        thus thesis by A5,A6;
      end;
      consider f being Function such that
A7:   dom f = T-level n &
     for x being object st x in T-level n holds P[x,f.x] from
      CLASSES1:sch 1(A4);
A8:   X c= rng f
      proof
        let x be object;
        assume x in X;
        then consider w being Element of T such that
A9:     x = succ w and
A10:    len w = n;
A11:    w in T-level n by A10;
        then ex w9 being Element of T st w = w9 & f.w = succ w9 by A7;
        hence thesis by A7,A9,A11,FUNCT_1:def 3;
      end;
      card rng f c= card dom f by CARD_1:12;
      then rng f is finite by A2,A7;
      hence thesis by A8;
    end;
A12: for Y being set st Y in X holds Y is finite
    proof
      let Y be set;
      assume Y in X;
      then ex w being Element of T st Y = succ w & len w = n;
      hence thesis;
    end;
    T-level (n+1) = union { succ w where w is Element of T : len w = n }
    by Th45;
    hence thesis by A3,A12,FINSET_1:7;
  end;
A13: Q[0] by Th44;
  thus for n holds Q[n] from NAT_1:sch 2(A13,A1);
end;
