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))


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 contains 2 SCCs with 6 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

APP2(app2(minus, app2(s, x)), app2(s, y)) -> APP2(app2(minus, x), y)
Used argument filtering: APP2(x1, x2)  =  x2
app2(x1, x2)  =  app1(x2)
s  =  s
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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(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.