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 [1,13] 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)) → G(b)
G(x) → G(x)
F(g(a)) → F(s(g(b)))

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
      ↳ EdgeDeletionProof

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)
F(g(a)) → F(s(g(b)))

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

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 [13,14,18] contains 1 SCC with 1 less node.

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


The following pairs can be oriented strictly and are deleted.


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

G(x) → F(g(x))
G(x) → G(x)
Used ordering: Combined order from the following AFS and order.
G(x1)  =  G(x1)
F(x1)  =  F(x1)
g(x1)  =  x1
a  =  a
b  =  b
f(x1)  =  x1
s(x1)  =  s

Lexicographic Path Order [19].
Precedence:
[G1, F1, a, s] > b


The following usable rules [14] were oriented:

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



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

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

G(x) → F(g(x))
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 [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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.