reserve k, m, n, p, K, N for Nat;
reserve i for Integer;
reserve x, y, eps for Real;
reserve seq, seq1, seq2 for Real_Sequence;
reserve sq for FinSequence of REAL;

theorem Th5:
  (n choose (k+1))=((n-k)/(k+1))*(n choose k)
proof
  per cases;
  suppose
A1: k+1<=n;
    then reconsider l = n-(k+1) as Element of NAT by INT_1:5;
    l+1 = n-k;
    then reconsider l1 = n-k as Element of NAT;
    k <= k+1 by NAT_1:11;
    then
A2: k <= n by A1,XXREAL_0:2;
    thus (n choose (k+1)) = n!/((k+1)!*(l!)) by A1,NEWTON:def 3
      .= n!/((k!*(k+1))*(l!)) by NEWTON:15
      .= n!/((k!*(k+1))*((l!)*(l+1)/(l+1))) by XCMPLX_1:89
      .= n!/((k!*(k+1))*((l+1)!/(l+1))) by NEWTON:15
      .= (l1/(k+1))*(n!/((k!)*(l1!))) by XCMPLX_1:233
      .= ((n-k)/(k+1))*(n choose k) by A2,NEWTON:def 3;
  end;
  suppose
A3: k+1>n & k<=n;
    then k>=n by NAT_1:13;
    then k=n by A3,XXREAL_0:1;
    hence thesis by A3,NEWTON:def 3;
  end;
  suppose
A4: k+1>n & k>n;
    hence (n choose (k+1)) = ((n-k)/(k+1))*0 by NEWTON:def 3
      .= ((n-k)/(k+1))*(n choose k) by A4,NEWTON:def 3;
  end;
end;
