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
            ↳ QDPOrderProof
          ↳ 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 use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


G(h(x)) → G(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(h(x1)) = 1 + (4)x_1   
POL(G(x1)) = (4)x_1   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none



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

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.

↳ 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 [25,35]:

POL(a) = 4   
POL(g(x1)) = 1   
POL(h(x1)) = x_1   
POL(F(x1, x2)) = (3)x_1   
The value of delta used in the strict ordering is 9.
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.