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

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

f(a, h(x)) → f(g(x), h(x))
h(g(x)) → h(a)
g(h(x)) → g(x)
h(h(x)) → x

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(a, h(x)) → f(g(x), h(x))
h(g(x)) → h(a)
g(h(x)) → g(x)
h(h(x)) → 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(a, h(x)) → F(g(x), h(x))
H(g(x)) → H(a)
G(h(x)) → G(x)
F(a, h(x)) → G(x)

The TRS R consists of the following rules:

f(a, h(x)) → f(g(x), h(x))
h(g(x)) → h(a)
g(h(x)) → g(x)
h(h(x)) → 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(a, h(x)) → F(g(x), h(x))
H(g(x)) → H(a)
G(h(x)) → G(x)
F(a, h(x)) → G(x)

The TRS R consists of the following rules:

f(a, h(x)) → f(g(x), h(x))
h(g(x)) → h(a)
g(h(x)) → g(x)
h(h(x)) → 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 2 less nodes.

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

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

G(h(x)) → G(x)

The TRS R consists of the following rules:

f(a, h(x)) → f(g(x), h(x))
h(g(x)) → h(a)
g(h(x)) → g(x)
h(h(x)) → 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:

G(h(x)) → G(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
            ↳ QDPOrderProof

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

F(a, h(x)) → F(g(x), h(x))

The TRS R consists of the following rules:

f(a, h(x)) → f(g(x), h(x))
h(g(x)) → h(a)
g(h(x)) → g(x)
h(h(x)) → x

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.


F(a, h(x)) → F(g(x), h(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(F(x1, x2)) = x1   
POL(a) = 1   
POL(g(x1)) = 0   
POL(h(x1)) = 0   

The following usable rules [17] were oriented:

g(h(x)) → g(x)



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

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

f(a, h(x)) → f(g(x), h(x))
h(g(x)) → h(a)
g(h(x)) → g(x)
h(h(x)) → 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.