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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
*2(+2(y, z), x) -> +2(*2(x, y), *2(x, z))
*2(*2(x, y), z) -> *2(x, *2(y, z))
+2(+2(x, y), z) -> +2(x, +2(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:

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
*2(+2(y, z), x) -> +2(*2(x, y), *2(x, z))
*2(*2(x, y), z) -> *2(x, *2(y, z))
+2(+2(x, y), z) -> +2(x, +2(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.

+12(+2(x, y), z) -> +12(y, z)
+12(+2(x, y), z) -> +12(x, +2(y, z))
Used argument filtering: +12(x1, x2)  =  x1
+2(x1, x2)  =  +2(x1, x2)
Used ordering: 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:

*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
*2(+2(y, z), x) -> +2(*2(x, y), *2(x, z))
*2(*2(x, y), z) -> *2(x, *2(y, z))
+2(+2(x, y), z) -> +2(x, +2(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.

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

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

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

The TRS R consists of the following rules:

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

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