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:

a(a(divides, 0), a(s, y)) → true
a(a(divides, a(s, x)), a(s, y)) → a(a(a(div2, x), a(s, y)), y)
a(a(a(div2, x), y), 0) → a(a(divides, x), y)
a(a(a(div2, 0), y), a(s, z)) → false
a(a(a(div2, a(s, x)), y), a(s, z)) → a(a(a(div2, x), y), z)
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), x) → a(not2, a(f, x))
a(not2, true) → false
a(not2, false) → true
a(sieve, nil) → nil
a(sieve, a(a(cons, x), xs)) → a(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))

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:

a(a(divides, 0), a(s, y)) → true
a(a(divides, a(s, x)), a(s, y)) → a(a(a(div2, x), a(s, y)), y)
a(a(a(div2, x), y), 0) → a(a(divides, x), y)
a(a(a(div2, 0), y), a(s, z)) → false
a(a(a(div2, a(s, x)), y), a(s, z)) → a(a(a(div2, x), y), z)
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), x) → a(not2, a(f, x))
a(not2, true) → false
a(not2, false) → true
a(sieve, nil) → nil
a(sieve, a(a(cons, x), xs)) → a(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))

The set Q consists of the following terms:

a(a(divides, 0), a(s, x0))
a(a(divides, a(s, x0)), a(s, x1))
a(a(a(div2, x0), x1), 0)
a(a(a(div2, 0), x0), a(s, x1))
a(a(a(div2, a(s, x0)), x1), a(s, x2))
a(a(filter, x0), nil)
a(a(filter, x0), a(a(cons, x1), x2))
a(a(a(if, true), x0), x1)
a(a(a(if, false), x0), x1)
a(a(not, x0), x1)
a(not2, true)
a(not2, false)
a(sieve, nil)
a(sieve, a(a(cons, 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:

A(a(divides, a(s, x)), a(s, y)) → A(a(a(div2, x), a(s, y)), y)
A(a(divides, a(s, x)), a(s, y)) → A(a(div2, x), a(s, y))
A(a(divides, a(s, x)), a(s, y)) → A(div2, x)
A(a(a(div2, x), y), 0) → A(a(divides, x), y)
A(a(a(div2, x), y), 0) → A(divides, x)
A(a(a(div2, a(s, x)), y), a(s, z)) → A(a(a(div2, x), y), z)
A(a(a(div2, a(s, x)), y), a(s, z)) → A(a(div2, x), y)
A(a(a(div2, a(s, x)), y), a(s, z)) → A(div2, x)
A(a(filter, f), a(a(cons, x), xs)) → A(a(a(if, a(f, x)), x), a(a(filter, f), xs))
A(a(filter, f), a(a(cons, x), xs)) → A(a(if, a(f, x)), x)
A(a(filter, f), a(a(cons, x), xs)) → A(if, a(f, x))
A(a(filter, f), a(a(cons, x), xs)) → A(f, x)
A(a(filter, f), a(a(cons, x), xs)) → A(a(filter, f), xs)
A(a(a(if, true), x), xs) → A(a(cons, x), xs)
A(a(a(if, true), x), xs) → A(cons, x)
A(a(not, f), x) → A(not2, a(f, x))
A(a(not, f), x) → A(f, x)
A(sieve, a(a(cons, x), xs)) → A(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))
A(sieve, a(a(cons, x), xs)) → A(sieve, a(a(filter, a(not, a(divides, x))), xs))
A(sieve, a(a(cons, x), xs)) → A(a(filter, a(not, a(divides, x))), xs)
A(sieve, a(a(cons, x), xs)) → A(filter, a(not, a(divides, x)))
A(sieve, a(a(cons, x), xs)) → A(not, a(divides, x))
A(sieve, a(a(cons, x), xs)) → A(divides, x)

