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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE

(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) 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(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))

The set Q consists of the following terms:

app(app(app(compose, x0), x1), x2)
app(reverse, x0)
app(app(reverse2, nil), x0)
app(app(reverse2, app(app(cons, x0), x1)), x2)
app(hd, app(app(cons, x0), x1))
app(tl, app(app(cons, x0), x1))
last
init

(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(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))

The set Q consists of the following terms:

app(app(app(compose, x0), x1), x2)
app(reverse, x0)
app(app(reverse2, nil), x0)
app(app(reverse2, app(app(cons, x0), x1)), x2)
app(hd, app(app(cons, x0), x1))
app(tl, app(app(cons, x0), x1))
last
init

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) 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))

The set Q consists of the following terms:

app(app(app(compose, x0), x1), x2)
app(reverse, x0)
app(app(reverse2, nil), x0)
app(app(reverse2, app(app(cons, x0), x1)), x2)
app(hd, app(app(cons, x0), x1))
app(tl, app(app(cons, x0), x1))
last
init

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

(8) 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

reverse21(cons(x, xs), l) → reverse21(xs, cons(x, l))

The a-transformed usable rules are
none


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.
reverse21(x1, x2)  =  x1
cons(x1, x2)  =  cons(x2)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(9) 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))

The set Q consists of the following terms:

app(app(app(compose, x0), x1), x2)
app(reverse, x0)
app(app(reverse2, nil), x0)
app(app(reverse2, app(app(cons, x0), x1)), x2)
app(hd, app(app(cons, x0), x1))
app(tl, app(app(cons, x0), x1))
last
init

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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))

The set Q consists of the following terms:

app(app(app(compose, x0), x1), x2)
app(reverse, x0)
app(app(reverse2, nil), x0)
app(app(reverse2, app(app(cons, x0), x1)), x2)
app(hd, app(app(cons, x0), x1))
app(tl, app(app(cons, x0), x1))
last
init

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

(13) 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

Lexicographic Path Order [LPO].
Precedence:
[app2, compose, reverse, nil] > [reverse2, cons]


The following usable rules [FROCOS05] were oriented: none

(14) 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))

The set Q consists of the following terms:

app(app(app(compose, x0), x1), x2)
app(reverse, x0)
app(app(reverse2, nil), x0)
app(app(reverse2, app(app(cons, x0), x1)), x2)
app(hd, app(app(cons, x0), x1))
app(tl, app(app(cons, x0), x1))
last
init

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE