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:

app2(app2(*, x), app2(app2(+, y), z)) -> app2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
app2(app2(*, app2(app2(+, y), z)), x) -> app2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
app2(app2(*, app2(app2(*, x), y)), z) -> app2(app2(*, x), app2(app2(*, y), z))
app2(app2(+, app2(app2(+, x), y)), z) -> app2(app2(+, x), app2(app2(+, y), z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(*, x), app2(app2(+, y), z)) -> app2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
app2(app2(*, app2(app2(+, y), z)), x) -> app2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
app2(app2(*, app2(app2(*, x), y)), z) -> app2(app2(*, x), app2(app2(*, y), z))
app2(app2(+, app2(app2(+, x), y)), z) -> app2(app2(+, x), app2(app2(+, y), z))

Q is empty.

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

APP2(app2(+, app2(app2(+, x), y)), z) -> APP2(+, y)
APP2(app2(+, app2(app2(+, x), y)), z) -> APP2(app2(+, x), app2(app2(+, y), z))
APP2(app2(*, app2(app2(*, x), y)), z) -> APP2(*, y)
APP2(app2(*, app2(app2(+, y), z)), x) -> APP2(app2(*, x), z)
APP2(app2(*, app2(app2(+, y), z)), x) -> APP2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
APP2(app2(*, x), app2(app2(+, y), z)) -> APP2(app2(*, x), z)
APP2(app2(*, x), app2(app2(+, y), z)) -> APP2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
APP2(app2(*, app2(app2(+, y), z)), x) -> APP2(+, app2(app2(*, x), y))
APP2(app2(*, app2(app2(*, x), y)), z) -> APP2(app2(*, y), z)
APP2(app2(*, x), app2(app2(+, y), z)) -> APP2(+, app2(app2(*, x), y))
APP2(app2(+, app2(app2(+, x), y)), z) -> APP2(app2(+, y), z)
APP2(app2(*, app2(app2(+, y), z)), x) -> APP2(*, x)
APP2(app2(*, app2(app2(+, y), z)), x) -> APP2(app2(*, x), y)
APP2(app2(*, x), app2(app2(+, y), z)) -> APP2(app2(*, x), y)
APP2(app2(*, app2(app2(*, x), y)), z) -> APP2(app2(*, x), app2(app2(*, y), z))

The TRS R consists of the following rules:

app2(app2(*, x), app2(app2(+, y), z)) -> app2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
app2(app2(*, app2(app2(+, y), z)), x) -> app2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
app2(app2(*, app2(app2(*, x), y)), z) -> app2(app2(*, x), app2(app2(*, y), z))
app2(app2(+, app2(app2(+, x), y)), z) -> app2(app2(+, x), app2(app2(+, 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:

APP2(app2(+, app2(app2(+, x), y)), z) -> APP2(+, y)
APP2(app2(+, app2(app2(+, x), y)), z) -> APP2(app2(+, x), app2(app2(+, y), z))
APP2(app2(*, app2(app2(*, x), y)), z) -> APP2(*, y)
APP2(app2(*, app2(app2(+, y), z)), x) -> APP2(app2(*, x), z)
APP2(app2(*, app2(app2(+, y), z)), x) -> APP2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
APP2(app2(*, x), app2(app2(+, y), z)) -> APP2(app2(*, x), z)
APP2(app2(*, x), app2(app2(+, y), z)) -> APP2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
APP2(app2(*, app2(app2(+, y), z)), x) -> APP2(+, app2(app2(*, x), y))
APP2(app2(*, app2(app2(*, x), y)), z) -> APP2(app2(*, y), z)
APP2(app2(*, x), app2(app2(+, y), z)) -> APP2(+, app2(app2(*, x), y))
APP2(app2(+, app2(app2(+, x), y)), z) -> APP2(app2(+, y), z)
APP2(app2(*, app2(app2(+, y), z)), x) -> APP2(*, x)
APP2(app2(*, app2(app2(+, y), z)), x) -> APP2(app2(*, x), y)
APP2(app2(*, x), app2(app2(+, y), z)) -> APP2(app2(*, x), y)
APP2(app2(*, app2(app2(*, x), y)), z) -> APP2(app2(*, x), app2(app2(*, y), z))

The TRS R consists of the following rules:

app2(app2(*, x), app2(app2(+, y), z)) -> app2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
app2(app2(*, app2(app2(+, y), z)), x) -> app2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
app2(app2(*, app2(app2(*, x), y)), z) -> app2(app2(*, x), app2(app2(*, y), z))
app2(app2(+, app2(app2(+, x), y)), z) -> app2(app2(+, x), app2(app2(+, 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 7 less nodes.

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

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

APP2(app2(+, app2(app2(+, x), y)), z) -> APP2(app2(+, x), app2(app2(+, y), z))
APP2(app2(+, app2(app2(+, x), y)), z) -> APP2(app2(+, y), z)

The TRS R consists of the following rules:

app2(app2(*, x), app2(app2(+, y), z)) -> app2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
app2(app2(*, app2(app2(+, y), z)), x) -> app2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
app2(app2(*, app2(app2(*, x), y)), z) -> app2(app2(*, x), app2(app2(*, y), z))
app2(app2(+, app2(app2(+, x), y)), z) -> app2(app2(+, x), app2(app2(+, 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.

APP2(app2(+, app2(app2(+, x), y)), z) -> APP2(app2(+, x), app2(app2(+, y), z))
APP2(app2(+, app2(app2(+, x), y)), z) -> APP2(app2(+, y), z)
Used argument filtering: APP2(x1, x2)  =  x1
app2(x1, x2)  =  app2(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:

app2(app2(*, x), app2(app2(+, y), z)) -> app2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
app2(app2(*, app2(app2(+, y), z)), x) -> app2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
app2(app2(*, app2(app2(*, x), y)), z) -> app2(app2(*, x), app2(app2(*, y), z))
app2(app2(+, app2(app2(+, x), y)), z) -> app2(app2(+, x), app2(app2(+, 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:

APP2(app2(*, app2(app2(+, y), z)), x) -> APP2(app2(*, x), z)
APP2(app2(*, app2(app2(+, y), z)), x) -> APP2(app2(*, x), y)
APP2(app2(*, x), app2(app2(+, y), z)) -> APP2(app2(*, x), z)
APP2(app2(*, x), app2(app2(+, y), z)) -> APP2(app2(*, x), y)
APP2(app2(*, app2(app2(*, x), y)), z) -> APP2(app2(*, y), z)
APP2(app2(*, app2(app2(*, x), y)), z) -> APP2(app2(*, x), app2(app2(*, y), z))

The TRS R consists of the following rules:

app2(app2(*, x), app2(app2(+, y), z)) -> app2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
app2(app2(*, app2(app2(+, y), z)), x) -> app2(app2(+, app2(app2(*, x), y)), app2(app2(*, x), z))
app2(app2(*, app2(app2(*, x), y)), z) -> app2(app2(*, x), app2(app2(*, y), z))
app2(app2(+, app2(app2(+, x), y)), z) -> app2(app2(+, x), app2(app2(+, y), z))

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