The TRS R consists of the following rules:

a(a(divides, 0), a(s, y)) → true
a(a(divides, a(s, x)), a(s, y)) → a(a(a(div2, x), a(s, y)), y)
a(a(a(div2, x), y), 0) → a(a(divides, x), y)
a(a(a(div2, 0), y), a(s, z)) → false
a(a(a(div2, a(s, x)), y), a(s, z)) → a(a(a(div2, x), y), z)
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), x) → a(not2, a(f, x))
a(not2, true) → false
a(not2, false) → true
a(sieve, nil) → nil
a(sieve, a(a(cons, x), xs)) → a(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))

The set Q consists of the following terms:

a(a(divides, 0), a(s, x0))
a(a(divides, a(s, x0)), a(s, x1))
a(a(a(div2, x0), x1), 0)
a(a(a(div2, 0), x0), a(s, x1))
a(a(a(div2, a(s, x0)), x1), a(s, x2))
a(a(filter, x0), nil)
a(a(filter, x0), a(a(cons, x1), x2))
a(a(a(if, true), x0), x1)
a(a(a(if, false), x0), x1)
a(a(not, x0), x1)
a(not2, true)
a(not2, false)
a(sieve, nil)
a(sieve, a(a(cons, 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 2 SCCs with 15 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

A(a(a(div2, x), y), 0) → A(a(divides, x), y)
A(a(divides, a(s, x)), a(s, y)) → A(a(a(div2, x), a(s, y)), y)
A(a(a(div2, a(s, x)), y), a(s, z)) → A(a(a(div2, x), y), z)

The TRS R consists of the following rules:

a(a(divides, 0), a(s, y)) → true
a(a(divides, a(s, x)), a(s, y)) → a(a(a(div2, x), a(s, y)), y)
a(a(a(div2, x), y), 0) → a(a(divides, x), y)
a(a(a(div2, 0), y), a(s, z)) → false
a(a(a(div2, a(s, x)), y), a(s, z)) → a(a(a(div2, x), y), z)
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), x) → a(not2, a(f, x))
a(not2, true) → false
a(not2, false) → true
a(sieve, nil) → nil
a(sieve, a(a(cons, x), xs)) → a(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))

The set Q consists of the following terms:

a(a(divides, 0), a(s, x0))
a(a(divides, a(s, x0)), a(s, x1))
a(a(a(div2, x0), x1), 0)
a(a(a(div2, 0), x0), a(s, x1))
a(a(a(div2, a(s, x0)), x1), a(s, x2))
a(a(filter, x0), nil)
a(a(filter, x0), a(a(cons, x1), x2))
a(a(a(if, true), x0), x1)
a(a(a(if, false), x0), x1)
a(a(not, x0), x1)
a(not2, true)
a(not2, false)
a(sieve, nil)
a(sieve, a(a(cons, 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:

A(a(a(div2, x), y), 0) → A(a(divides, x), y)
A(a(divides, a(s, x)), a(s, y)) → A(a(a(div2, x), a(s, y)), y)
A(a(a(div2, a(s, x)), y), a(s, z)) → A(a(a(div2, x), y), z)

R is empty.
The set Q consists of the following terms:

a(a(divides, 0), a(s, x0))
a(a(divides, a(s, x0)), a(s, x1))
a(a(a(div2, x0), x1), 0)
a(a(a(div2, 0), x0), a(s, x1))
a(a(a(div2, a(s, x0)), x1), a(s, x2))
a(a(filter, x0), nil)
a(a(filter, x0), a(a(cons, x1), x2))
a(a(a(if, true), x0), x1)
a(a(a(if, false), x0), x1)
a(a(not, x0), x1)
a(not2, true)
a(not2, false)
a(sieve, nil)
a(sieve, a(a(cons, x0), x1))

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

(10) ATransformationProof (EQUIVALENT transformation)

We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem.

