theorem Th6: len f > 0 implies (con f)/.1 = f/.1
  proof
   assume
A1: len f > 0;
    then A2: 1 <= len f by NAT_1:25;
    then len con f >= 1 by Def2;
    hence (con f)/.1 = (con f).1 by FINSEQ_4:15
    .= f.1 by Def2,A1
    .= f/.1 by FINSEQ_4:15,A2;
  end;
