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(f, 0) -> app2(s, 0)
app2(f, app2(s, x)) -> app2(app2(minus, app2(s, x)), app2(g, app2(f, x)))
app2(g, 0) -> 0
app2(g, app2(s, x)) -> app2(app2(minus, app2(s, x)), app2(f, app2(g, x)))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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(f, 0) -> app2(s, 0)
app2(f, app2(s, x)) -> app2(app2(minus, app2(s, x)), app2(g, app2(f, x)))
app2(g, 0) -> 0
app2(g, app2(s, x)) -> app2(app2(minus, app2(s, x)), app2(f, app2(g, x)))

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(app2(minus, x), 0) -> x
app2(app2(minus, app2(s, x)), app2(s, y)) -> app2(app2(minus, x), y)
app2(f, 0) -> app2(s, 0)
app2(f, app2(s, x)) -> app2(app2(minus, app2(s, x)), app2(g, app2(f, x)))
app2(g, 0) -> 0
app2(g, app2(s, x)) -> app2(app2(minus, app2(s, x)), app2(f, app2(g, x)))

The set Q consists of the following terms:

app2(app2(minus, x0), 0)
app2(app2(minus, app2(s, x0)), app2(s, x1))
app2(f, 0)
app2(f, app2(s, x0))
app2(g, 0)
app2(g, app2(s, x0))


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(f, app2(s, x)) -> APP2(f, x)
APP2(f, app2(s, x)) -> APP2(minus, app2(s, x))
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(minus, x)
APP2(f, 0) -> APP2(s, 0)
APP2(g, app2(s, x)) -> APP2(g, x)
APP2(g, app2(s, x)) -> APP2(f, app2(g, x))
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)
APP2(g, app2(s, x)) -> APP2(minus, app2(s, x))
APP2(g, app2(s, x)) -> APP2(app2(minus, app2(s, x)), app2(f, app2(g, x)))
APP2(f, app2(s, x)) -> APP2(g, app2(f, x))
APP2(f, app2(s, x)) -> APP2(app2(minus, app2(s, x)), app2(g, app2(f, 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(f, 0) -> app2(s, 0)
app2(f, app2(s, x)) -> app2(app2(minus, app2(s, x)), app2(g, app2(f, x)))
app2(g, 0) -> 0
app2(g, app2(s, x)) -> app2(app2(minus, app2(s, x)), app2(f, app2(g, x)))

The set Q consists of the following terms:

app2(app2(minus, x0), 0)
app2(app2(minus, app2(s, x0)), app2(s, x1))
app2(f, 0)
app2(f, app2(s, x0))
app2(g, 0)
app2(g, app2(s, x0))

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(f, app2(s, x)) -> APP2(f, x)
APP2(f, app2(s, x)) -> APP2(minus, app2(s, x))
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(minus, x)
APP2(f, 0) -> APP2(s, 0)
APP2(g, app2(s, x)) -> APP2(g, x)
APP2(g, app2(s, x)) -> APP2(f, app2(g, x))
APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)
APP2(g, app2(s, x)) -> APP2(minus, app2(s, x))
APP2(g, app2(s, x)) -> APP2(app2(minus, app2(s, x)), app2(f, app2(g, x)))
APP2(f, app2(s, x)) -> APP2(g, app2(f, x))
APP2(f, app2(s, x)) -> APP2(app2(minus, app2(s, x)), app2(g, app2(f, 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(f, 0) -> app2(s, 0)
app2(f, app2(s, x)) -> app2(app2(minus, app2(s, x)), app2(g, app2(f, x)))
app2(g, 0) -> 0
app2(g, app2(s, x)) -> app2(app2(minus, app2(s, x)), app2(f, app2(g, x)))

The set Q consists of the following terms:

app2(app2(minus, x0), 0)
app2(app2(minus, app2(s, x0)), app2(s, x1))
app2(f, 0)
app2(f, app2(s, x0))
app2(g, 0)
app2(g, app2(s, x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 6 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ 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(f, 0) -> app2(s, 0)
app2(f, app2(s, x)) -> app2(app2(minus, app2(s, x)), app2(g, app2(f, x)))
app2(g, 0) -> 0
app2(g, app2(s, x)) -> app2(app2(minus, app2(s, x)), app2(f, app2(g, x)))

The set Q consists of the following terms:

app2(app2(minus, x0), 0)
app2(app2(minus, app2(s, x0)), app2(s, x1))
app2(f, 0)
app2(f, app2(s, x0))
app2(g, 0)
app2(g, app2(s, x0))

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)  =  APP1(x2)
app2(x1, x2)  =  app1(x2)
minus  =  minus
s  =  s

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

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ 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(f, 0) -> app2(s, 0)
app2(f, app2(s, x)) -> app2(app2(minus, app2(s, x)), app2(g, app2(f, x)))
app2(g, 0) -> 0
app2(g, app2(s, x)) -> app2(app2(minus, app2(s, x)), app2(f, app2(g, x)))

The set Q consists of the following terms:

app2(app2(minus, x0), 0)
app2(app2(minus, app2(s, x0)), app2(s, x1))
app2(f, 0)
app2(f, app2(s, x0))
app2(g, 0)
app2(g, app2(s, x0))

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

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

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

APP2(f, app2(s, x)) -> APP2(f, x)
APP2(g, app2(s, x)) -> APP2(g, x)
APP2(g, app2(s, x)) -> APP2(f, app2(g, x))
APP2(f, app2(s, x)) -> APP2(g, app2(f, 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(f, 0) -> app2(s, 0)
app2(f, app2(s, x)) -> app2(app2(minus, app2(s, x)), app2(g, app2(f, x)))
app2(g, 0) -> 0
app2(g, app2(s, x)) -> app2(app2(minus, app2(s, x)), app2(f, app2(g, x)))

The set Q consists of the following terms:

app2(app2(minus, x0), 0)
app2(app2(minus, app2(s, x0)), app2(s, x1))
app2(f, 0)
app2(f, app2(s, x0))
app2(g, 0)
app2(g, app2(s, x0))

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