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(twice, f), x) -> app2(f, app2(f, x))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(fmap, nil), x) -> nil
app2(app2(fmap, app2(app2(cons, f), t_f)), x) -> app2(app2(cons, app2(f, x)), app2(app2(fmap, t_f), x))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

app2(app2(twice, f), x) -> app2(f, app2(f, x))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(fmap, nil), x) -> nil
app2(app2(fmap, app2(app2(cons, f), t_f)), x) -> app2(app2(cons, app2(f, x)), app2(app2(fmap, t_f), x))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

app2(app2(twice, f), x) -> app2(f, app2(f, x))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(fmap, nil), x) -> nil
app2(app2(fmap, app2(app2(cons, f), t_f)), x) -> app2(app2(cons, app2(f, x)), app2(app2(fmap, t_f), x))

The set Q consists of the following terms:

app2(app2(twice, x0), x1)
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(app2(fmap, nil), x0)
app2(app2(fmap, app2(app2(cons, x0), t_f)), x1)


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

APP2(app2(fmap, app2(app2(cons, f), t_f)), x) -> APP2(f, x)
APP2(app2(twice, f), x) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(app2(cons, app2(f, h)), app2(app2(map, f), t))
APP2(app2(fmap, app2(app2(cons, f), t_f)), x) -> APP2(app2(cons, app2(f, x)), app2(app2(fmap, t_f), x))
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(f, h)
APP2(app2(fmap, app2(app2(cons, f), t_f)), x) -> APP2(fmap, t_f)
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(app2(map, f), t)
APP2(app2(twice, f), x) -> APP2(f, app2(f, x))
APP2(app2(fmap, app2(app2(cons, f), t_f)), x) -> APP2(app2(fmap, t_f), x)
APP2(app2(fmap, app2(app2(cons, f), t_f)), x) -> APP2(cons, app2(f, x))
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(cons, app2(f, h))

The TRS R consists of the following rules:

app2(app2(twice, f), x) -> app2(f, app2(f, x))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(fmap, nil), x) -> nil
app2(app2(fmap, app2(app2(cons, f), t_f)), x) -> app2(app2(cons, app2(f, x)), app2(app2(fmap, t_f), x))

The set Q consists of the following terms:

app2(app2(twice, x0), x1)
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(app2(fmap, nil), x0)
app2(app2(fmap, app2(app2(cons, x0), t_f)), x1)

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

APP2(app2(fmap, app2(app2(cons, f), t_f)), x) -> APP2(f, x)
APP2(app2(twice, f), x) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(app2(cons, app2(f, h)), app2(app2(map, f), t))
APP2(app2(fmap, app2(app2(cons, f), t_f)), x) -> APP2(app2(cons, app2(f, x)), app2(app2(fmap, t_f), x))
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(f, h)
APP2(app2(fmap, app2(app2(cons, f), t_f)), x) -> APP2(fmap, t_f)
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(app2(map, f), t)
APP2(app2(twice, f), x) -> APP2(f, app2(f, x))
APP2(app2(fmap, app2(app2(cons, f), t_f)), x) -> APP2(app2(fmap, t_f), x)
APP2(app2(fmap, app2(app2(cons, f), t_f)), x) -> APP2(cons, app2(f, x))
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(cons, app2(f, h))

The TRS R consists of the following rules:

app2(app2(twice, f), x) -> app2(f, app2(f, x))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(fmap, nil), x) -> nil
app2(app2(fmap, app2(app2(cons, f), t_f)), x) -> app2(app2(cons, app2(f, x)), app2(app2(fmap, t_f), x))

The set Q consists of the following terms:

app2(app2(twice, x0), x1)
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(app2(fmap, nil), x0)
app2(app2(fmap, app2(app2(cons, x0), t_f)), x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 6 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof

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

APP2(app2(fmap, app2(app2(cons, f), t_f)), x) -> APP2(f, x)
APP2(app2(twice, f), x) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(f, h)
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(app2(map, f), t)
APP2(app2(twice, f), x) -> APP2(f, app2(f, x))

The TRS R consists of the following rules:

app2(app2(twice, f), x) -> app2(f, app2(f, x))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(fmap, nil), x) -> nil
app2(app2(fmap, app2(app2(cons, f), t_f)), x) -> app2(app2(cons, app2(f, x)), app2(app2(fmap, t_f), x))

The set Q consists of the following terms:

app2(app2(twice, x0), x1)
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(app2(fmap, nil), x0)
app2(app2(fmap, app2(app2(cons, x0), t_f)), x1)

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(fmap, app2(app2(cons, f), t_f)), x) -> APP2(f, x)
APP2(app2(twice, f), x) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(f, h)
APP2(app2(twice, f), x) -> APP2(f, app2(f, x))
Used argument filtering: APP2(x1, x2)  =  x1
app2(x1, x2)  =  app2(x1, x2)
fmap  =  fmap
cons  =  cons
t_f  =  t_f
twice  =  twice
map  =  map
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof
                ↳ QDP
                  ↳ QDPAfsSolverProof

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

APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(app2(map, f), t)

The TRS R consists of the following rules:

app2(app2(twice, f), x) -> app2(f, app2(f, x))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(fmap, nil), x) -> nil
app2(app2(fmap, app2(app2(cons, f), t_f)), x) -> app2(app2(cons, app2(f, x)), app2(app2(fmap, t_f), x))

The set Q consists of the following terms:

app2(app2(twice, x0), x1)
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(app2(fmap, nil), x0)
app2(app2(fmap, app2(app2(cons, x0), t_f)), x1)

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(map, f), app2(app2(cons, h), t)) -> APP2(app2(map, f), t)
Used argument filtering: APP2(x1, x2)  =  x2
app2(x1, x2)  =  app1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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(twice, f), x) -> app2(f, app2(f, x))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(fmap, nil), x) -> nil
app2(app2(fmap, app2(app2(cons, f), t_f)), x) -> app2(app2(cons, app2(f, x)), app2(app2(fmap, t_f), x))

The set Q consists of the following terms:

app2(app2(twice, x0), x1)
app2(app2(map, x0), nil)
app2(app2(map, x0), app2(app2(cons, x1), x2))
app2(app2(fmap, nil), x0)
app2(app2(fmap, app2(app2(cons, x0), t_f)), x1)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.