reserve i for Element of NAT;

theorem
  for G being strict solvable Group, H being strict Subgroup of G
  holds H is solvable
proof
  let G be strict solvable Group;
  let H be strict Subgroup of G;
  consider F being FinSequence of Subgroups G such that
A1: len F>0 and
A2: F.1=(Omega).G and
A3: F.(len F)=(1).G and
A4: for i st i in dom F & i+1 in dom F for G1,G2 being strict Subgroup
  of G st G1=F.i & G2=F.(i+1) holds G2 is strict normal Subgroup of G1 & for N
  being normal Subgroup of G1 st N=G2 holds G1./.N is commutative by Def1;
  defpred P[set,set] means ex I being strict Subgroup of G st I=F.$1 & $2=I /\
  H;
A5: for k be Nat st k in Seg len F ex x being Element of Subgroups H st P[k, x]
  proof
    let k be Nat;
    assume k in Seg len F;
    then k in dom F by FINSEQ_1:def 3;
    then F.k in Subgroups G by FINSEQ_2:11;
    then reconsider I=F.k as strict Subgroup of G by GROUP_3:def 1;
    reconsider x=I /\ H as strict Subgroup of H by GROUP_2:88;
    reconsider y=x as Element of Subgroups H by GROUP_3:def 1;
    take y;
    take I;
    thus thesis;
  end;
  consider R being FinSequence of Subgroups H such that
A6: dom R=Seg len F & for i be Nat st i in Seg len F holds P[i,R.i] from
  FINSEQ_1:sch 5(A5);
A7: for i st i in dom R & i+1 in dom R for H1,H2 being strict Subgroup of H
  st H1=R.i & H2=R.(i+1) holds H2 is strict normal Subgroup of H1 & for N being
  normal Subgroup of H1 st N=H2 holds H1./.N is commutative
  proof
    let i;
    assume that
A8: i in dom R and
A9: i+1 in dom R;
    consider J being strict Subgroup of G such that
A10: J=F.(i+1) and
A11: R.(i+1)=J /\ H by A6,A9;
    consider I being strict Subgroup of G such that
A12: I=F.i and
A13: R.i=I /\ H by A6,A8;
    let H1,H2 be strict Subgroup of H;
    assume that
A14: H1=R.i and
A15: H2=R.(i+1);
A16: i in dom F & i+1 in dom F by A6,A8,A9,FINSEQ_1:def 3;
    then reconsider J1=J as strict normal Subgroup of I by A4,A12,A10;
A17: for N being strict normal Subgroup of H1 st N=H2 holds H1./.N is
    commutative
    proof
      let N be strict normal Subgroup of H1;
      assume N=H2;
      then consider G3 being Subgroup of I./.J1 such that
A18:  H1 ./. N, G3 are_isomorphic by A14,A15,A13,A11,Th4;
      consider h being Homomorphism of H1./. N,G3 such that
A19:  h is bijective by A18,GROUP_6:def 11;
A20:  h is one-to-one by A19,FUNCT_2:def 4;
A21:  I ./.J1 is commutative by A4,A12,A10,A16;
      now
        let a,b be Element of H1./.N;
        consider a9 being Element of G3 such that
A22:    a9=h.a;
        consider b9 being Element of G3 such that
A23:    b9=h.b;
        the multF of G3 is commutative by A21,GROUP_3:2;
        then
A24:    a9*b9=b9*a9 by BINOP_1:def 2;
        thus (the multF of H1./.N).(a,b)=h".(h.(a*b)) by A20,FUNCT_2:26
          .=h".(h.b*h.a) by A22,A23,A24,GROUP_6:def 6
          .=h".(h.(b*a)) by GROUP_6:def 6
          .=(the multF of H1./.N).(b,a) by A20,FUNCT_2:26;
      end;
      then the multF of (H1./.N) is commutative by BINOP_1:def 2;
      hence thesis by GROUP_3:2;
    end;
    H2=J1/\H by A15,A11;
    hence thesis by A14,A13,A17,Th1;
  end;
A25: len R=len F by A6,FINSEQ_1:def 3;
A26: len R >0 by A1,A6,FINSEQ_1:def 3;
A27: R.1=(Omega).H
  proof
    1<=len R by A26,NAT_1:14;
    then 1 in Seg len F by A25,FINSEQ_1:1;
    then ex I being strict Subgroup of G st I=F.1 & R.1=I /\ H by A6;
    hence thesis by A2,GROUP_2:86;
  end;
  take R;
  R.(len R)= (1).H
  proof
    ex I being strict Subgroup of G st I=F.(len R) & R.(len R )=I /\ H by A1,A6
,A25,FINSEQ_1:3;
    hence R.(len R)=(1).G by A3,A25,GROUP_2:85
      .=(1).H by GROUP_2:63;
  end;
  hence thesis by A1,A6,A27,A7,FINSEQ_1:def 3;
end;
