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:

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)  =  INC(x1)
s(x1)  =  s(x1)
half(x1)  =  half(x1)
0  =  0
inc(x1)  =  x1
zero(x1)  =  zero(x1)
true  =  true
false  =  false
p(x1)  =  x1
bits(x1)  =  x1
bitIter(x1, x2)  =  x2
if(x1, x2, x3)  =  x3

Lexicographic path order with status [LPO].
Precedence:

INC₁ > 0
half₁ > s₁ > false > 0
zero₁ > true > 0

Status:

zero₁: [1]
half₁: [1]
INC₁: [1]
true: []
false: []
s₁: [1]
0: []

The following usable rules [FROCOS05] were oriented:

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)

(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)  =  HALF(x1)
s(x1)  =  s(x1)
half(x1)  =  half(x1)
0  =  0
inc(x1)  =  x1
zero(x1)  =  zero(x1)
true  =  true
false  =  false
p(x1)  =  x1
bits(x1)  =  x1
bitIter(x1, x2)  =  x2
if(x1, x2, x3)  =  x3

Lexicographic path order with status [LPO].
Precedence:

HALF₁ > 0
s₁ > half₁ > 0
zero₁ > true > 0
zero₁ > false > half₁ > 0

Status:

zero₁: [1]
half₁: [1]
HALF₁: [1]
true: []
false: []
s₁: [1]
0: []

The following usable rules [FROCOS05] were oriented:

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)

(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.