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


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YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRSToCSRProof [EQUIVALENT, 0 ms]
(2) CSR
(3) CSRInnermostProof [EQUIVALENT, 0 ms]
(4) CSR
(5) CSDependencyPairsProof [EQUIVALENT, 0 ms]
(6) QCSDP
(7) QCSDependencyGraphProof [EQUIVALENT, 0 ms]
(8) QCSDP
(9) QCSDPSubtermProof [EQUIVALENT, 24 ms]
(10) QCSDP
(11) PIsEmptyProof [EQUIVALENT, 0 ms]
(12) YES


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   active(primes) -> mark(sieve(from(s(s(0)))))
   active(from(X)) -> mark(cons(X, from(s(X))))
   active(head(cons(X, Y))) -> mark(X)
   active(tail(cons(X, Y))) -> mark(Y)
   active(if(true, X, Y)) -> mark(X)
   active(if(false, X, Y)) -> mark(Y)
   active(filter(s(s(X)), cons(Y, Z))) -> mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
   active(sieve(cons(X, Y))) -> mark(cons(X, filter(X, sieve(Y))))
   active(sieve(X)) -> sieve(active(X))
   active(from(X)) -> from(active(X))
   active(s(X)) -> s(active(X))
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(head(X)) -> head(active(X))
   active(tail(X)) -> tail(active(X))
   active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
   active(filter(X1, X2)) -> filter(active(X1), X2)
   active(filter(X1, X2)) -> filter(X1, active(X2))
   active(divides(X1, X2)) -> divides(active(X1), X2)
   active(divides(X1, X2)) -> divides(X1, active(X2))
   sieve(mark(X)) -> mark(sieve(X))
   from(mark(X)) -> mark(from(X))
   s(mark(X)) -> mark(s(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   head(mark(X)) -> mark(head(X))
   tail(mark(X)) -> mark(tail(X))
   if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
   filter(mark(X1), X2) -> mark(filter(X1, X2))
   filter(X1, mark(X2)) -> mark(filter(X1, X2))
   divides(mark(X1), X2) -> mark(divides(X1, X2))
   divides(X1, mark(X2)) -> mark(divides(X1, X2))
   proper(primes) -> ok(primes)
   proper(sieve(X)) -> sieve(proper(X))
   proper(from(X)) -> from(proper(X))
   proper(s(X)) -> s(proper(X))
   proper(0) -> ok(0)
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(head(X)) -> head(proper(X))
   proper(tail(X)) -> tail(proper(X))
   proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
   proper(true) -> ok(true)
   proper(false) -> ok(false)
   proper(filter(X1, X2)) -> filter(proper(X1), proper(X2))
   proper(divides(X1, X2)) -> divides(proper(X1), proper(X2))
   sieve(ok(X)) -> ok(sieve(X))
   from(ok(X)) -> ok(from(X))
   s(ok(X)) -> ok(s(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   head(ok(X)) -> ok(head(X))
   tail(ok(X)) -> ok(tail(X))
   if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
   filter(ok(X1), ok(X2)) -> ok(filter(X1, X2))
   divides(ok(X1), ok(X2)) -> ok(divides(X1, X2))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

Q is empty.

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

(1) QTRSToCSRProof (EQUIVALENT)
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:

   active(primes) -> mark(sieve(from(s(s(0)))))
   active(from(X)) -> mark(cons(X, from(s(X))))
   active(head(cons(X, Y))) -> mark(X)
   active(tail(cons(X, Y))) -> mark(Y)
   active(if(true, X, Y)) -> mark(X)
   active(if(false, X, Y)) -> mark(Y)
   active(filter(s(s(X)), cons(Y, Z))) -> mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
   active(sieve(cons(X, Y))) -> mark(cons(X, filter(X, sieve(Y))))
   active(sieve(X)) -> sieve(active(X))
   active(from(X)) -> from(active(X))
   active(s(X)) -> s(active(X))
   active(cons(X1, X2)) -> cons(active(X1), X2)
   active(head(X)) -> head(active(X))
   active(tail(X)) -> tail(active(X))
   active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
   active(filter(X1, X2)) -> filter(active(X1), X2)
   active(filter(X1, X2)) -> filter(X1, active(X2))
   active(divides(X1, X2)) -> divides(active(X1), X2)
   active(divides(X1, X2)) -> divides(X1, active(X2))
   sieve(mark(X)) -> mark(sieve(X))
   from(mark(X)) -> mark(from(X))
   s(mark(X)) -> mark(s(X))
   cons(mark(X1), X2) -> mark(cons(X1, X2))
   head(mark(X)) -> mark(head(X))
   tail(mark(X)) -> mark(tail(X))
   if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
   filter(mark(X1), X2) -> mark(filter(X1, X2))
   filter(X1, mark(X2)) -> mark(filter(X1, X2))
   divides(mark(X1), X2) -> mark(divides(X1, X2))
   divides(X1, mark(X2)) -> mark(divides(X1, X2))
   proper(primes) -> ok(primes)
   proper(sieve(X)) -> sieve(proper(X))
   proper(from(X)) -> from(proper(X))
   proper(s(X)) -> s(proper(X))
   proper(0) -> ok(0)
   proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
   proper(head(X)) -> head(proper(X))
   proper(tail(X)) -> tail(proper(X))
   proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
   proper(true) -> ok(true)
   proper(false) -> ok(false)
   proper(filter(X1, X2)) -> filter(proper(X1), proper(X2))
   proper(divides(X1, X2)) -> divides(proper(X1), proper(X2))
   sieve(ok(X)) -> ok(sieve(X))
   from(ok(X)) -> ok(from(X))
   s(ok(X)) -> ok(s(X))
   cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
   head(ok(X)) -> ok(head(X))
   tail(ok(X)) -> ok(tail(X))
   if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
   filter(ok(X1), ok(X2)) -> ok(filter(X1, X2))
   divides(ok(X1), ok(X2)) -> ok(divides(X1, X2))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

