reserve x,y,x1,x2,z for set,
  n,m,k for Nat,
  t1 for (DecoratedTree of [: NAT,NAT :]),
  w,s,t,u for FinSequence of NAT,
  D for non empty set;
reserve s9,w9,v9 for Element of NAT*;

theorem Th15:
  for Z being Tree,o being Element of Z st o <> Root Z holds Z|o,{
  o^s9: o^s9 in Z } are_equipotent & not Root Z in { o^w9 : o^w9 in Z }
proof
  let Z be Tree,o be Element of Z such that
A1: o <> Root Z;
  set A = { o^s9 : o^s9 in Z };
  thus Z|o,A are_equipotent
  proof
    defpred P[object,object] means for s st $1 = s holds $2 = o^s;
A2: for x being object st x in Z|o ex y being object st P[x,y]
    proof
      let x be object;
      assume x in Z|o;
      then reconsider s = x as FinSequence of NAT by TREES_1:19;
      take o^s;
      let w;
      assume x = w;
      hence thesis;
    end;
    ex f being Function st dom f = Z|o &
for x being object st x in Z|o holds P[x,f.x]
    from CLASSES1:sch 1(A2);
    then consider f being Function such that
A3: dom f = Z|o and
A4: for x being object st x in Z|o for s st x = s holds f.x = o^s;
    now
      let x be object;
      thus x in rng f implies x in A
      proof
        assume x in rng f;
        then consider x1 being object such that
A5:     x1 in dom f and
A6:     x = f.x1 by FUNCT_1:def 3;
        reconsider x1 as FinSequence of NAT by A3,A5,TREES_1:19;
        reconsider x1 as Element of NAT* by FINSEQ_1:def 11;
        o^x1 in Z & x = o^x1 by A3,A4,A5,A6,TREES_1:def 6;
        hence thesis;
      end;
      assume x in A;
      then consider v9 such that
A7:   x = o^v9 and
A8:   o^v9 in Z;
      v9 in Z|o by A8,TREES_1:def 6;
      then
A9:   x = f.v9 by A4,A7;
      v9 in dom f by A3,A8,TREES_1:def 6;
      hence x in rng f by A9,FUNCT_1:def 3;
    end;
    then
A10: rng f = A by TARSKI:2;
    now
      let x1,x2 be object;
      assume that
A11:  x1 in dom f and
A12:  x2 in dom f and
A13:  f.x1 = f.x2;
      reconsider s1 = x1, s2 = x2 as FinSequence of NAT by A3,A11,A12,
TREES_1:19;
      o^s1 = f.x2 by A3,A4,A11,A13
        .= o^s2 by A3,A4,A12;
      hence x1 = x2 by FINSEQ_1:33;
    end;
    then f is one-to-one by FUNCT_1:def 4;
    hence thesis by A3,A10,WELLORD2:def 4;
  end;
  assume not thesis;
  then ex v9 st Root Z = o^v9 & o^v9 in Z;
  hence contradiction by A1;
end;
