Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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 [1,15] 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) → F(g(x, y), g(x, y), g(x, y), h(x))
H(g(x, y)) → H(x)
F(0, 1, g(x, y), z) → 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
      ↳ DependencyGraphProof

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))
H(g(x, y)) → H(x)
F(0, 1, g(x, y), z) → 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 [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof
          ↳ QDP

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

H(g(x, y)) → H(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ NonTerminationProof

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.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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)


s = F(g(0, 1), g(0, 1), g(x, y), z) evaluates to t =F(g(x, y), g(x, y), g(x, y), h(x))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

F(g(0, 1), g(0, 1), g(0, 1), h(0)) → F(g(0, 1), 1, g(0, 1), h(0))
with rule g(0, 1) → 1 at position [1] and matcher [ ]

F(g(0, 1), 1, g(0, 1), h(0)) → F(0, 1, g(0, 1), h(0))
with rule g(0, 1) → 0 at position [0] and matcher [ ]

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

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.