theorem Th2:
  len f > 0 implies len Ant(f)+1 = len f & len Ant(f) < len f
proof
  assume len f > 0;
  then consider i being Nat such that
A1: len f = i+1 by NAT_1:6;
  reconsider i as Element of NAT by ORDINAL1:def 12;
  Ant(f) = f|(Seg i) by A1,Def1;
  then dom Ant(f) = dom f /\ Seg i by RELAT_1:61;
  then Seg (len Ant(f)) = dom f /\ Seg i by FINSEQ_1:def 3;
  then
A2: Seg (len Ant(f)) = Seg (len f) /\ Seg i by FINSEQ_1:def 3;
  i <= len f by A1,NAT_1:11;
  then
A3: Seg i c= Seg (len f) by FINSEQ_1:5;
  hence len Ant(f)+1 = len f by A1,A2,FINSEQ_1:6,XBOOLE_1:28;
  len Ant(f) = i by A2,A3,FINSEQ_1:6,XBOOLE_1:28;
  hence thesis by A1,NAT_1:13;
end;
