
theorem Th19:
  for A being Universal_Algebra, B being Subset of A for n being Nat
  holds B|^(n+1) = (B|^n) \/ {Den(o,A).p
  where o is (Element of dom the charact of A),
  p is Element of (the carrier of A)*: p in dom Den(o,A) & rng p c= B|^n}
proof
  let A be Universal_Algebra;
  let B be Subset of A;
  let n be Nat;
  deffunc Rec(set,set) = $2 \/
  {Den(o,A).p where o is (Element of dom the charact of A),
  p is Element of (the carrier of A)*: p in dom Den(o,A) & rng p c= $2};
  consider F1 being sequence of  bool the carrier of A such that
A1: B|^(n) = F1.(n) and
A2: F1.0 = B and
A3: for n being Nat holds F1.(n+1) = Rec(n,F1.n) by Def8;
  consider F2 being sequence of  bool the carrier of A such that
A4: B|^(n+1) = F2.(n+1) and
A5: F2.0 = B and
A6: for n being Nat holds F2.(n+1) = Rec(n,F2.n) by Def8;
A7: dom F1 = NAT by FUNCT_2:def 1;
A8: F1.0 = B by A2;
A9: for n being Nat holds F1.(n+1) = Rec(n,F1.n) by A3;
A10: dom F2 = NAT by FUNCT_2:def 1;
A11: F2.0 = B by A5;
A12: for n being Nat holds F2.(n+1) = Rec(n,F2.n) by A6;
  F1 = F2 from NAT_1:sch 15(A7,A8,A9,A10,A11,A12);
  hence thesis by A1,A3,A4;
end;