(11) Obligation:

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

div21(x, y, 0) → divides1(x, y)
divides1(s(x), s(y)) → div21(x, s(y), y)
div21(s(x), y, s(z)) → div21(x, y, z)

R is empty.
The set Q consists of the following terms:

divides(0, s(x0))
divides(s(x0), s(x1))
div2(x0, x1, 0)
div2(0, x0, s(x1))
div2(s(x0), x1, s(x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
if(true, x0, x1)
if(false, x0, x1)
not(x0, x1)
not2(true)
not2(false)
sieve(nil)
sieve(cons(x0, x1))

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

(12) 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].

divides(0, s(x0))
divides(s(x0), s(x1))
div2(x0, x1, 0)
div2(0, x0, s(x1))
div2(s(x0), x1, s(x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
if(true, x0, x1)
if(false, x0, x1)
not(x0, x1)
not2(true)
not2(false)
sieve(nil)
sieve(cons(x0, x1))

(13) Obligation:

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

div21(x, y, 0) → divides1(x, y)
divides1(s(x), s(y)) → div21(x, s(y), y)
div21(s(x), y, s(z)) → div21(x, y, z)

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

(14) 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:

• divides1(s(x), s(y)) → div21(x, s(y), y)
  The graph contains the following edges 1 > 1, 2 >= 2, 2 > 3

• div21(s(x), y, s(z)) → div21(x, y, z)
  The graph contains the following edges 1 > 1, 2 >= 2, 3 > 3

• div21(x, y, 0) → divides1(x, y)
  The graph contains the following edges 1 >= 1, 2 >= 2

(15) TRUE

(16) Obligation:

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

A(a(filter, f), a(a(cons, x), xs)) → A(a(filter, f), xs)
A(a(filter, f), a(a(cons, x), xs)) → A(f, x)
A(a(not, f), x) → A(f, x)
A(sieve, a(a(cons, x), xs)) → A(sieve, a(a(filter, a(not, a(divides, x))), xs))
A(sieve, a(a(cons, x), xs)) → A(a(filter, a(not, a(divides, x))), xs)

The TRS R consists of the following rules:

a(a(divides, 0), a(s, y)) → true
a(a(divides, a(s, x)), a(s, y)) → a(a(a(div2, x), a(s, y)), y)
a(a(a(div2, x), y), 0) → a(a(divides, x), y)
a(a(a(div2, 0), y), a(s, z)) → false
a(a(a(div2, a(s, x)), y), a(s, z)) → a(a(a(div2, x), y), z)
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), x) → a(not2, a(f, x))
a(not2, true) → false
a(not2, false) → true
a(sieve, nil) → nil
a(sieve, a(a(cons, x), xs)) → a(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))

The set Q consists of the following terms:

a(a(divides, 0), a(s, x0))
a(a(divides, a(s, x0)), a(s, x1))
a(a(a(div2, x0), x1), 0)
a(a(a(div2, 0), x0), a(s, x1))
a(a(a(div2, a(s, x0)), x1), a(s, x2))
a(a(filter, x0), nil)
a(a(filter, x0), a(a(cons, x1), x2))
a(a(a(if, true), x0), x1)
a(a(a(if, false), x0), x1)
a(a(not, x0), x1)
a(not2, true)
a(not2, false)
a(sieve, nil)
a(sieve, a(a(cons, x0), x1))

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

(17) QDPApplicativeOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].First, we preprocessed all pairs by applying the argument filter which replaces every head symbol by its second argument. Then we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

cons(x, xs) → xs
cons(x, xs) → x
x → x
cons(x, xs) → filter(not(divides(x)), xs)
cons(x, xs) → xs

The a-transformed usable rules are

if(true, x, xs) → cons(x, xs)
filter(f, nil) → nil
if(false, x, xs) → xs
filter(f, cons(x, xs)) → if(notProper, x, filter(f, xs))

The following pairs can be oriented strictly and are deleted.

A(a(filter, f), a(a(cons, x), xs)) → A(a(filter, f), xs)
A(a(filter, f), a(a(cons, x), xs)) → A(f, x)
A(sieve, a(a(cons, x), xs)) → A(a(filter, a(not, a(divides, x))), xs)

The remaining pairs can at least be oriented weakly.

