reserve i,n,m for Nat,
        r,s for Real,
        A for non empty closed_interval Subset of REAL;

theorem Th10:
  (r >=0 implies alternating_series abs(Leibniz_Series_of r)
      = Leibniz_Series_of r) &
  (r < 0 implies (-1)(#) alternating_series abs(Leibniz_Series_of r)
     = Leibniz_Series_of r)
proof
  set rL=Leibniz_Series_of r, A=abs rL,aA=alternating_series A;
A1:dom A = dom rL & dom rL=NAT by VALUED_1:def 11,FUNCT_2:def 1;
A2: aA.n = (-1)|^n * (|.r.| |^(2*n+1) / (2*n+1))
  proof
    --1 = 1;
    then |.-1.| = 1 by ABSVALUE:def 1;
    then
A3: |.(-1)|^n.| = 1 |^n by TAYLOR_2:1;
A4: A.n = |.rL.n.| by A1,VALUED_1:def 11,ORDINAL1:def 12;
    rL.n = (-1)|^n * (r|^(2*n+1))/ (2*n+1) by Def2;
    then  A.n = |.(-1)|^n *(r|^(2*n+1)).|/ |.(2*n+1).|
       by COMPLEX1:67,A4
             .= 1*|.(r|^(2*n+1)).|/ |.(2*n+1).| by A3,COMPLEX1:65
             .= 1*(|.r.| |^(2*n+1)) / |.(2*n+1).| by TAYLOR_2:1
             .= |.r.| |^(2*n+1) / (2*n+1) by ABSVALUE:def 1;
    hence thesis by Def1;
  end;
  thus r>=0 implies aA=rL
  proof
    assume r>=0;
    then
A5: |.r.| = r by ABSVALUE:def 1;
    now let n;
      thus aA.n = (-1)|^n * (r |^(2*n+1) / (2*n+1)) by A2,A5
               .=(-1)|^n * r |^(2*n+1) / (2*n+1)
               .= rL.n by Def2;
    end;
    hence thesis;
  end;
  assume r <0;
  then
A6: |.r.|=-r by ABSVALUE:def 1;
  now let n;
    |.r.| |^(2*n+1) = - (r |^ (2*n+1)) by A6,POWER:2;
    then aA.n = (-1)|^n * ((- (r |^ (2*n+1))) / (2*n+1)) by A2
             .= - (-1)|^n * (r |^ (2*n+1)) / (2*n+1)
             .=-rL.n by Def2;
    hence rL.n = (-1)*aA.n
              .=((-1)(#)aA).n by SEQ_1:9;
  end;
  hence thesis;
end;
