reserve x,y,z for set;
reserve f,f1,f2,f3 for FinSequence,
  p,p1,p2,p3 for set,
  i,k for Nat;
reserve D for non empty set,
  p,p1,p2,p3 for Element of D,
  f,f1,f2 for FinSequence of D;
reserve D for non empty set;
reserve p, q for FinSequence,
  X, Y, x, y for set,
  D for non empty set,
  i, j, k, l, m, n, r for Nat;

theorem
  1<=m & m<=n & n<=len p implies ((m,n)-cut p).1 = p.m &
  ((m,n)-cut p).len ((m,n)-cut p)=p.n
proof
  set c = ((m,n)-cut p);
  assume that
A1: 1<=m and
A2: m<=n and
A3: n<=len p;
A4: m<=n+1 by A2,NAT_1:12;
  then
A5: len c +m = n+1 by A1,A3,Lm2;
A6: now
    assume len c = 0;
    then n+1<=n+0 by A2,A5;
    hence contradiction by XREAL_1:6;
  end;
  then 0+1<=len c by NAT_1:13;
  then consider i being Nat such that
  0<=i and
A7: i<len c and
A8: len c =i+1 by Th1;
  0+1=1;
  hence c.1 = p.(m+0) by A1,A3,A4,A6,Lm2
    .= p.m;
  m+i = n by A5,A8;
  hence thesis by A1,A3,A4,A7,A8,Lm2;
end;