A(a(not, f), x) → A(f, x)
A(sieve, a(a(cons, x), xs)) → A(sieve, a(a(filter, a(not, a(divides, x))), xs))

Used ordering: Polynomial interpretation [POLO]:

POL(cons(x₁, x₂)) = 1 + x₁ + x₂   
POL(divides(x₁)) = 1   
POL(false) = 0   
POL(filter(x₁, x₂)) = x₁ + x₂   
POL(if(x₁, x₂, x₃)) = 1 + x₂ + x₃   
POL(nil) = 0   
POL(not(x₁)) = x₁   
POL(notProper) = 0   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(filter, f), nil) → nil
a(a(a(if, false), x), xs) → xs
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))

(18) Obligation:

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

A(a(not, f), x) → A(f, x)
A(sieve, a(a(cons, x), xs)) → A(sieve, a(a(filter, a(not, a(divides, x))), xs))

The TRS R consists of the following rules:

a(a(divides, 0), a(s, y)) → true
a(a(divides, a(s, x)), a(s, y)) → a(a(a(div2, x), a(s, y)), y)
a(a(a(div2, x), y), 0) → a(a(divides, x), y)
a(a(a(div2, 0), y), a(s, z)) → false
a(a(a(div2, a(s, x)), y), a(s, z)) → a(a(a(div2, x), y), z)
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), x) → a(not2, a(f, x))
a(not2, true) → false
a(not2, false) → true
a(sieve, nil) → nil
a(sieve, a(a(cons, x), xs)) → a(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))

The set Q consists of the following terms:

a(a(divides, 0), a(s, x0))
a(a(divides, a(s, x0)), a(s, x1))
a(a(a(div2, x0), x1), 0)
a(a(a(div2, 0), x0), a(s, x1))
a(a(a(div2, a(s, x0)), x1), a(s, x2))
a(a(filter, x0), nil)
a(a(filter, x0), a(a(cons, x1), x2))
a(a(a(if, true), x0), x1)
a(a(a(if, false), x0), x1)
a(a(not, x0), x1)
a(not2, true)
a(not2, false)
a(sieve, nil)
a(sieve, a(a(cons, x0), x1))

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

(19) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.

(20) Complex Obligation (AND)

(21) Obligation:

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

A(sieve, a(a(cons, x), xs)) → A(sieve, a(a(filter, a(not, a(divides, x))), xs))

The TRS R consists of the following rules:

a(a(divides, 0), a(s, y)) → true
a(a(divides, a(s, x)), a(s, y)) → a(a(a(div2, x), a(s, y)), y)
a(a(a(div2, x), y), 0) → a(a(divides, x), y)
a(a(a(div2, 0), y), a(s, z)) → false
a(a(a(div2, a(s, x)), y), a(s, z)) → a(a(a(div2, x), y), z)
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), x) → a(not2, a(f, x))
a(not2, true) → false
a(not2, false) → true
a(sieve, nil) → nil
a(sieve, a(a(cons, x), xs)) → a(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))

The set Q consists of the following terms:

a(a(divides, 0), a(s, x0))
a(a(divides, a(s, x0)), a(s, x1))
a(a(a(div2, x0), x1), 0)
a(a(a(div2, 0), x0), a(s, x1))
a(a(a(div2, a(s, x0)), x1), a(s, x2))
a(a(filter, x0), nil)
a(a(filter, x0), a(a(cons, x1), x2))
a(a(a(if, true), x0), x1)
a(a(a(if, false), x0), x1)
a(a(not, x0), x1)
a(not2, true)
a(not2, false)
a(sieve, nil)
a(sieve, a(a(cons, x0), x1))

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

(22) QDPApplicativeOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

sieve1(cons(x, xs)) → sieve1(filter(not(divides(x)), xs))

The a-transformed usable rules are

if(true, x, xs) → cons(x, xs)
filter(f, nil) → nil
if(false, x, xs) → xs
filter(f, cons(x, xs)) → if(notProper, x, filter(f, xs))

The following pairs can be oriented strictly and are deleted.

