
theorem Th22:
  for A being Universal_Algebra for B1,B2 being Subset of A st B1 c= B2
  for n being Nat holds B1|^n c= B2|^n
proof
  let A be Universal_Algebra;
  let B1,B2 be Subset of A such that
A1: B1 c= B2;
  defpred P[Nat] means B1|^$1 c= B2|^$1;
  B1|^0 = B1 by Th18;
  then
A2: P[0] by A1,Th18;
A3: now
    let n be Nat;
    assume
A4: P[n];
    thus P[n+1]
    proof
      let x be object;
      assume that
A5:   x in B1|^(n+1) and
A6:   x nin B2|^(n+1);
      reconsider a = x as Element of A by A5;
      a nin B1|^n by A4,A6,Th20;
      then consider o being (Element of dom the charact of A),
      p being Element of (the carrier of A)* such that
A7:   a = Den(o,A).p and
A8:   p in dom Den(o,A) and
A9:   rng p c= B1|^n by A5,Th20;
      rng p c= B2|^n by A4,A9,XBOOLE_1:1;
      hence contradiction by A6,A7,A8,Th20;
    end;
  end;
  for n being Nat holds P[n] from NAT_1:sch 2(A2,A3);
  hence thesis;
end;
