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

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
  ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 PisEmptyProof (⇔)
    ↳21 TRUE
  ↳22 QDP

(0) Obligation:

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

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

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:

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

(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:

CHECK(s(s(s(x)))) → CHECK(s(x))
HALF(s(s(x))) → HALF(x)
PLUS(s(x), y) → PLUS(x, y)
TIMES(x, y) → TIMESITER(x, y, 0)
TIMESITER(x, y, z) → IF(check(x), x, y, z, plus(z, y))
TIMESITER(x, y, z) → CHECK(x)
TIMESITER(x, y, z) → PLUS(z, y)
IF(odd, x, y, z, u) → TIMESITER(p(x), y, u)
IF(odd, x, y, z, u) → P(x)
IF(even, x, y, z, u) → PLUS(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))
IF(even, x, y, z, u) → TIMESITER(half(x), y, half(z))
IF(even, x, y, z, u) → HALF(x)
IF(even, x, y, z, u) → HALF(z)
IF(even, x, y, z, u) → TIMESITER(half(x), y, half(s(z)))
IF(even, x, y, z, u) → HALF(s(z))

The TRS R consists of the following rules:

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 8 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PLUS(s(x), y) → PLUS(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
[PLUS2, s1]


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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:

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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.


HALF(s(s(x))) → HALF(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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

(17) Obligation:

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

CHECK(s(s(s(x)))) → CHECK(s(x))

The TRS R consists of the following rules:

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CHECK(s(s(s(x)))) → CHECK(s(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
CHECK1 > s1


The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

TIMESITER(x, y, z) → IF(check(x), x, y, z, plus(z, y))
IF(odd, x, y, z, u) → TIMESITER(p(x), y, u)
IF(even, x, y, z, u) → TIMESITER(half(x), y, half(z))
IF(even, x, y, z, u) → TIMESITER(half(x), y, half(s(z)))

The TRS R consists of the following rules:

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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