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

(0) Obligation:

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

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
inc(0) → 0
inc(s(x)) → s(inc(x))
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(x)) → x
bits(x) → bitIter(x, 0)
bitIter(x, y) → if(zero(x), x, inc(y))
if(true, x, y) → p(y)
if(false, x, y) → bitIter(half(x), 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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
inc(0) → 0
inc(s(x)) → s(inc(x))
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(x)) → x
bits(x) → bitIter(x, 0)
bitIter(x, y) → if(zero(x), x, inc(y))
if(true, x, y) → p(y)
if(false, x, y) → bitIter(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(x0, x1)
if(true, x0, x1)
if(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:

HALF(s(s(x))) → HALF(x)
INC(s(x)) → INC(x)
BITS(x) → BITITER(x, 0)
BITITER(x, y) → IF(zero(x), x, inc(y))
BITITER(x, y) → ZERO(x)
BITITER(x, y) → INC(y)
IF(true, x, y) → P(y)
IF(false, x, y) → BITITER(half(x), y)
IF(false, x, y) → HALF(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
inc(0) → 0
inc(s(x)) → s(inc(x))
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(x)) → x
bits(x) → bitIter(x, 0)
bitIter(x, y) → if(zero(x), x, inc(y))
if(true, x, y) → p(y)
if(false, x, y) → bitIter(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(x0, x1)
if(true, x0, x1)
if(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 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

INC(s(x)) → INC(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
inc(0) → 0
inc(s(x)) → s(inc(x))
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(x)) → x
bits(x) → bitIter(x, 0)
bitIter(x, y) → if(zero(x), x, inc(y))
if(true, x, y) → p(y)
if(false, x, y) → bitIter(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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.


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

Recursive path order with status [RPO].
Quasi-Precedence:
trivial

Status:
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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
inc(0) → 0
inc(s(x)) → s(inc(x))
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(x)) → x
bits(x) → bitIter(x, 0)
bitIter(x, y) → if(zero(x), x, inc(y))
if(true, x, y) → p(y)
if(false, x, y) → bitIter(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
inc(0) → 0
inc(s(x)) → s(inc(x))
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(x)) → x
bits(x) → bitIter(x, 0)
bitIter(x, y) → if(zero(x), x, inc(y))
if(true, x, y) → p(y)
if(false, x, y) → bitIter(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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: Combined order from the following AFS and order.
HALF(x1)  =  x1
s(x1)  =  s(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
trivial

Status:
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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
inc(0) → 0
inc(s(x)) → s(inc(x))
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(x)) → x
bits(x) → bitIter(x, 0)
bitIter(x, y) → if(zero(x), x, inc(y))
if(true, x, y) → p(y)
if(false, x, y) → bitIter(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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:

IF(false, x, y) → BITITER(half(x), y)
BITITER(x, y) → IF(zero(x), x, inc(y))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
inc(0) → 0
inc(s(x)) → s(inc(x))
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(x)) → x
bits(x) → bitIter(x, 0)
bitIter(x, y) → if(zero(x), x, inc(y))
if(true, x, y) → p(y)
if(false, x, y) → bitIter(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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