Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

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:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(x1, x2))

(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:

REV(cons(x, l)) → REV2(x, l)
LAST(x, l) → IF(empty(l), x, l)
LAST(x, l) → EMPTY(l)
IF(false, x, l) → LAST(head(l), tail(l))
IF(false, x, l) → HEAD(l)
IF(false, x, l) → TAIL(l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))
REV2(x, cons(y, l)) → REV2(y, l)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

IF(false, x, l) → LAST(head(l), tail(l))
LAST(x, l) → IF(empty(l), x, l)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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

(8) Obligation:

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

REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))
REV(cons(x, l)) → REV2(x, l)
REV2(x, cons(y, l)) → REV2(y, l)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))
REV(cons(x, l)) → REV2(x, l)
REV2(x, cons(y, l)) → REV2(y, l)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
REV2(x1, x2)  =  REV2(x2)
cons(x1, x2)  =  cons(x2)
REV(x1)  =  x1
rev2(x1, x2)  =  x2
nil  =  nil
rev(x1)  =  x1
rev1(x1, x2)  =  x2

Recursive Path Order [RPO].
Precedence:

cons₁ > REV2₁

The following usable rules [FROCOS05] were oriented:

rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))

(10) Obligation:

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

empty(nil) → true
empty(cons(x, l)) → false
head(cons(x, l)) → x
tail(nil) → nil
tail(cons(x, l)) → l
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
last(x, l) → if(empty(l), x, l)
if(true, x, l) → x
if(false, x, l) → last(head(l), tail(l))
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
rev(nil)
rev(cons(x0, x1))
last(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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