theorem
  1<=m & m<=len p implies (m,m)-cut p = <*p.m*>
proof
  assume that
A1: 1<=m and
A2: m<=len p;
  set mp = (m,m)-cut p;
A3: len mp + m = m + 1 by A1,A2,Def1;
  then mp.(0+1) = p.(m+0) by A1,A2,Def1
    .= p.m;
  hence thesis by A3,FINSEQ_1:40;
end;
