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(app2(f, 0), 1), x) -> app2(app2(app2(f, app2(s, x)), x), x)
app2(app2(app2(f, x), y), app2(s, z)) -> app2(s, app2(app2(app2(f, 0), 1), z))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

app2(app2(app2(f, 0), 1), x) -> app2(app2(app2(f, app2(s, x)), x), x)
app2(app2(app2(f, x), y), app2(s, z)) -> app2(s, app2(app2(app2(f, 0), 1), z))

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(app2(f, 0), 1), x) -> APP2(f, app2(s, x))
APP2(app2(app2(f, 0), 1), x) -> APP2(app2(app2(f, app2(s, x)), x), x)
APP2(app2(app2(f, x), y), app2(s, z)) -> APP2(app2(f, 0), 1)
APP2(app2(app2(f, 0), 1), x) -> APP2(app2(f, app2(s, x)), x)
APP2(app2(app2(f, x), y), app2(s, z)) -> APP2(f, 0)
APP2(app2(app2(f, x), y), app2(s, z)) -> APP2(app2(app2(f, 0), 1), z)
APP2(app2(app2(f, x), y), app2(s, z)) -> APP2(s, app2(app2(app2(f, 0), 1), z))
APP2(app2(app2(f, 0), 1), x) -> APP2(s, x)

The TRS R consists of the following rules:

app2(app2(app2(f, 0), 1), x) -> app2(app2(app2(f, app2(s, x)), x), x)
app2(app2(app2(f, x), y), app2(s, z)) -> app2(s, app2(app2(app2(f, 0), 1), z))

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(app2(f, 0), 1), x) -> APP2(f, app2(s, x))
APP2(app2(app2(f, 0), 1), x) -> APP2(app2(app2(f, app2(s, x)), x), x)
APP2(app2(app2(f, x), y), app2(s, z)) -> APP2(app2(f, 0), 1)
APP2(app2(app2(f, 0), 1), x) -> APP2(app2(f, app2(s, x)), x)
APP2(app2(app2(f, x), y), app2(s, z)) -> APP2(f, 0)
APP2(app2(app2(f, x), y), app2(s, z)) -> APP2(app2(app2(f, 0), 1), z)
APP2(app2(app2(f, x), y), app2(s, z)) -> APP2(s, app2(app2(app2(f, 0), 1), z))
APP2(app2(app2(f, 0), 1), x) -> APP2(s, x)

The TRS R consists of the following rules:

app2(app2(app2(f, 0), 1), x) -> app2(app2(app2(f, app2(s, x)), x), x)
app2(app2(app2(f, x), y), app2(s, z)) -> app2(s, app2(app2(app2(f, 0), 1), z))

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 6 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof

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

APP2(app2(app2(f, 0), 1), x) -> APP2(app2(app2(f, app2(s, x)), x), x)
APP2(app2(app2(f, x), y), app2(s, z)) -> APP2(app2(app2(f, 0), 1), z)

The TRS R consists of the following rules:

app2(app2(app2(f, 0), 1), x) -> app2(app2(app2(f, app2(s, x)), x), x)
app2(app2(app2(f, x), y), app2(s, z)) -> app2(s, app2(app2(app2(f, 0), 1), z))

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(app2(f, x), y), app2(s, z)) -> APP2(app2(app2(f, 0), 1), z)
The remaining pairs can at least by weakly be oriented.

APP2(app2(app2(f, 0), 1), x) -> APP2(app2(app2(f, app2(s, x)), x), x)
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x2)
app2(x1, x2)  =  app1(x2)
f  =  f
0  =  0
1  =  1
s  =  s

Lexicographic Path Order [19].
Precedence:
[APP1, app1] > f > s
0 > s
1 > s


The following usable rules [14] were oriented: none



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

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

APP2(app2(app2(f, 0), 1), x) -> APP2(app2(app2(f, app2(s, x)), x), x)

The TRS R consists of the following rules:

app2(app2(app2(f, 0), 1), x) -> app2(app2(app2(f, app2(s, x)), x), x)
app2(app2(app2(f, x), y), app2(s, z)) -> app2(s, app2(app2(app2(f, 0), 1), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.