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))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), 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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(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)
LASTBIT(s(s(x))) → LASTBIT(x)
CONV(x) → CONVITER(x, cons(0, nil))
CONVITER(x, l) → IF(zero(x), x, l)
CONVITER(x, l) → ZERO(x)
IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))
IF(false, x, l) → HALF(x)
IF(false, x, l) → LASTBIT(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(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 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LASTBIT(s(s(x))) → LASTBIT(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(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.


LASTBIT(s(s(x))) → LASTBIT(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LASTBIT(x1)  =  x1
s(x1)  =  s(x1)
half(x1)  =  half(x1)
0  =  0
lastbit(x1)  =  lastbit(x1)
zero(x1)  =  x1
true  =  true
false  =  false
conv(x1)  =  conv
conviter(x1, x2)  =  conviter(x2)
cons(x1, x2)  =  cons
nil  =  nil
if(x1, x2, x3)  =  if(x3)

Recursive Path Order [RPO].
Precedence:
half1 > s1 > false > lastbit1 > cons
0 > true > cons
conv > [conviter1, if1] > lastbit1 > cons
nil > cons


The following usable rules [FROCOS05] were oriented:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

(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))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(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))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(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)
half(x1)  =  half(x1)
0  =  0
lastbit(x1)  =  lastbit(x1)
zero(x1)  =  x1
true  =  true
false  =  false
conv(x1)  =  conv
conviter(x1, x2)  =  conviter(x2)
cons(x1, x2)  =  cons
nil  =  nil
if(x1, x2, x3)  =  if(x3)

Recursive Path Order [RPO].
Precedence:
half1 > s1 > false > lastbit1 > cons
0 > true > cons
conv > [conviter1, if1] > lastbit1 > cons
nil > cons


The following usable rules [FROCOS05] were oriented:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

(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))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(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, l) → CONVITER(half(x), cons(lastbit(x), l))
CONVITER(x, l) → IF(zero(x), x, l)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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