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:

f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))

Q is empty.

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

G1(x) -> F1(g1(x))
F1(g1(a)) -> G1(b)
G1(x) -> G1(x)
F1(g1(a)) -> F1(s1(g1(b)))

The TRS R consists of the following rules:

f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(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:

G1(x) -> F1(g1(x))
F1(g1(a)) -> G1(b)
G1(x) -> G1(x)
F1(g1(a)) -> F1(s1(g1(b)))

The TRS R consists of the following rules:

f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPAfsSolverProof

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

G1(x) -> F1(g1(x))
F1(g1(a)) -> G1(b)
G1(x) -> G1(x)

The TRS R consists of the following rules:

f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F1(g1(a)) -> G1(b)
Used argument filtering: G1(x1)  =  x1
F1(x1)  =  x1
g1(x1)  =  x1
a  =  a
b  =  b
f1(x1)  =  f
Used ordering: Quasi Precedence: a > [b, f]


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
QDP
              ↳ DependencyGraphProof

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

G1(x) -> F1(g1(x))
G1(x) -> G1(x)

The TRS R consists of the following rules:

f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ DependencyGraphProof
QDP

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

G1(x) -> G1(x)

The TRS R consists of the following rules:

f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))

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