We consider termination of the LCTRS with only rule scheme Calc:

  Signature: f    :: Int -> Int -> Int -> Int -> Int
             sqrt :: Int -> Int

  Rules: sqrt(x) -> f(x, 0, 1, 1)
         f(x, z, u, w) -> f(x, z + 1, u + 2, w + u + 2) | x >= w /\ u >= 0
         f(x, z, u, w) -> z | w > x

The system is accessible function passing by a sort ordering that equates all sorts.

We start by computing the initial DP problem D1 = (P1, R, f, c), where:

  P1. (1) sqrt#(x) => f#(x, 0, 1, 1)
      (2) f#(x, z, u, w) => f#(x, z + 1, u + 2, w + u + 2) | x >= w /\ u >= 0

***** We apply the Graph Processor on D1 = (P1, R, f, c).

We compute a graph approximation with the following edges:

  1: 2 
  2: 2 

There is only one SCC, so all DPs not inside the SCC can be removed.

Processor output: { D2 = (P2, R, f, c) }, where:

  P2. (1) f#(x, z, u, w) => f#(x, z + 1, u + 2, w + u + 2) | x >= w /\ u >= 0

***** We apply the Integer Function Processor on D2 = (P2, R, f, c).

We use the following integer mapping:

  J(f#) = arg_1 - arg_4

We thus have:

  (1) x >= w /\ u >= 0 |= x - w > x - (w + u + 2) (and x - w >= 0)

All DPs are strictly oriented, and may be removed.  Hence, this DP problem is finite.

Processor output: { }.