Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(x, y), 0)
AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(ap(g, x), y), ap(ap(x, y), 0))
AP(ap(ap(g, x), y), ap(s, z)) → AP(x, y)

The TRS R consists of the following rules:

ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(x, y), 0)
AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(ap(g, x), y), ap(ap(x, y), 0))
AP(ap(ap(g, x), y), ap(s, z)) → AP(x, y)

The TRS R consists of the following rules:

ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(ap(g, x), y), ap(ap(x, y), 0))
AP(ap(ap(g, x), y), ap(s, z)) → AP(x, y)

The TRS R consists of the following rules:

ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


AP(ap(ap(g, x), y), ap(s, z)) → AP(x, y)
The remaining pairs can at least be oriented weakly.

AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(ap(g, x), y), ap(ap(x, y), 0))
Used ordering: Polynomial interpretation [25,35]:

POL(ap(x1, x2)) = 3/2 + (1/4)x_1 + (4)x_2   
POL(f) = 0   
POL(AP(x1, x2)) = (1/2)x_1   
POL(g) = 0   
POL(s) = 0   
POL(0) = 1   
The value of delta used in the strict ordering is 15/16.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(ap(g, x), y), ap(ap(x, y), 0))

The TRS R consists of the following rules:

ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.