reserve R for commutative Ring;
reserve A for non degenerated commutative Ring;
reserve I,J,q for Ideal of A;
reserve p for prime Ideal of A;
reserve M,M1,M2 for Ideal of A/q;

theorem Th43:
    for I be proper Ideal of A, x be Element of (sqrt I)
    ex m be Nat
    st m in {n where n is Element of NAT: not(x|^n in I)} & x|^(m+1) in I
    proof
      let I be proper Ideal of A, x be Element of sqrt I;
      set C = {n where n is Element of NAT: not(x|^n in I)};
      x in sqrt I; then
      x in {a where a is Element of A: ex n being Element of NAT st
            a|^n in I} by IDEAL_1:def 24; then
      consider x1 be Element of A such that
A1:   x1 = x and
A2:   ex n being Element of NAT st x1|^n in I;
      consider n1 being Element of NAT such that
A3:   x1|^n1 in I by A2;
      reconsider n1 as Element of NAT;
A4:   not(n1 in {n where n is Element of NAT: not(x|^n in I)})
      proof
        assume n1 in {n where n is Element of NAT: not(x|^n in I)}; then
        consider n0 be Element of NAT such that
A6:     n0 = n1 & not(x|^n0 in I);
        thus contradiction by A3,A1,A6;
      end;
      assume
A7:   not(ex m be Nat st m in {n where n is Element of NAT: not(x|^n in I)}
          & x|^(m+1) in I);
A8:   0 in {n where n is Element of NAT: not(x|^n in I)}
      proof
        set n = 0;
        x|^n = 1_A by BINOM:8 .= 1.A; then
        not x|^n in I by IDEAL_1:19;
        hence thesis;
      end;
      {n where n is Element of NAT: not(x|^n in I)} = NAT
      proof
        defpred P[Nat] means
             $1 in {n where n is Element of NAT: not(x|^n in I)};
A12:    for n be Nat st P[n] holds P[n+1]
        proof
          let n be Nat;
          assume P[n]; then
          not (x|^(n+1) in I) by A7;
          hence thesis;
        end;
A15:    P[0] by A8;
        for n be Nat holds P[n] from NAT_1:sch 2(A15,A12); then
     H: NAT c= {n where n is Element of NAT: not(x|^n in I)};
        {n where n is Element of NAT: not(x|^n in I)} c= NAT
           proof
           now let o be object;
             assume o in {n where n is Element of NAT: not(x|^n in I)};
             then consider n be Element of NAT such that
             I: n = o & not(x|^n in I);
             thus o in NAT by I;
             end;
           hence thesis;
           end;
        hence thesis by H;
      end;
      hence contradiction by A4;
    end;
