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:

f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)

Q is empty.

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

F(0, 1, g(x, y), z) → H(x)
H(g(x, y)) → H(x)
F(0, 1, g(x, y), z) → F(g(x, y), g(x, y), g(x, y), h(x))

The TRS R consists of the following rules:

f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

F(0, 1, g(x, y), z) → H(x)
H(g(x, y)) → H(x)
F(0, 1, g(x, y), z) → F(g(x, y), g(x, y), g(x, y), h(x))

The TRS R consists of the following rules:

f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

F(0, 1, g(x, y), z) → H(x)
H(g(x, y)) → H(x)
F(0, 1, g(x, y), z) → F(g(x, y), g(x, y), g(x, y), h(x))

The TRS R consists of the following rules:

f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

H(g(x, y)) → H(x)

The TRS R consists of the following rules:

f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)

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


The following pairs can be oriented strictly and are deleted.


H(g(x, y)) → H(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
H(x1)  =  x1
g(x1, x2)  =  g(x1)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP

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

F(0, 1, g(x, y), z) → F(g(x, y), g(x, y), g(x, y), h(x))

The TRS R consists of the following rules:

f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)

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