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:

ap2(f, x) -> x
ap2(ap2(ap2(g, x), y), ap2(s, z)) -> ap2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

ap2(f, x) -> x
ap2(ap2(ap2(g, x), y), ap2(s, z)) -> ap2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

ap2(f, x) -> x
ap2(ap2(ap2(g, x), y), ap2(s, z)) -> ap2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))

The set Q consists of the following terms:

ap2(f, x0)
ap2(ap2(ap2(g, x0), x1), ap2(s, x2))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))
AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(x, y)
AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(ap2(x, y), 0)

The TRS R consists of the following rules:

ap2(f, x) -> x
ap2(ap2(ap2(g, x), y), ap2(s, z)) -> ap2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))

The set Q consists of the following terms:

ap2(f, x0)
ap2(ap2(ap2(g, x0), x1), ap2(s, x2))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))
AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(x, y)
AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(ap2(x, y), 0)

The TRS R consists of the following rules:

ap2(f, x) -> x
ap2(ap2(ap2(g, x), y), ap2(s, z)) -> ap2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))

The set Q consists of the following terms:

ap2(f, x0)
ap2(ap2(ap2(g, x0), x1), ap2(s, x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

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

AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))
AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(x, y)

The TRS R consists of the following rules:

ap2(f, x) -> x
ap2(ap2(ap2(g, x), y), ap2(s, z)) -> ap2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))

The set Q consists of the following terms:

ap2(f, x0)
ap2(ap2(ap2(g, x0), x1), ap2(s, x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(x, y)
The remaining pairs can at least by weakly be oriented.

AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))
Used ordering: Combined order from the following AFS and order.
AP2(x1, x2)  =  AP1(x1)
ap2(x1, x2)  =  ap2(x1, x2)
g  =  g
s  =  s
0  =  0
f  =  f

Lexicographic Path Order [19].
Precedence:
AP1 > ap2
[g, s, 0]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP

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

AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))

The TRS R consists of the following rules:

ap2(f, x) -> x
ap2(ap2(ap2(g, x), y), ap2(s, z)) -> ap2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))

The set Q consists of the following terms:

ap2(f, x0)
ap2(ap2(ap2(g, x0), x1), ap2(s, x2))

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