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(not, app2(not, x)) -> x
app2(not, app2(app2(or, x), y)) -> app2(app2(and, app2(not, x)), app2(not, y))
app2(not, app2(app2(and, x), y)) -> app2(app2(or, app2(not, x)), app2(not, y))
app2(app2(and, x), app2(app2(or, y), z)) -> app2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), z))
app2(app2(and, app2(app2(or, y), z)), x) -> app2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(not, app2(not, x)) -> x
app2(not, app2(app2(or, x), y)) -> app2(app2(and, app2(not, x)), app2(not, y))
app2(not, app2(app2(and, x), y)) -> app2(app2(or, app2(not, x)), app2(not, y))
app2(app2(and, x), app2(app2(or, y), z)) -> app2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), z))
app2(app2(and, app2(app2(or, y), z)), x) -> app2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), z))

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(and, app2(app2(or, y), z)), x) -> APP2(app2(and, x), z)
APP2(not, app2(app2(or, x), y)) -> APP2(not, y)
APP2(app2(and, app2(app2(or, y), z)), x) -> APP2(and, x)
APP2(app2(and, app2(app2(or, y), z)), x) -> APP2(app2(and, x), y)
APP2(not, app2(app2(or, x), y)) -> APP2(app2(and, app2(not, x)), app2(not, y))
APP2(app2(and, x), app2(app2(or, y), z)) -> APP2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), z))
APP2(not, app2(app2(or, x), y)) -> APP2(not, x)
APP2(app2(and, app2(app2(or, y), z)), x) -> APP2(or, app2(app2(and, x), y))
APP2(not, app2(app2(or, x), y)) -> APP2(and, app2(not, x))
APP2(not, app2(app2(and, x), y)) -> APP2(or, app2(not, x))
APP2(app2(and, x), app2(app2(or, y), z)) -> APP2(or, app2(app2(and, x), y))
APP2(app2(and, app2(app2(or, y), z)), x) -> APP2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), z))
APP2(not, app2(app2(and, x), y)) -> APP2(not, x)
APP2(not, app2(app2(and, x), y)) -> APP2(app2(or, app2(not, x)), app2(not, y))
APP2(not, app2(app2(and, x), y)) -> APP2(not, y)
APP2(app2(and, x), app2(app2(or, y), z)) -> APP2(app2(and, x), y)
APP2(app2(and, x), app2(app2(or, y), z)) -> APP2(app2(and, x), z)

The TRS R consists of the following rules:

app2(not, app2(not, x)) -> x
app2(not, app2(app2(or, x), y)) -> app2(app2(and, app2(not, x)), app2(not, y))
app2(not, app2(app2(and, x), y)) -> app2(app2(or, app2(not, x)), app2(not, y))
app2(app2(and, x), app2(app2(or, y), z)) -> app2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), z))
app2(app2(and, app2(app2(or, y), z)), x) -> app2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), 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(and, app2(app2(or, y), z)), x) -> APP2(app2(and, x), z)
APP2(not, app2(app2(or, x), y)) -> APP2(not, y)
APP2(app2(and, app2(app2(or, y), z)), x) -> APP2(and, x)
APP2(app2(and, app2(app2(or, y), z)), x) -> APP2(app2(and, x), y)
APP2(not, app2(app2(or, x), y)) -> APP2(app2(and, app2(not, x)), app2(not, y))
APP2(app2(and, x), app2(app2(or, y), z)) -> APP2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), z))
APP2(not, app2(app2(or, x), y)) -> APP2(not, x)
APP2(app2(and, app2(app2(or, y), z)), x) -> APP2(or, app2(app2(and, x), y))
APP2(not, app2(app2(or, x), y)) -> APP2(and, app2(not, x))
APP2(not, app2(app2(and, x), y)) -> APP2(or, app2(not, x))
APP2(app2(and, x), app2(app2(or, y), z)) -> APP2(or, app2(app2(and, x), y))
APP2(app2(and, app2(app2(or, y), z)), x) -> APP2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), z))
APP2(not, app2(app2(and, x), y)) -> APP2(not, x)
APP2(not, app2(app2(and, x), y)) -> APP2(app2(or, app2(not, x)), app2(not, y))
APP2(not, app2(app2(and, x), y)) -> APP2(not, y)
APP2(app2(and, x), app2(app2(or, y), z)) -> APP2(app2(and, x), y)
APP2(app2(and, x), app2(app2(or, y), z)) -> APP2(app2(and, x), z)

The TRS R consists of the following rules:

app2(not, app2(not, x)) -> x
app2(not, app2(app2(or, x), y)) -> app2(app2(and, app2(not, x)), app2(not, y))
app2(not, app2(app2(and, x), y)) -> app2(app2(or, app2(not, x)), app2(not, y))
app2(app2(and, x), app2(app2(or, y), z)) -> app2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), z))
app2(app2(and, app2(app2(or, y), z)), x) -> app2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 9 less nodes.

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

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

APP2(app2(and, app2(app2(or, y), z)), x) -> APP2(app2(and, x), z)
APP2(app2(and, app2(app2(or, y), z)), x) -> APP2(app2(and, x), y)
APP2(app2(and, x), app2(app2(or, y), z)) -> APP2(app2(and, x), y)
APP2(app2(and, x), app2(app2(or, y), z)) -> APP2(app2(and, x), z)

The TRS R consists of the following rules:

app2(not, app2(not, x)) -> x
app2(not, app2(app2(or, x), y)) -> app2(app2(and, app2(not, x)), app2(not, y))
app2(not, app2(app2(and, x), y)) -> app2(app2(or, app2(not, x)), app2(not, y))
app2(app2(and, x), app2(app2(or, y), z)) -> app2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), z))
app2(app2(and, app2(app2(or, y), z)), x) -> app2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), z))

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

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

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

APP2(not, app2(app2(or, x), y)) -> APP2(not, x)
APP2(not, app2(app2(or, x), y)) -> APP2(not, y)
APP2(not, app2(app2(and, x), y)) -> APP2(not, x)
APP2(not, app2(app2(and, x), y)) -> APP2(not, y)

The TRS R consists of the following rules:

app2(not, app2(not, x)) -> x
app2(not, app2(app2(or, x), y)) -> app2(app2(and, app2(not, x)), app2(not, y))
app2(not, app2(app2(and, x), y)) -> app2(app2(or, app2(not, x)), app2(not, y))
app2(app2(and, x), app2(app2(or, y), z)) -> app2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), z))
app2(app2(and, app2(app2(or, y), z)), x) -> app2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), z))

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(not, app2(app2(or, x), y)) -> APP2(not, x)
APP2(not, app2(app2(or, x), y)) -> APP2(not, y)
APP2(not, app2(app2(and, x), y)) -> APP2(not, x)
APP2(not, app2(app2(and, x), y)) -> APP2(not, 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)  =  x2
not  =  not
app2(x1, x2)  =  app2(x1, x2)
or  =  or
and  =  and

Lexicographic Path Order [19].
Precedence:
app2 > not
or > not


The following usable rules [14] were oriented: none



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

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

app2(not, app2(not, x)) -> x
app2(not, app2(app2(or, x), y)) -> app2(app2(and, app2(not, x)), app2(not, y))
app2(not, app2(app2(and, x), y)) -> app2(app2(or, app2(not, x)), app2(not, y))
app2(app2(and, x), app2(app2(or, y), z)) -> app2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), z))
app2(app2(and, app2(app2(or, y), z)), x) -> app2(app2(or, app2(app2(and, x), y)), app2(app2(and, x), 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.