reserve X for Banach_Algebra,
  w,z,z1,z2 for Element of X,
  k,l,m,n,n1,n2 for Nat,
  seq,seq1,seq2,s,s9 for sequence of X,
  rseq for Real_Sequence;

theorem Th14:
  z rExpSeq.(n+1) = (1/(n+1) * z) * (z rExpSeq.n) & z rExpSeq.0=1.
  X & ||.(z rExpSeq).n.|| <= (||.z.|| rExpSeq ).n
proof
  defpred X[Nat] means
   ||. ((z rExpSeq)).$1 .|| <=( ||.z.|| rExpSeq).$1;
A1: (z rExpSeq).0=1/(0!) * z #N 0 by Def2
    .=1/1 *1.X by LOPBAN_3:def 9,NEWTON:12
    .=1.X by RLVECT_1:def 8;
A2: now
    let n be Nat;
    thus (z rExpSeq).(n+1)=1/((n+1) !)*z #N (n+1) by Def2
      .=1/((n+1)!)*((z GeoSeq).n * z) by LOPBAN_3:def 9
      .=1/(n! * (n+1))*(z #N n * z) by NEWTON:15
      .=((1*1)/((n!) * (n+1)))*(z* z #N n ) by Lm1
      .=((1/(n!)) * (1/(n+1)) )*(z* z #N n ) by XCMPLX_1:76
      .=( (1/(n+1))*z)*((1/(n!)) * z #N n ) by LOPBAN_3:38
      .=((1 /(n+1)) *z)*(z rExpSeq.n) by Def2;
  end;
A3: for n be Nat st X[n] holds X[n+1]
  proof
    let n such that
A4: X[n];
    0<= ||. (1/(n+1) *z) .|| by NORMSP_1:4;
    then
A5: ||. (1/(n+1) *z) .|| * ||. (z rExpSeq.n ) .|| <= ||. (1/(n+1) *z) .||
    * (||.z.|| rExpSeq ).n by A4,XREAL_1:64;
A6: ||. (1/(n+1) *z)* (z rExpSeq.n ) .|| <= ||. (1/(n+1) *z) .|| * ||. (z
    rExpSeq.n ) .|| by LOPBAN_3:38;
A7: 1/(n+1) * ||.z.|| * (||.z.|| rExpSeq ).n = 1/(n+1) * ( (||.z.||
    rExpSeq ).n * ||.z.|| )
      .= 1/(n+1) * ( ((||.z.||)|^n)/(n!) *||.z.|| ) by SIN_COS:def 5
      .=1/(n+1) * ( ((||.z.||)|^n)*||.z.|| /(n!) ) by XCMPLX_1:74
      .=1/(n+1) * ( (||.z.||)|^(n+1) /(n!) ) by NEWTON:6
      .= ((||.z.||)|^ (n+1)) /((n)!*(n+1)) by XCMPLX_1:103
      .= ((||.z.||)|^ (n+1)) /((n+1)!) by NEWTON:15
      .=( ||.z.|| rExpSeq ).(n+1) by SIN_COS:def 5;
A8: ||. (1/(n+1) *z) .|| =|.1/(n+1).| * ||.z.|| by NORMSP_1:def 1
      .= 1/(n+1) * ||.z.|| by ABSVALUE:def 1;
    ||. ((z rExpSeq)).(n+1) .|| =||. (1/(n+1) *z)* (z rExpSeq.n ) .|| by A2;
    hence thesis by A6,A8,A5,A7,XXREAL_0:2;
  end;
  ( ||.z.|| rExpSeq ).0 = ((||.z.||) |^ 0) / (0!) by SIN_COS:def 5
    .= 1/(0!) by NEWTON:4
    .= 1/((Prod_real_n).0) by SIN_COS:def 3
    .= 1/1 by SIN_COS:def 2
    .= 1;
  then
A9: X[0] by A1,LOPBAN_3:38;
  for n holds X[n] from NAT_1:sch 2(A9,A3);
  hence thesis by A2,A1;
end;
