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(app2(times, x), 0) -> 0
app2(app2(times, x), app2(s, y)) -> app2(app2(plus, app2(app2(times, x), y)), x)
app2(app2(plus, x), 0) -> x
app2(app2(plus, 0), x) -> x
app2(app2(plus, x), app2(s, y)) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(times, x), 0) -> 0
app2(app2(times, x), app2(s, y)) -> app2(app2(plus, app2(app2(times, x), y)), x)
app2(app2(plus, x), 0) -> x
app2(app2(plus, 0), x) -> x
app2(app2(plus, x), app2(s, y)) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

Q is empty.

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

APP2(app2(times, x), app2(s, y)) -> APP2(app2(times, x), 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), y)
APP2(app2(times, x), app2(s, y)) -> APP2(plus, app2(app2(times, x), y))
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
APP2(app2(plus, x), app2(s, y)) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(plus, x), app2(s, y)) -> APP2(app2(plus, x), y)
APP2(app2(times, x), app2(s, y)) -> APP2(app2(plus, app2(app2(times, x), y)), x)

The TRS R consists of the following rules:

app2(app2(times, x), 0) -> 0
app2(app2(times, x), app2(s, y)) -> app2(app2(plus, app2(app2(times, x), y)), x)
app2(app2(plus, x), 0) -> x
app2(app2(plus, 0), x) -> x
app2(app2(plus, x), app2(s, y)) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), 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(times, x), app2(s, y)) -> APP2(app2(times, x), 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), y)
APP2(app2(times, x), app2(s, y)) -> APP2(plus, app2(app2(times, x), y))
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
APP2(app2(plus, x), app2(s, y)) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(plus, x), app2(s, y)) -> APP2(app2(plus, x), y)
APP2(app2(times, x), app2(s, y)) -> APP2(app2(plus, app2(app2(times, x), y)), x)

The TRS R consists of the following rules:

app2(app2(times, x), 0) -> 0
app2(app2(times, x), app2(s, y)) -> app2(app2(plus, app2(app2(times, x), y)), x)
app2(app2(plus, x), 0) -> x
app2(app2(plus, 0), x) -> x
app2(app2(plus, x), app2(s, y)) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ 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, x), app2(s, y)) -> APP2(app2(plus, x), y)

The TRS R consists of the following rules:

app2(app2(times, x), 0) -> 0
app2(app2(times, x), app2(s, y)) -> app2(app2(plus, app2(app2(times, x), y)), x)
app2(app2(plus, x), 0) -> x
app2(app2(plus, 0), x) -> x
app2(app2(plus, x), app2(s, y)) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

Q is empty.
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(plus, x), app2(s, y)) -> APP2(app2(plus, x), y)
Used argument filtering: APP2(x1, x2)  =  x2
app2(x1, x2)  =  app1(x2)
s  =  s
Used ordering: Quasi Precedence: trivial 


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

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

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

The TRS R consists of the following rules:

app2(app2(times, x), 0) -> 0
app2(app2(times, x), app2(s, y)) -> app2(app2(plus, app2(app2(times, x), y)), x)
app2(app2(plus, x), 0) -> x
app2(app2(plus, 0), x) -> x
app2(app2(plus, x), app2(s, y)) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

Q is empty.
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(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
Used argument filtering: APP2(x1, x2)  =  x1
app2(x1, x2)  =  app1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ QDP

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

app2(app2(times, x), 0) -> 0
app2(app2(times, x), app2(s, y)) -> app2(app2(plus, app2(app2(times, x), y)), x)
app2(app2(plus, x), 0) -> x
app2(app2(plus, 0), x) -> x
app2(app2(plus, x), app2(s, y)) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), 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
            ↳ QDPAfsSolverProof

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

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

The TRS R consists of the following rules:

app2(app2(times, x), 0) -> 0
app2(app2(times, x), app2(s, y)) -> app2(app2(plus, app2(app2(times, x), y)), x)
app2(app2(plus, x), 0) -> x
app2(app2(plus, 0), x) -> x
app2(app2(plus, x), app2(s, y)) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

Q is empty.
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(times, x), app2(s, y)) -> APP2(app2(times, x), y)
Used argument filtering: APP2(x1, x2)  =  x2
app2(x1, x2)  =  app1(x2)
s  =  s
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ PisEmptyProof

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

app2(app2(times, x), 0) -> 0
app2(app2(times, x), app2(s, y)) -> app2(app2(plus, app2(app2(times, x), y)), x)
app2(app2(plus, x), 0) -> x
app2(app2(plus, 0), x) -> x
app2(app2(plus, x), app2(s, y)) -> app2(s, app2(app2(plus, x), y))
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), 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.