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, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))

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, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

(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:

P(s(s(x))) → P(s(x))
INTLIST(x) → IF_INTLIST(empty(x), x)
INTLIST(x) → EMPTY(x)
IF_INTLIST(false, x) → HEAD(x)
IF_INTLIST(false, x) → INTLIST(tail(x))
IF_INTLIST(false, x) → TAIL(x)
INT(x, y) → IF_INT(zero(x), zero(y), x, y)
INT(x, y) → ZERO(x)
INT(x, y) → ZERO(y)
IF_INT(true, b, x, y) → IF1(b, x, y)
IF_INT(false, b, x, y) → IF2(b, x, y)
IF1(false, x, y) → INT(s(0), y)
IF2(false, x, y) → INTLIST(int(p(x), p(y)))
IF2(false, x, y) → INT(p(x), p(y))
IF2(false, x, y) → P(x)
IF2(false, x, y) → P(y)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 8 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

IF_INTLIST(false, x) → INTLIST(tail(x))
INTLIST(x) → IF_INTLIST(empty(x), x)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

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

(8) Obligation:

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

P(s(s(x))) → P(s(x))

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

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.

P(s(s(x))) → P(s(x))

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

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:

empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

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

(13) Obligation:

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

IF2(false, x, y) → INT(p(x), p(y))
INT(x, y) → IF_INT(zero(x), zero(y), x, y)
IF_INT(true, b, x, y) → IF1(b, x, y)
IF1(false, x, y) → INT(s(0), y)
IF_INT(false, b, x, y) → IF2(b, x, y)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(x, y)) → false
tail(nil) → nil
tail(cons(x, y)) → y
head(cons(x, y)) → x
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
intlist(x) → if_intlist(empty(x), x)
if_intlist(true, x) → nil
if_intlist(false, x) → cons(s(head(x)), intlist(tail(x)))
int(x, y) → if_int(zero(x), zero(y), x, y)
if_int(true, b, x, y) → if1(b, x, y)
if_int(false, b, x, y) → if2(b, x, y)
if1(true, x, y) → cons(0, nil)
if1(false, x, y) → cons(0, int(s(0), y))
if2(true, x, y) → nil
if2(false, x, y) → intlist(int(p(x), p(y)))

The set Q consists of the following terms:

empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
zero(0)
zero(s(x0))
p(0)
p(s(0))
p(s(s(x0)))
intlist(x0)
if_intlist(true, x0)
if_intlist(false, x0)
int(x0, x1)
if_int(true, x0, x1, x2)
if_int(false, x0, x1, x2)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

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