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:

f2(s1(0), g1(x)) -> f2(x, g1(x))
g1(s1(x)) -> g1(x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f2(s1(0), g1(x)) -> f2(x, g1(x))
g1(s1(x)) -> g1(x)

Q is empty.

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

G1(s1(x)) -> G1(x)
F2(s1(0), g1(x)) -> F2(x, g1(x))

The TRS R consists of the following rules:

f2(s1(0), g1(x)) -> f2(x, g1(x))
g1(s1(x)) -> 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(s1(x)) -> G1(x)
F2(s1(0), g1(x)) -> F2(x, g1(x))

The TRS R consists of the following rules:

f2(s1(0), g1(x)) -> f2(x, g1(x))
g1(s1(x)) -> g1(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs.

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

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

G1(s1(x)) -> G1(x)

The TRS R consists of the following rules:

f2(s1(0), g1(x)) -> f2(x, g1(x))
g1(s1(x)) -> 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.

G1(s1(x)) -> G1(x)
Used argument filtering: G1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

f2(s1(0), g1(x)) -> f2(x, g1(x))
g1(s1(x)) -> g1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP

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

F2(s1(0), g1(x)) -> F2(x, g1(x))

The TRS R consists of the following rules:

f2(s1(0), g1(x)) -> f2(x, g1(x))
g1(s1(x)) -> g1(x)

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