/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 59 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) ATransformationProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QReductionProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) QDPApplicativeOrderProof [EQUIVALENT, 31 ms]
        (18) QDP
        (19) UsableRulesProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (22) YES


----------------------------------------

(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)
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)
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)
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)
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)
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)
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)
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)
YES

----------------------------------------

(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)
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 [LPAR04,JAR06] 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 

   filter(f, cons(x, xs)) -> if(notProper, x, filter(f, xs))
   if(true, x, xs) -> cons(x, xs)
   filter(f, nil) -> nil
   if(false, x, xs) -> 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(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 remaining pairs can at least be oriented weakly.

   A(a(not, f), x) -> A(f, x)
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( filter_2(x_1, x_2) ) = max{0, 2x_2 - 1}
POL( cons_2(x_1, x_2) ) = 2x_1 + 2x_2 + 1
POL( if_3(x_1, ..., x_3) ) = 2x_2 + 2x_3 + 1
POL( notProper ) = 0
POL( true ) = 0
POL( nil ) = 2
POL( false ) = 0
POL( not_1(x_1) ) = max{0, -2}
POL( divides_1(x_1) ) = x_1 + 1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   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(filter, f), nil) -> nil
   a(a(a(if, false), x), xs) -> xs


----------------------------------------

(18)
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.
----------------------------------------

(19) UsableRulesProof (EQUIVALENT)
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.
----------------------------------------

(20)
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.
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(21) QDPSizeChangeProof (EQUIVALENT)
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


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(22)
YES