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


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


NO
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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 disproven:

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 18 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) TransformationProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) TransformationProof [EQUIVALENT, 0 ms]
        (15) QDP
        (16) NonTerminationLoopProof [COMPLETE, 0 ms]
        (17) NO
    (18) QDP
        (19) UsableRulesProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QReductionProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (24) YES
    (25) QDP
        (26) UsableRulesProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) QReductionProof [EQUIVALENT, 0 ms]
        (29) QDP
        (30) QDPOrderProof [EQUIVALENT, 9 ms]
        (31) QDP
        (32) PisEmptyProof [EQUIVALENT, 0 ms]
        (33) YES


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

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

   filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
   filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
   sieve(cons(0, Y)) -> cons(0, sieve(Y))
   sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
   nats(N) -> cons(N, nats(s(N)))
   zprimes -> sieve(nats(s(s(0))))

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:

   filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
   filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
   sieve(cons(0, Y)) -> cons(0, sieve(Y))
   sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
   nats(N) -> cons(N, nats(s(N)))
   zprimes -> sieve(nats(s(s(0))))

The set Q consists of the following terms:

   filter(cons(x0, x1), 0, x2)
   filter(cons(x0, x1), s(x2), x3)
   sieve(cons(0, x0))
   sieve(cons(s(x0), x1))
   nats(x0)
   zprimes


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

(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:

   FILTER(cons(X, Y), 0, M) -> FILTER(Y, M, M)
   FILTER(cons(X, Y), s(N), M) -> FILTER(Y, N, M)
   SIEVE(cons(0, Y)) -> SIEVE(Y)
   SIEVE(cons(s(N), Y)) -> SIEVE(filter(Y, N, N))
   SIEVE(cons(s(N), Y)) -> FILTER(Y, N, N)
   NATS(N) -> NATS(s(N))
   ZPRIMES -> SIEVE(nats(s(s(0))))
   ZPRIMES -> NATS(s(s(0)))

The TRS R consists of the following rules:

   filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
   filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
   sieve(cons(0, Y)) -> cons(0, sieve(Y))
   sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
   nats(N) -> cons(N, nats(s(N)))
   zprimes -> sieve(nats(s(s(0))))

The set Q consists of the following terms:

   filter(cons(x0, x1), 0, x2)
   filter(cons(x0, x1), s(x2), x3)
   sieve(cons(0, x0))
   sieve(cons(s(x0), x1))
   nats(x0)
   zprimes

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 3 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   NATS(N) -> NATS(s(N))

The TRS R consists of the following rules:

   filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
   filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
   sieve(cons(0, Y)) -> cons(0, sieve(Y))
   sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
   nats(N) -> cons(N, nats(s(N)))
   zprimes -> sieve(nats(s(s(0))))

The set Q consists of the following terms:

   filter(cons(x0, x1), 0, x2)
   filter(cons(x0, x1), s(x2), x3)
   sieve(cons(0, x0))
   sieve(cons(s(x0), x1))
   nats(x0)
   zprimes

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:

   NATS(N) -> NATS(s(N))

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

   filter(cons(x0, x1), 0, x2)
   filter(cons(x0, x1), s(x2), x3)
   sieve(cons(0, x0))
   sieve(cons(s(x0), x1))
   nats(x0)
   zprimes

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

(10) 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].

   filter(cons(x0, x1), 0, x2)
   filter(cons(x0, x1), s(x2), x3)
   sieve(cons(0, x0))
   sieve(cons(s(x0), x1))
   nats(x0)
   zprimes


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

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   NATS(N) -> NATS(s(N))

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

(12) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule NATS(N) -> NATS(s(N)) we obtained the following new rules [LPAR04]:

   (NATS(s(z0)) -> NATS(s(s(z0))),NATS(s(z0)) -> NATS(s(s(z0))))


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

(13)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   NATS(s(z0)) -> NATS(s(s(z0)))

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

(14) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule NATS(s(z0)) -> NATS(s(s(z0))) we obtained the following new rules [LPAR04]:

   (NATS(s(s(z0))) -> NATS(s(s(s(z0)))),NATS(s(s(z0))) -> NATS(s(s(s(z0)))))


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

(15)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   NATS(s(s(z0))) -> NATS(s(s(s(z0))))

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

(16) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = NATS(s(s(z0))) evaluates to  t =NATS(s(s(s(z0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [z0 / s(z0)]
* Semiunifier: [ ]

--------------------------------------------------------------------------------
Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from NATS(s(s(z0))) to NATS(s(s(s(z0)))).




