theorem Th13:
  seq is convergent implies abs seq is convergent
proof
  assume seq is convergent;
  then consider g1 such that
A1: for p st 0<p ex n st for m st n<=m holds |.seq.m-g1.|<p;
  reconsider g1 as Real;
  take g=|.g1.|;
  let p;
  assume 0<p;
  then consider n1 such that
A2: for m st n1<=m holds |.seq.m-g1.|<p by A1;
  take n=n1;
  let m;
  |.g1-seq.m.|=|.-((seq.m)-g1).| .=|.seq.m-g1.| by COMPLEX1:52;
  then g-|.seq.m.|<=|.seq.m-g1.| by COMPLEX1:59;
  then |.seq.m.|-g<=|.seq.m-g1.| & -|.seq.m-g1.|<= -(g-|.seq.m.|) by
COMPLEX1:59,XREAL_1:24;
  then |.|.seq.m.|-g.|<=|.seq.m-g1.| by ABSVALUE:5;
  then
A3: |.((abs seq).m)-g.| <=|.seq.m-g1.| by SEQ_1:12;
  assume n<=m;
  then |.seq.m-g1.|<p by A2;
  hence thesis by A3,XXREAL_0:2;
end;