Q is empty.
Special symbols used for the transformation (see [GM04]):
top: top_1, active: active_1, mark: mark_1, ok: ok_1, proper: proper_1
The replacement map contains the following entries:

primes: empty set
sieve: {1}
from: {1}
s: {1}
0: empty set
cons: {1}
head: {1}
tail: {1}
if: {1}
true: empty set
false: empty set
filter: {1, 2}
divides: {1, 2}
The QTRS contained all rules created by the complete Giesl-Middeldorp transformation. Therefore, the inverse transformation is complete (and sound).
----------------------------------------

(2)
Obligation:
Context-sensitive rewrite system:
The TRS R consists of the following rules:

   primes -> sieve(from(s(s(0))))
   from(X) -> cons(X, from(s(X)))
   head(cons(X, Y)) -> X
   tail(cons(X, Y)) -> Y
   if(true, X, Y) -> X
   if(false, X, Y) -> Y
   filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
   sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))

The replacement map contains the following entries:

primes: empty set
sieve: {1}
from: {1}
s: {1}
0: empty set
cons: {1}
head: {1}
tail: {1}
if: {1}
true: empty set
false: empty set
filter: {1, 2}
divides: {1, 2}

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(3) CSRInnermostProof (EQUIVALENT)
The CSR is orthogonal. By [CS_Inn] we can switch to innermost.
----------------------------------------

(4)
Obligation:
Context-sensitive rewrite system:
The TRS R consists of the following rules:

   primes -> sieve(from(s(s(0))))
   from(X) -> cons(X, from(s(X)))
   head(cons(X, Y)) -> X
   tail(cons(X, Y)) -> Y
   if(true, X, Y) -> X
   if(false, X, Y) -> Y
   filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
   sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))

The replacement map contains the following entries:

primes: empty set
sieve: {1}
from: {1}
s: {1}
0: empty set
cons: {1}
head: {1}
tail: {1}
if: {1}
true: empty set
false: empty set
filter: {1, 2}
divides: {1, 2}


Innermost Strategy.

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(5) CSDependencyPairsProof (EQUIVALENT)
Using Improved CS-DPs [LPAR08] we result in the following initial Q-CSDP problem.
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(6)
Obligation:
Q-restricted context-sensitive dependency pair problem:
The symbols in {sieve_1, from_1, s_1, head_1, tail_1, filter_2, divides_2, SIEVE_1, FROM_1, FILTER_2, TAIL_1} are replacing on all positions.
For all symbols f in {cons_2, if_3, IF_3} we have mu(f) = {1}.
The symbols in {U_1} are not replacing on any position.

The ordinary context-sensitive dependency pairs DP_o are:
   PRIMES -> SIEVE(from(s(s(0))))
   PRIMES -> FROM(s(s(0)))
   FILTER(s(s(X)), cons(Y, Z)) -> IF(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))

The collapsing dependency pairs are DP_c:
   TAIL(cons(X, Y)) -> Y
   IF(true, X, Y) -> X
   IF(false, X, Y) -> Y


The hidden terms of R are:

   from(s(x0))
   filter(s(s(x0)), x1)
   filter(x0, sieve(x1))
   sieve(x0)

Every hiding context is built from:
   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@2e1e4dad
   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@6bcdc2a4
   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@41eae19a
   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@472eb744
   aprove.DPFramework.CSDPProblem.QCSDPProblem$1@1b81f929

Hence, the new unhiding pairs DP_u are :
   TAIL(cons(X, Y)) -> U(Y)
   IF(true, X, Y) -> U(X)
   IF(false, X, Y) -> U(Y)
   U(s(x_0)) -> U(x_0)
   U(from(x_0)) -> U(x_0)
   U(filter(x_0, x_1)) -> U(x_0)
   U(filter(x_0, x_1)) -> U(x_1)
   U(sieve(x_0)) -> U(x_0)
   U(cons(x_0, x_1)) -> U(x_0)
   U(from(s(x0))) -> FROM(s(x0))
   U(filter(s(s(x0)), x1)) -> FILTER(s(s(x0)), x1)
   U(filter(x0, sieve(x1))) -> FILTER(x0, sieve(x1))
   U(sieve(x0)) -> SIEVE(x0)