A(sieve, a(a(cons, x), xs)) → A(sieve, a(a(filter, a(not, a(divides, x))), xs))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [POLO]:

POL(cons(x₁, x₂)) = 1 + x₂   
POL(divides(x₁)) = 0   
POL(false) = 0   
POL(filter(x₁, x₂)) = x₂   
POL(if(x₁, x₂, x₃)) = 1 + x₃   
POL(nil) = 0   
POL(not(x₁)) = 0   
POL(notProper) = 0   
POL(sieve1(x₁)) = x₁   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(filter, f), nil) → nil
a(a(a(if, false), x), xs) → xs
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))

(23) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a(a(divides, 0), a(s, y)) → true
a(a(divides, a(s, x)), a(s, y)) → a(a(a(div2, x), a(s, y)), y)
a(a(a(div2, x), y), 0) → a(a(divides, x), y)
a(a(a(div2, 0), y), a(s, z)) → false
a(a(a(div2, a(s, x)), y), a(s, z)) → a(a(a(div2, x), y), z)
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), x) → a(not2, a(f, x))
a(not2, true) → false
a(not2, false) → true
a(sieve, nil) → nil
a(sieve, a(a(cons, x), xs)) → a(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))

The set Q consists of the following terms:

a(a(divides, 0), a(s, x0))
a(a(divides, a(s, x0)), a(s, x1))
a(a(a(div2, x0), x1), 0)
a(a(a(div2, 0), x0), a(s, x1))
a(a(a(div2, a(s, x0)), x1), a(s, x2))
a(a(filter, x0), nil)
a(a(filter, x0), a(a(cons, x1), x2))
a(a(a(if, true), x0), x1)
a(a(a(if, false), x0), x1)
a(a(not, x0), x1)
a(not2, true)
a(not2, false)
a(sieve, nil)
a(sieve, a(a(cons, x0), x1))

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

(24) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(25) TRUE

(26) Obligation:

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

A(a(not, f), x) → A(f, x)

The TRS R consists of the following rules:

a(a(divides, 0), a(s, y)) → true
a(a(divides, a(s, x)), a(s, y)) → a(a(a(div2, x), a(s, y)), y)
a(a(a(div2, x), y), 0) → a(a(divides, x), y)
a(a(a(div2, 0), y), a(s, z)) → false
a(a(a(div2, a(s, x)), y), a(s, z)) → a(a(a(div2, x), y), z)
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), x) → a(not2, a(f, x))
a(not2, true) → false
a(not2, false) → true
a(sieve, nil) → nil
a(sieve, a(a(cons, x), xs)) → a(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))

The set Q consists of the following terms:

a(a(divides, 0), a(s, x0))
a(a(divides, a(s, x0)), a(s, x1))
a(a(a(div2, x0), x1), 0)
a(a(a(div2, 0), x0), a(s, x1))
a(a(a(div2, a(s, x0)), x1), a(s, x2))
a(a(filter, x0), nil)
a(a(filter, x0), a(a(cons, x1), x2))
a(a(a(if, true), x0), x1)
a(a(a(if, false), x0), x1)
a(a(not, x0), x1)
a(not2, true)
a(not2, false)
a(sieve, nil)
a(sieve, a(a(cons, x0), x1))

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

(27) 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.

(28) Obligation:

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

A(a(not, f), x) → A(f, x)

R is empty.
The set Q consists of the following terms:

a(a(divides, 0), a(s, x0))
a(a(divides, a(s, x0)), a(s, x1))
a(a(a(div2, x0), x1), 0)
a(a(a(div2, 0), x0), a(s, x1))
a(a(a(div2, a(s, x0)), x1), a(s, x2))
a(a(filter, x0), nil)
a(a(filter, x0), a(a(cons, x1), x2))
a(a(a(if, true), x0), x1)
a(a(a(if, false), x0), x1)
a(a(not, x0), x1)
a(not2, true)
a(not2, false)
a(sieve, nil)
a(sieve, a(a(cons, x0), x1))

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

(29) 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:

• A(a(not, f), x) → A(f, x)
  The graph contains the following edges 1 > 1, 2 >= 2

(30) TRUE