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(app(compose, f), g), x) → app(g, app(f, x))
app(reverse, l) → app(app(reverse2, l), nil)
app(app(reverse2, nil), l) → l
app(app(reverse2, app(app(cons, x), xs)), l) → app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) → x
app(tl, app(app(cons, x), xs)) → xs
last → app(app(compose, hd), reverse)
init → app(app(compose, reverse), app(app(compose, tl), reverse))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

APP(app(app(compose, f), g), x) → APP(g, app(f, x))
APP(app(app(compose, f), g), x) → APP(f, x)
APP(reverse, l) → APP(app(reverse2, l), nil)
APP(reverse, l) → APP(reverse2, l)
APP(app(reverse2, app(app(cons, x), xs)), l) → APP(app(reverse2, xs), app(app(cons, x), l))
APP(app(reverse2, app(app(cons, x), xs)), l) → APP(reverse2, xs)
APP(app(reverse2, app(app(cons, x), xs)), l) → APP(app(cons, x), l)
LAST → APP(app(compose, hd), reverse)
LAST → APP(compose, hd)
INIT → APP(app(compose, reverse), app(app(compose, tl), reverse))
INIT → APP(compose, reverse)
INIT → APP(app(compose, tl), reverse)
INIT → APP(compose, tl)

The TRS R consists of the following rules:

app(app(app(compose, f), g), x) → app(g, app(f, x))
app(reverse, l) → app(app(reverse2, l), nil)
app(app(reverse2, nil), l) → l
app(app(reverse2, app(app(cons, x), xs)), l) → app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) → x
app(tl, app(app(cons, x), xs)) → xs
last → app(app(compose, hd), reverse)
init → app(app(compose, reverse), app(app(compose, tl), reverse))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 10 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(app(reverse2, app(app(cons, x), xs)), l) → APP(app(reverse2, xs), app(app(cons, x), l))

The TRS R consists of the following rules:

app(app(app(compose, f), g), x) → app(g, app(f, x))
app(reverse, l) → app(app(reverse2, l), nil)
app(app(reverse2, nil), l) → l
app(app(reverse2, app(app(cons, x), xs)), l) → app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) → x
app(tl, app(app(cons, x), xs)) → xs
last → app(app(compose, hd), reverse)
init → app(app(compose, reverse), app(app(compose, tl), reverse))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

APP(app(reverse2, app(app(cons, x), xs)), l) → APP(app(reverse2, xs), app(app(cons, x), l))

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(x2)
reverse2  =  reverse2
cons  =  cons

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

app(app(app(compose, f), g), x) → app(g, app(f, x))
app(reverse, l) → app(app(reverse2, l), nil)
app(app(reverse2, nil), l) → l
app(app(reverse2, app(app(cons, x), xs)), l) → app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) → x
app(tl, app(app(cons, x), xs)) → xs
last → app(app(compose, hd), reverse)
init → app(app(compose, reverse), app(app(compose, tl), reverse))

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

APP(app(app(compose, f), g), x) → APP(f, x)
APP(app(app(compose, f), g), x) → APP(g, app(f, x))

The TRS R consists of the following rules:

app(app(app(compose, f), g), x) → app(g, app(f, x))
app(reverse, l) → app(app(reverse2, l), nil)
app(app(reverse2, nil), l) → l
app(app(reverse2, app(app(cons, x), xs)), l) → app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) → x
app(tl, app(app(cons, x), xs)) → xs
last → app(app(compose, hd), reverse)
init → app(app(compose, reverse), app(app(compose, tl), reverse))

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

APP(app(app(compose, f), g), x) → APP(f, x)
APP(app(app(compose, f), g), x) → APP(g, 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)  =  x1
app(x1, x2)  =  app(x1, x2)
compose  =  compose
reverse  =  reverse
reverse2  =  reverse2
nil  =  nil
cons  =  cons
hd  =  hd
tl  =  tl

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

app(app(app(compose, f), g), x) → app(g, app(f, x))
app(reverse, l) → app(app(reverse2, l), nil)
app(app(reverse2, nil), l) → l
app(app(reverse2, app(app(cons, x), xs)), l) → app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) → x
app(tl, app(app(cons, x), xs)) → xs
last → app(app(compose, hd), reverse)
init → app(app(compose, reverse), app(app(compose, tl), reverse))

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE