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

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

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

G(g(x)) → H(g(x))
G(g(x)) → G(h(g(x)))
H(h(x)) → H(f(h(x), 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 deleted some edges using various graph approximations

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

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

G(g(x)) → G(h(g(x)))
G(g(x)) → H(g(x))
H(h(x)) → H(f(h(x), 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 [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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 [13].


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: Combined order from the following AFS and order.
G(x1)  =  G(x1)
g(x1)  =  g(x1)
h(x1)  =  h
f(x1, x2)  =  f(x1, x2)

Recursive Path Order [2].
Precedence:
G1 > h
g1 > h

The following usable rules [14] were oriented:

h(h(x)) → h(f(h(x), x))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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.