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))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

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)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

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)))
conv(0)
conv(s(x0))

(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(s(x)) → CONV(half(s(x)))
CONV(s(x)) → HALF(s(x))
CONV(s(x)) → LASTBIT(s(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)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

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)))
conv(0)
conv(s(x0))

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 2 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)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

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)))
conv(0)
conv(s(x0))

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)  =  LASTBIT(x1)
s(x1)  =  s(x1)
half(x1)  =  half(x1)
0  =  0
lastbit(x1)  =  lastbit(x1)
conv(x1)  =  conv(x1)
cons(x1, x2)  =  cons(x2)
nil  =  nil

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

LASTBIT₁ > [s₁, half₁, 0]
[conv₁, nil] > lastbit₁ > [s₁, half₁, 0]
[conv₁, nil] > cons₁ > [s₁, half₁, 0]

Status:

LASTBIT₁: multiset
s₁: multiset
half₁: multiset
0: multiset
lastbit₁: [1]
conv₁: [1]
cons₁: multiset
nil: multiset

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)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

(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)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

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)))
conv(0)
conv(s(x0))

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)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

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)))
conv(0)
conv(s(x0))

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
lastbit(x1)  =  lastbit(x1)
conv(x1)  =  conv(x1)
cons(x1, x2)  =  cons(x2)
nil  =  nil

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

HALF₁ > [s₁, half₁, 0]
[conv₁, nil] > lastbit₁ > [s₁, half₁, 0]
[conv₁, nil] > cons₁ > [s₁, half₁, 0]

Status:

HALF₁: multiset
s₁: multiset
half₁: multiset
0: multiset
lastbit₁: [1]
conv₁: [1]
cons₁: multiset
nil: multiset

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)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

(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)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

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)))
conv(0)
conv(s(x0))

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:

CONV(s(x)) → CONV(half(s(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)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))

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)))
conv(0)
conv(s(x0))

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