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

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

The TRS R consists of the following rules:

f(a, x) → f(g(x), 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
      ↳ EdgeDeletionProof

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

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

The TRS R consists of the following rules:

f(a, x) → f(g(x), 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 deleted some edges using various graph approximations

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

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

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

The TRS R consists of the following rules:

f(a, x) → f(g(x), 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 [13,14,18] contains 2 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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, x) → f(g(x), 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 [13].


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: Combined order from the following AFS and order.
G(x1)  =  x1
h(x1)  =  h(x1)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



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

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

f(a, x) → f(g(x), 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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof

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

F(a, x) → F(g(x), x)

The TRS R consists of the following rules:

f(a, x) → f(g(x), 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 [13].


The following pairs can be oriented strictly and are deleted.


F(a, x) → F(g(x), x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  F(x1, x2)
a  =  a
g(x1)  =  g
h(x1)  =  h

Recursive Path Order [2].
Precedence:
F2 > g
a > g

The following usable rules [14] were oriented:

g(h(x)) → g(x)



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

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

f(a, x) → f(g(x), 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.