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(ack, 0), y) -> app2(succ, y)
app2(app2(ack, app2(succ, x)), y) -> app2(app2(ack, x), app2(succ, 0))
app2(app2(ack, app2(succ, x)), app2(succ, y)) -> app2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(ack, 0), y) -> app2(succ, y)
app2(app2(ack, app2(succ, x)), y) -> app2(app2(ack, x), app2(succ, 0))
app2(app2(ack, app2(succ, x)), app2(succ, y)) -> app2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))

Q is empty.

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

APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
APP2(app2(ack, app2(succ, x)), y) -> APP2(ack, x)
APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, app2(succ, x)), y)
APP2(app2(ack, app2(succ, x)), y) -> APP2(succ, 0)
APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(ack, x)
APP2(app2(ack, app2(succ, x)), y) -> APP2(app2(ack, x), app2(succ, 0))
APP2(app2(ack, 0), y) -> APP2(succ, y)

The TRS R consists of the following rules:

app2(app2(ack, 0), y) -> app2(succ, y)
app2(app2(ack, app2(succ, x)), y) -> app2(app2(ack, x), app2(succ, 0))
app2(app2(ack, app2(succ, x)), app2(succ, y)) -> app2(app2(ack, x), app2(app2(ack, app2(succ, 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(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
APP2(app2(ack, app2(succ, x)), y) -> APP2(ack, x)
APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, app2(succ, x)), y)
APP2(app2(ack, app2(succ, x)), y) -> APP2(succ, 0)
APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(ack, x)
APP2(app2(ack, app2(succ, x)), y) -> APP2(app2(ack, x), app2(succ, 0))
APP2(app2(ack, 0), y) -> APP2(succ, y)

The TRS R consists of the following rules:

app2(app2(ack, 0), y) -> app2(succ, y)
app2(app2(ack, app2(succ, x)), y) -> app2(app2(ack, x), app2(succ, 0))
app2(app2(ack, app2(succ, x)), app2(succ, y)) -> app2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 4 less nodes.

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

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

APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, app2(succ, x)), y)
APP2(app2(ack, app2(succ, x)), y) -> APP2(app2(ack, x), app2(succ, 0))

The TRS R consists of the following rules:

app2(app2(ack, 0), y) -> app2(succ, y)
app2(app2(ack, app2(succ, x)), y) -> app2(app2(ack, x), app2(succ, 0))
app2(app2(ack, app2(succ, x)), app2(succ, y)) -> app2(app2(ack, x), app2(app2(ack, app2(succ, 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(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
APP2(app2(ack, app2(succ, x)), y) -> APP2(app2(ack, x), app2(succ, 0))
Used argument filtering: APP2(x1, x2)  =  x1
app2(x1, x2)  =  app2(x1, x2)
ack  =  ack
succ  =  succ
0  =  0
Used ordering: Precedence:
app2 > 0
ack > succ



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

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

APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, app2(succ, x)), y)

The TRS R consists of the following rules:

app2(app2(ack, 0), y) -> app2(succ, y)
app2(app2(ack, app2(succ, x)), y) -> app2(app2(ack, x), app2(succ, 0))
app2(app2(ack, app2(succ, x)), app2(succ, y)) -> app2(app2(ack, x), app2(app2(ack, app2(succ, 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(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, app2(succ, x)), y)
Used argument filtering: APP2(x1, x2)  =  x2
app2(x1, x2)  =  app1(x2)
succ  =  succ
Used ordering: Precedence:
trivial



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

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

app2(app2(ack, 0), y) -> app2(succ, y)
app2(app2(ack, app2(succ, x)), y) -> app2(app2(ack, x), app2(succ, 0))
app2(app2(ack, app2(succ, x)), app2(succ, y)) -> app2(app2(ack, x), app2(app2(ack, app2(succ, 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.