Termination w.r.t. Q of the given QTRS could be proven:

(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) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

LASTBIT(s(s(x))) → LASTBIT(x)

R is empty.
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) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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)

(11) Obligation:

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

LASTBIT(s(s(x))) → LASTBIT(x)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• LASTBIT(s(s(x))) → LASTBIT(x)
  The graph contains the following edges 1 > 1

(13) TRUE

(14) 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.

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

HALF(s(s(x))) → HALF(x)

R is empty.
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.

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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)

(18) Obligation:

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

HALF(s(s(x))) → HALF(x)

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• HALF(s(s(x))) → HALF(x)
  The graph contains the following edges 1 > 1

(20) TRUE

(21) 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.

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) 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:

zero(0) → true
zero(s(x)) → false
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)

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.

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

(25) 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:

zero(0) → true
zero(s(x)) → false
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)

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))

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONVITER(x, l) → IF(zero(x), x, l)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(IF(x₁, x₂, x₃)) = (1/2)x₁ + (7/4)x₂ + (1/2)x₃   
POL(false) = 1/2   
POL(CONVITER(x₁, x₂)) = 1/4 + (2)x₁ + (1/2)x₂   
POL(half(x₁)) = (1/2)x₁   
POL(cons(x₁, x₂)) = (1/2)x₂   
POL(lastbit(x₁)) = 1/2   
POL(zero(x₁)) = 1/4 + (1/2)x₁   
POL(0) = 0   
POL(true) = 1/4   
POL(s(x₁)) = 1/2 + (9/4)x₁   

The value of delta used in the strict ordering is 1/8.
The following usable rules [FROCOS05] were oriented:

zero(0) → true
zero(s(x)) → false
half(s(s(x))) → s(half(x))
half(0) → 0
half(s(0)) → 0

(27) Obligation:

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

IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))

The TRS R consists of the following rules:

zero(0) → true
zero(s(x)) → false
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)

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))

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

(28) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(29) TRUE