Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(g(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(g(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(x) → F(g(x))
F(g(a)) → F(s(g(b)))
F(g(a)) → G(b)
G(x) → G(x)

The TRS R consists of the following rules:

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(g(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(x) → F(g(x))
F(g(a)) → F(s(g(b)))
F(g(a)) → G(b)
G(x) → G(x)

The TRS R consists of the following rules:

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(g(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 1 less node.

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

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

G(x) → F(g(x))
F(g(a)) → G(b)
G(x) → G(x)

The TRS R consists of the following rules:

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(g(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(x) → F(g(x))
The remaining pairs can at least be oriented weakly.

F(g(a)) → G(b)
G(x) → G(x)
Used ordering: Polynomial interpretation [25,35]:

POL(a) = 3   
POL(f(x1)) = (1/4)x_1   
POL(g(x1)) = 1/2 + (1/2)x_1   
POL(G(x1)) = 7/4 + (1/4)x_1   
POL(b) = 0   
POL(s(x1)) = 5/4   
POL(F(x1)) = 3/4 + (1/2)x_1   
The value of delta used in the strict ordering is 3/4.
The following usable rules [17] were oriented:

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(g(x))



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

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

F(g(a)) → G(b)
G(x) → G(x)

The TRS R consists of the following rules:

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(g(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 1 less node.

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

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

G(x) → G(x)

The TRS R consists of the following rules:

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(g(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.