theorem Th105:
  n = m + 1 & i in Seg n implies len Sgm(Seg n \ {i}) = m
proof
  assume that
A1: n = m + 1 and
A2: i in Seg n;
  set X = Seg n \ {i};
  i in {i} by TARSKI:def 1;
  then not i in X by XBOOLE_0:def 5;
  then card X + 1 = card (X \/ {i}) by CARD_2:41
    .= card ((Seg n) \/ {i}) by XBOOLE_1:39
    .= card (Seg n) by A2,ZFMISC_1:40
    .= m + 1 by A1,FINSEQ_1:57;
  hence thesis by Th37;
end;
