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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(a) → g(h(a))
h(g(x)) → g(h(f(x)))
k(x, h(x), a) → h(x)
k(f(x), y, x) → f(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(a)
H(g(x)) → H(f(x))
H(g(x)) → F(x)

The TRS R consists of the following rules:

f(a) → g(h(a))
h(g(x)) → g(h(f(x)))
k(x, h(x), a) → h(x)
k(f(x), y, x) → f(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(a)
H(g(x)) → H(f(x))
H(g(x)) → F(x)

The TRS R consists of the following rules:

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

H(g(x)) → H(f(x))

The TRS R consists of the following rules:

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


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

POL(f(x1)) = 2 + (9/4)x_1   
POL(a) = 2   
POL(g(x1)) = 4 + (5/2)x_1   
POL(h(x1)) = (1/2)x_1   
POL(H(x1)) = (13/4)x_1   
The value of delta used in the strict ordering is 13/2.
The following usable rules [17] were oriented:

f(a) → g(h(a))



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

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

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