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(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(minus, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, app2(app2(minus, x), y)), app2(double, y))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(minus, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))
APP2(double, app2(s, x)) -> APP2(s, app2(double, x))
APP2(app2(plus, app2(s, x)), y) -> APP2(double, y)
APP2(app2(plus, app2(s, x)), y) -> APP2(minus, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, y)
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, app2(app2(minus, x), y))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), app2(s, y))
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)
APP2(double, app2(s, x)) -> APP2(s, app2(s, app2(double, x)))
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
APP2(double, app2(s, x)) -> APP2(double, x)

The TRS R consists of the following rules:

app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))

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(minus, app2(s, x)), app2(s, y)) -> APP2(minus, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, app2(app2(minus, x), y)), app2(double, y))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(minus, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))
APP2(double, app2(s, x)) -> APP2(s, app2(double, x))
APP2(app2(plus, app2(s, x)), y) -> APP2(double, y)
APP2(app2(plus, app2(s, x)), y) -> APP2(minus, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, y)
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, app2(app2(minus, x), y))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), app2(s, y))
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)
APP2(double, app2(s, x)) -> APP2(s, app2(s, app2(double, x)))
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
APP2(double, app2(s, x)) -> APP2(double, x)

The TRS R consists of the following rules:

app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 11 less nodes.

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

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

APP2(double, app2(s, x)) -> APP2(double, x)

The TRS R consists of the following rules:

app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(double, app2(s, x)) -> APP2(double, x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x2)
double  =  double
app2(x1, x2)  =  app1(x2)
s  =  s

Lexicographic Path Order [19].
Precedence:
[APP1, app1] > double
s > double


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))

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
            ↳ QDPOrderProof
          ↳ QDP

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

APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)

The TRS R consists of the following rules:

app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  x1
app2(x1, x2)  =  app2(x1, x2)
minus  =  minus
s  =  s

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))

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
          ↳ QDP

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

APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, app2(app2(minus, x), y)), app2(double, y))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), app2(s, y))

The TRS R consists of the following rules:

app2(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(double, 0) -> 0
app2(double, app2(s, x)) -> app2(s, app2(s, app2(double, x)))
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(app2(plus, x), app2(s, y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, app2(app2(minus, x), y)), app2(double, y)))

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