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:

app2(D, t) -> 1
app2(D, constant) -> 0
app2(D, app2(app2(+, x), y)) -> app2(app2(+, app2(D, x)), app2(D, y))
app2(D, app2(app2(*, x), y)) -> app2(app2(+, app2(app2(*, y), app2(D, x))), app2(app2(*, x), app2(D, y)))
app2(D, app2(app2(-, x), y)) -> app2(app2(-, app2(D, x)), app2(D, y))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

app2(D, t) -> 1
app2(D, constant) -> 0
app2(D, app2(app2(+, x), y)) -> app2(app2(+, app2(D, x)), app2(D, y))
app2(D, app2(app2(*, x), y)) -> app2(app2(+, app2(app2(*, y), app2(D, x))), app2(app2(*, x), app2(D, y)))
app2(D, app2(app2(-, x), y)) -> app2(app2(-, app2(D, x)), app2(D, y))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

app2(D, t) -> 1
app2(D, constant) -> 0
app2(D, app2(app2(+, x), y)) -> app2(app2(+, app2(D, x)), app2(D, y))
app2(D, app2(app2(*, x), y)) -> app2(app2(+, app2(app2(*, y), app2(D, x))), app2(app2(*, x), app2(D, y)))
app2(D, app2(app2(-, x), y)) -> app2(app2(-, app2(D, x)), app2(D, y))

The set Q consists of the following terms:

app2(D, t)
app2(D, constant)
app2(D, app2(app2(+, x0), x1))
app2(D, app2(app2(*, x0), x1))
app2(D, app2(app2(-, x0), x1))


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

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

The TRS R consists of the following rules:

app2(D, t) -> 1
app2(D, constant) -> 0
app2(D, app2(app2(+, x), y)) -> app2(app2(+, app2(D, x)), app2(D, y))
app2(D, app2(app2(*, x), y)) -> app2(app2(+, app2(app2(*, y), app2(D, x))), app2(app2(*, x), app2(D, y)))
app2(D, app2(app2(-, x), y)) -> app2(app2(-, app2(D, x)), app2(D, y))

The set Q consists of the following terms:

app2(D, t)
app2(D, constant)
app2(D, app2(app2(+, x0), x1))
app2(D, app2(app2(*, x0), x1))
app2(D, app2(app2(-, x0), x1))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

app2(D, t) -> 1
app2(D, constant) -> 0
app2(D, app2(app2(+, x), y)) -> app2(app2(+, app2(D, x)), app2(D, y))
app2(D, app2(app2(*, x), y)) -> app2(app2(+, app2(app2(*, y), app2(D, x))), app2(app2(*, x), app2(D, y)))
app2(D, app2(app2(-, x), y)) -> app2(app2(-, app2(D, x)), app2(D, y))

The set Q consists of the following terms:

app2(D, t)
app2(D, constant)
app2(D, app2(app2(+, x0), x1))
app2(D, app2(app2(*, x0), x1))
app2(D, app2(app2(-, x0), x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 9 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof

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

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

The TRS R consists of the following rules:

app2(D, t) -> 1
app2(D, constant) -> 0
app2(D, app2(app2(+, x), y)) -> app2(app2(+, app2(D, x)), app2(D, y))
app2(D, app2(app2(*, x), y)) -> app2(app2(+, app2(app2(*, y), app2(D, x))), app2(app2(*, x), app2(D, y)))
app2(D, app2(app2(-, x), y)) -> app2(app2(-, app2(D, x)), app2(D, y))

The set Q consists of the following terms:

app2(D, t)
app2(D, constant)
app2(D, app2(app2(+, x0), x1))
app2(D, app2(app2(*, x0), x1))
app2(D, app2(app2(-, x0), x1))

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(D, app2(app2(-, x), y)) -> APP2(D, y)
APP2(D, app2(app2(-, x), y)) -> APP2(D, x)
APP2(D, app2(app2(*, x), y)) -> APP2(D, x)
APP2(D, app2(app2(+, x), y)) -> APP2(D, x)
APP2(D, app2(app2(+, x), y)) -> APP2(D, y)
APP2(D, app2(app2(*, x), y)) -> APP2(D, y)
Used argument filtering: APP2(x1, x2)  =  x2
app2(x1, x2)  =  app2(x1, x2)
-  =  -
*  =  *
+  =  +
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof
                ↳ QDP
                  ↳ PisEmptyProof

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

app2(D, t) -> 1
app2(D, constant) -> 0
app2(D, app2(app2(+, x), y)) -> app2(app2(+, app2(D, x)), app2(D, y))
app2(D, app2(app2(*, x), y)) -> app2(app2(+, app2(app2(*, y), app2(D, x))), app2(app2(*, x), app2(D, y)))
app2(D, app2(app2(-, x), y)) -> app2(app2(-, app2(D, x)), app2(D, y))

The set Q consists of the following terms:

app2(D, t)
app2(D, constant)
app2(D, app2(app2(+, x0), x1))
app2(D, app2(app2(*, x0), x1))
app2(D, app2(app2(-, x0), x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.