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

theorem
  for t being finite-branching DecoratedTree, p being Node of t,
  q being Node of t|p holds succ(t,p^q) = succ(t|p,q)
proof
  let t be finite-branching DecoratedTree, p be Node of t, q be Node of t|p;
A1: dom (t|p) = (dom t)|p by TREES_2:def 10;
  then reconsider pq = p^q as Element of dom t by TREES_1:def 6;
  reconsider q as Element of dom (t|p);
  dom t = dom t with-replacement (p,(dom t)|p) by TREES_2:6;
  then succ pq,succ q are_equipotent by A1,TREES_2:37;
  then
A2: card succ q = card succ pq by CARD_1:5;
A3: ex r being Element of dom (t|p) st r = q & succ(t|p,q) = (t|p)*(r succ)
  by Def6;
  rng (q succ) c= dom (t|p);
  then
A4: dom succ(t|p,q) = dom (q succ) by A3,RELAT_1:27;
A5: ex q being Element of dom t st q = pq & succ(t,pq) = t*(q succ) by Def6;
  rng (pq succ) c= dom t;
  then
A6: dom succ(t,pq) = dom (pq succ) by A5,RELAT_1:27;
A7: len (q succ) = card succ q by Def5;
A8: len (pq succ) = card succ pq by Def5;
  then
A9: dom (pq succ) = dom (q succ) by A7,A2,FINSEQ_3:29;
  now
    let i be Nat;
    assume
A10: i in dom (q succ);
    then consider k being Nat such that
A11: i = k+1 and
A12: k < len (q succ) by Lm1;
A13: q^<*k*> in succ q by A7,A12,Th7;
     reconsider k as Element of NAT by ORDINAL1:def 12;
    thus succ(t,pq).i = t.((pq succ).i) by A5,A9,A6,A10,FUNCT_1:12
      .= t.(pq^<*k*>) by A8,A7,A2,A11,A12,Def5
      .= t.(p^(q^<*k*>)) by FINSEQ_1:32
      .= (t|p).(q^<*k*>) by A1,A13,TREES_2:def 10
      .= (t|p).((q succ).i) by A11,A12,Def5
      .= succ(t|p,q).i by A3,A4,A10,FUNCT_1:12;
  end;
  hence thesis by A9,A6,A4;
end;