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

(17)
NO

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

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FILTER(cons(X, Y), s(N), M) -> FILTER(Y, N, M)
   FILTER(cons(X, Y), 0, M) -> FILTER(Y, M, M)

The TRS R consists of the following rules:

   filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
   filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
   sieve(cons(0, Y)) -> cons(0, sieve(Y))
   sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
   nats(N) -> cons(N, nats(s(N)))
   zprimes -> sieve(nats(s(s(0))))

The set Q consists of the following terms:

   filter(cons(x0, x1), 0, x2)
   filter(cons(x0, x1), s(x2), x3)
   sieve(cons(0, x0))
   sieve(cons(s(x0), x1))
   nats(x0)
   zprimes

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:

   FILTER(cons(X, Y), s(N), M) -> FILTER(Y, N, M)
   FILTER(cons(X, Y), 0, M) -> FILTER(Y, M, M)

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

   filter(cons(x0, x1), 0, x2)
   filter(cons(x0, x1), s(x2), x3)
   sieve(cons(0, x0))
   sieve(cons(s(x0), x1))
   nats(x0)
   zprimes

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

(21) 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].

   filter(cons(x0, x1), 0, x2)
   filter(cons(x0, x1), s(x2), x3)
   sieve(cons(0, x0))
   sieve(cons(s(x0), x1))
   nats(x0)
   zprimes


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

(22)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FILTER(cons(X, Y), s(N), M) -> FILTER(Y, N, M)
   FILTER(cons(X, Y), 0, M) -> FILTER(Y, M, M)

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

(23) 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:
*FILTER(cons(X, Y), s(N), M) -> FILTER(Y, N, M)
The graph contains the following edges 1 > 1, 2 > 2, 3 >= 3


*FILTER(cons(X, Y), 0, M) -> FILTER(Y, M, M)
The graph contains the following edges 1 > 1, 3 >= 2, 3 >= 3


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

(24)
YES

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

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SIEVE(cons(s(N), Y)) -> SIEVE(filter(Y, N, N))
   SIEVE(cons(0, Y)) -> SIEVE(Y)

The TRS R consists of the following rules:

   filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
   filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
   sieve(cons(0, Y)) -> cons(0, sieve(Y))
   sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
   nats(N) -> cons(N, nats(s(N)))
   zprimes -> sieve(nats(s(s(0))))

The set Q consists of the following terms:

   filter(cons(x0, x1), 0, x2)
   filter(cons(x0, x1), s(x2), x3)
   sieve(cons(0, x0))
   sieve(cons(s(x0), x1))
   nats(x0)
   zprimes

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

(26) 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.
----------------------------------------

(27)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SIEVE(cons(s(N), Y)) -> SIEVE(filter(Y, N, N))
   SIEVE(cons(0, Y)) -> SIEVE(Y)

The TRS R consists of the following rules:

   filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
   filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))

The set Q consists of the following terms:

   filter(cons(x0, x1), 0, x2)
   filter(cons(x0, x1), s(x2), x3)
   sieve(cons(0, x0))
   sieve(cons(s(x0), x1))
   nats(x0)
   zprimes

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

(28) 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].

   sieve(cons(0, x0))
   sieve(cons(s(x0), x1))
   nats(x0)
   zprimes


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

(29)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SIEVE(cons(s(N), Y)) -> SIEVE(filter(Y, N, N))
   SIEVE(cons(0, Y)) -> SIEVE(Y)

The TRS R consists of the following rules:

   filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
   filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))

The set Q consists of the following terms:

   filter(cons(x0, x1), 0, x2)
   filter(cons(x0, x1), s(x2), x3)

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

(30) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   SIEVE(cons(s(N), Y)) -> SIEVE(filter(Y, N, N))
   SIEVE(cons(0, Y)) -> SIEVE(Y)
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
SIEVE(x1)  =  x1

cons(x1, x2)  =  cons(x2)

filter(x1, x2, x3)  =  x1


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   dummyConstant=1
   cons_1=1

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

   filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
   filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))


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

(31)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
   filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))

The set Q consists of the following terms:

   filter(cons(x0, x1), 0, x2)
   filter(cons(x0, x1), s(x2), x3)

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

(32) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(33)
YES