theorem Th88:
  k in Seg n & c = z.k implies (abs z).k = |.c.|
proof
  assume that
A1: k in Seg n and
A2: c = z.k;
  len abs z = n by CARD_1:def 7;
  then k in dom abs z by A1,FINSEQ_1:def 3;
  hence (abs z).k = abscomplex.c by A2,FUNCT_1:12
    .= |.c.| by Def5;
end;
