Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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

app(app(twice, f), x) → app(f, app(f, x))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(fmap, nil), x) → nil
app(app(fmap, app(app(cons, f), t_f)), x) → app(app(cons, app(f, x)), app(app(fmap, t_f), x))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

app(app(twice, f), x) → app(f, app(f, x))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(fmap, nil), x) → nil
app(app(fmap, app(app(cons, f), t_f)), x) → app(app(cons, app(f, x)), app(app(fmap, t_f), x))

The set Q consists of the following terms:

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

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

APP(app(twice, f), x) → APP(f, app(f, x))
APP(app(twice, f), x) → APP(f, x)
APP(app(map, f), app(app(cons, h), t)) → APP(app(cons, app(f, h)), app(app(map, f), t))
APP(app(map, f), app(app(cons, h), t)) → APP(cons, app(f, h))
APP(app(map, f), app(app(cons, h), t)) → APP(f, h)
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(app(cons, app(f, x)), app(app(fmap, t_f), x))
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(cons, app(f, x))
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(f, x)
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(app(fmap, t_f), x)
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(fmap, t_f)

The TRS R consists of the following rules:

app(app(twice, f), x) → app(f, app(f, x))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(fmap, nil), x) → nil
app(app(fmap, app(app(cons, f), t_f)), x) → app(app(cons, app(f, x)), app(app(fmap, t_f), x))

The set Q consists of the following terms:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 6 less nodes.

(6) Obligation:

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

APP(app(twice, f), x) → APP(f, x)
APP(app(twice, f), x) → APP(f, app(f, x))
APP(app(map, f), app(app(cons, h), t)) → APP(f, h)
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(f, x)

The TRS R consists of the following rules:

app(app(twice, f), x) → app(f, app(f, x))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(fmap, nil), x) → nil
app(app(fmap, app(app(cons, f), t_f)), x) → app(app(cons, app(f, x)), app(app(fmap, t_f), x))

The set Q consists of the following terms:

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

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

APP(app(twice, f), x) → APP(f, x)
APP(app(twice, f), x) → APP(f, app(f, x))
APP(app(map, f), app(app(cons, h), t)) → APP(f, h)
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(f, x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x1)
app(x1, x2)  =  app(x1, x2)
twice  =  twice
map  =  map
cons  =  cons
fmap  =  fmap
t_f  =  t_f
nil  =  nil

Recursive Path Order [RPO].
Precedence:

twice > APP₁
t_f > app₂ > [map, cons, fmap] > APP₁
t_f > app₂ > [map, cons, fmap] > nil

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

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

APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)

The TRS R consists of the following rules:

app(app(twice, f), x) → app(f, app(f, x))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(fmap, nil), x) → nil
app(app(fmap, app(app(cons, f), t_f)), x) → app(app(cons, app(f, x)), app(app(fmap, t_f), x))

The set Q consists of the following terms:

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

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

map1(f, cons(h, t)) → map1(f, t)

The a-transformed usable rules are
none

The following pairs can be oriented strictly and are deleted.

APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:

[map1₂, cons₂]

The following usable rules [FROCOS05] were oriented: none

(10) Obligation:

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

app(app(twice, f), x) → app(f, app(f, x))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(fmap, nil), x) → nil
app(app(fmap, app(app(cons, f), t_f)), x) → app(app(cons, app(f, x)), app(app(fmap, t_f), x))

The set Q consists of the following terms:

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

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

(11) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(12) TRUE