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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 PisEmptyProof (⇔)
        ↳14 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) 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: Polynomial interpretation [POLO]:

POL(APP(x1, x2)) = x1   
POL(app(x1, x2)) = 1 + x1 + x2   
POL(cons) = 0   
POL(reverse2) = 0   

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: Polynomial interpretation [POLO]:

POL(APP(x1, x2)) = x1   
POL(app(x1, x2)) = 1 + x1 + x2   
POL(compose) = 1   
POL(cons) = 0   
POL(hd) = 0   
POL(nil) = 0   
POL(reverse) = 0   
POL(reverse2) = 0   
POL(tl) = 0   

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