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(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)

Q is empty.

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

F1(+2(x, 0)) -> F1(x)
+12(x, +2(y, z)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)

The TRS R consists of the following rules:

f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)

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(+2(x, 0)) -> F1(x)
+12(x, +2(y, z)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)

The TRS R consists of the following rules:

f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)

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
          ↳ QDP

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

+12(x, +2(y, z)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)

The TRS R consists of the following rules:

f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)

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

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

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

F1(+2(x, 0)) -> F1(x)

The TRS R consists of the following rules:

f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)

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(+2(x, 0)) -> F1(x)
Used argument filtering: F1(x1)  =  x1
+2(x1, x2)  =  +1(x1)
0  =  0
Used ordering: Quasi Precedence: trivial 


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

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

f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)

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.