TRS Standard pair #487070853

details

property	value
status	complete
benchmark	ExIntrod_GM99_Z.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n074.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.78296 seconds
cpu usage	16.4776
user time	15.8638
system time	0.613805
max virtual memory	1.83473E7
max residence set size	1573600.0

stage attributes

key	value
starexec-result	NO

output

NO
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 21 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 44 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) TransformationProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) TransformationProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) DependencyGraphProof [EQUIVALENT, 0 ms]
        (15) QDP
        (16) TransformationProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) DependencyGraphProof [EQUIVALENT, 0 ms]
        (19) QDP
        (20) QDPOrderProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) QDPOrderProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) NonTerminationLoopProof [COMPLETE, 0 ms]
        (25) NO
    (26) QDP
        (27) UsableRulesProof [EQUIVALENT, 0 ms]
        (28) QDP
        (29) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (30) YES


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

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

   primes -> sieve(from(s(s(0))))
   from(X) -> cons(X, n__from(s(X)))
   head(cons(X, Y)) -> X
   tail(cons(X, Y)) -> activate(Y)
   if(true, X, Y) -> activate(X)
   if(false, X, Y) -> activate(Y)
   filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
   sieve(cons(X, Y)) -> cons(X, n__filter(X, sieve(activate(Y))))
   from(X) -> n__from(X)
   filter(X1, X2) -> n__filter(X1, X2)
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__filter(X1, X2)) -> filter(X1, X2)
   activate(n__cons(X1, X2)) -> cons(X1, X2)
   activate(X) -> X

Q is empty.

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

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PRIMES -> SIEVE(from(s(s(0))))
   PRIMES -> FROM(s(s(0)))
   FROM(X) -> CONS(X, n__from(s(X)))
   TAIL(cons(X, Y)) -> ACTIVATE(Y)
   IF(true, X, Y) -> ACTIVATE(X)
   IF(false, X, Y) -> ACTIVATE(Y)
   FILTER(s(s(X)), cons(Y, Z)) -> IF(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
   FILTER(s(s(X)), cons(Y, Z)) -> ACTIVATE(Z)
   FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)
   SIEVE(cons(X, Y)) -> CONS(X, n__filter(X, sieve(activate(Y))))
   SIEVE(cons(X, Y)) -> SIEVE(activate(Y))
   SIEVE(cons(X, Y)) -> ACTIVATE(Y)
   ACTIVATE(n__from(X)) -> FROM(X)
   ACTIVATE(n__filter(X1, X2)) -> FILTER(X1, X2)
   ACTIVATE(n__cons(X1, X2)) -> CONS(X1, X2)

The TRS R consists of the following rules:

   primes -> sieve(from(s(s(0))))
   from(X) -> cons(X, n__from(s(X)))
   head(cons(X, Y)) -> X
   tail(cons(X, Y)) -> activate(Y)
   if(true, X, Y) -> activate(X)
   if(false, X, Y) -> activate(Y)

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job log

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