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.

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(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 [13,14,18] contains 1 SCC with 4 less nodes.

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

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


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 remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP2(x1, x2)
app2(x1, x2)  =  app2(x1, x2)
ack  =  ack
succ  =  succ
0  =  0

Lexicographic Path Order [19].
Precedence:
[ack, 0] > succ > APP2 > app2


The following usable rules [14] were 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))
app2(app2(ack, 0), y) -> app2(succ, y)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ 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.