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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

g(h(g(x))) → g(x)
g(g(x)) → g(h(g(x)))
h(h(x)) → h(f(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 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

G(g(x)) → G(h(g(x)))

The TRS R consists of the following rules:

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

POL(G(x1)) = x1   
POL(f(x1, x2)) = 0   
POL(g(x1)) = 1 + x1   
POL(h(x1)) = 0   

The following usable rules [17] were oriented:

h(h(x)) → h(f(h(x), x))



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

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

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