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:

f1(0) -> 1
f1(s1(x)) -> g1(f1(x))
g1(x) -> +2(x, s1(x))
f1(s1(x)) -> +2(f1(x), s1(f1(x)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f1(0) -> 1
f1(s1(x)) -> g1(f1(x))
g1(x) -> +2(x, s1(x))
f1(s1(x)) -> +2(f1(x), s1(f1(x)))

Q is empty.

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

F1(s1(x)) -> G1(f1(x))
F1(s1(x)) -> F1(x)

The TRS R consists of the following rules:

f1(0) -> 1
f1(s1(x)) -> g1(f1(x))
g1(x) -> +2(x, s1(x))
f1(s1(x)) -> +2(f1(x), s1(f1(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:

F1(s1(x)) -> G1(f1(x))
F1(s1(x)) -> F1(x)

The TRS R consists of the following rules:

f1(0) -> 1
f1(s1(x)) -> g1(f1(x))
g1(x) -> +2(x, s1(x))
f1(s1(x)) -> +2(f1(x), s1(f1(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:

F1(s1(x)) -> F1(x)

The TRS R consists of the following rules:

f1(0) -> 1
f1(s1(x)) -> g1(f1(x))
g1(x) -> +2(x, s1(x))
f1(s1(x)) -> +2(f1(x), s1(f1(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(s1(x)) -> F1(x)
Used argument filtering: F1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


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

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

f1(0) -> 1
f1(s1(x)) -> g1(f1(x))
g1(x) -> +2(x, s1(x))
f1(s1(x)) -> +2(f1(x), s1(f1(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.