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.

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 contains 1 SCC with 6 less nodes.

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

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

APP2(app2(app2(f, x), y), app2(s, z)) -> APP2(app2(app2(f, 0), 1), z)
Used argument filtering: APP2(x1, x2)  =  x2
app2(x1, x2)  =  app1(x2)
s  =  s
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ 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 contains 0 SCCs with 1 less node.