The TRS R consists of the following rules:

   primes -> sieve(from(s(s(0))))
   from(X) -> cons(X, from(s(X)))
   head(cons(X, Y)) -> X
   tail(cons(X, Y)) -> Y
   if(true, X, Y) -> X
   if(false, X, Y) -> Y
   filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
   sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))

The set Q consists of the following terms:

   primes
   from(x0)
   head(cons(x0, x1))
   tail(cons(x0, x1))
   if(true, x0, x1)
   if(false, x0, x1)
   filter(s(s(x0)), cons(x1, x2))
   sieve(cons(x0, x1))


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(7) QCSDependencyGraphProof (EQUIVALENT)
The approximation of the Context-Sensitive Dependency Graph [LPAR08] contains 1 SCC with 8 less nodes.
The rules FILTER(s(s(z0)), cons(z1, z2)) -> IF(divides(s(s(z0)), z1), filter(s(s(z0)), z2), cons(z1, filter(z0, sieve(z1)))) and IF(true, x0, x1) -> U(x0) form no chain, because ECap^mu(IF(divides(s(s(z0)), z1), filter(s(s(z0)), z2), cons(z1, filter(z0, sieve(z1))))) = IF(divides(s(s(z0)), z1), filter(s(s(z0)), z2), cons(z1, filter(z0, sieve(z1)))) does not unify with IF(true, x0, x1). The rules FILTER(s(s(z0)), cons(z1, z2)) -> IF(divides(s(s(z0)), z1), filter(s(s(z0)), z2), cons(z1, filter(z0, sieve(z1)))) and IF(false, x0, x1) -> U(x1) form no chain, because ECap^mu(IF(divides(s(s(z0)), z1), filter(s(s(z0)), z2), cons(z1, filter(z0, sieve(z1))))) = IF(divides(s(s(z0)), z1), filter(s(s(z0)), z2), cons(z1, filter(z0, sieve(z1)))) does not unify with IF(false, x0, x1). 
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(8)
Obligation:
Q-restricted context-sensitive dependency pair problem:
The symbols in {sieve_1, from_1, s_1, head_1, tail_1, filter_2, divides_2} are replacing on all positions.
For all symbols f in {cons_2, if_3} we have mu(f) = {1}.
The symbols in {U_1} are not replacing on any position.

The TRS P consists of the following rules:

   U(s(x_0)) -> U(x_0)
   U(from(x_0)) -> U(x_0)
   U(filter(x_0, x_1)) -> U(x_0)
   U(filter(x_0, x_1)) -> U(x_1)
   U(sieve(x_0)) -> U(x_0)
   U(cons(x_0, x_1)) -> U(x_0)

The TRS R consists of the following rules:

   primes -> sieve(from(s(s(0))))
   from(X) -> cons(X, from(s(X)))
   head(cons(X, Y)) -> X
   tail(cons(X, Y)) -> Y
   if(true, X, Y) -> X
   if(false, X, Y) -> Y
   filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
   sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))

The set Q consists of the following terms:

   primes
   from(x0)
   head(cons(x0, x1))
   tail(cons(x0, x1))
   if(true, x0, x1)
   if(false, x0, x1)
   filter(s(s(x0)), cons(x1, x2))
   sieve(cons(x0, x1))


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(9) QCSDPSubtermProof (EQUIVALENT)
We use the subterm processor [DA_EMMES].


The following pairs can be oriented strictly and are deleted.

   U(s(x_0)) -> U(x_0)
   U(from(x_0)) -> U(x_0)
   U(filter(x_0, x_1)) -> U(x_0)
   U(filter(x_0, x_1)) -> U(x_1)
   U(sieve(x_0)) -> U(x_0)
   U(cons(x_0, x_1)) -> U(x_0)
The remaining pairs can at least be oriented weakly.
none
Used ordering:  Combined order from the following AFS and order.
U(x1)  =  x1


Subterm Order

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(10)
Obligation:
Q-restricted context-sensitive dependency pair problem:
The symbols in {sieve_1, from_1, s_1, head_1, tail_1, filter_2, divides_2} are replacing on all positions.
For all symbols f in {cons_2, if_3} we have mu(f) = {1}.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

   primes -> sieve(from(s(s(0))))
   from(X) -> cons(X, from(s(X)))
   head(cons(X, Y)) -> X
   tail(cons(X, Y)) -> Y
   if(true, X, Y) -> X
   if(false, X, Y) -> Y
   filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
   sieve(cons(X, Y)) -> cons(X, filter(X, sieve(Y)))

The set Q consists of the following terms:

   primes
   from(x0)
   head(cons(x0, x1))
   tail(cons(x0, x1))
   if(true, x0, x1)
   if(false, x0, x1)
   filter(s(s(x0)), cons(x1, x2))
   sieve(cons(x0, x1))


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(11) PIsEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
----------------------------------------

(12)
YES