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(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)

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(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(app(comp, x0), x1), x2)
app(twice, x0)

(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(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(app(comp, f), g), x) → APP(f, app(g, x))
APP(app(app(comp, f), g), x) → APP(g, x)
APP(twice, f) → APP(app(comp, f), f)
APP(twice, f) → APP(comp, f)

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(app(comp, x0), x1), x2)
app(twice, x0)

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 4 less nodes.

(6) Obligation:

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

APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(comp, f), g), x) → APP(f, app(g, x))
APP(app(app(comp, f), g), x) → APP(g, x)

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(app(comp, x0), x1), x2)
app(twice, x0)

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(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(comp, f), g), x) → APP(f, app(g, x))
APP(app(app(comp, f), g), x) → APP(g, x)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

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, x), xs)) → APP(app(map, f), xs)

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(app(comp, x0), x1), x2)
app(twice, x0)

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(x, xs)) → map1(f, xs)

The a-transformed usable rules are
none

The following pairs can be oriented strictly and are deleted.

APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
map1(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

trivial

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(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(app(comp, x0), x1), x2)
app(twice, x0)

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