details

property	value
status	complete
benchmark	Ex9_BLR02_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n140.star.cs.uiowa.edu
space	Transformed_CSR_innermost_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.4661 seconds
cpu usage	6.61136
user time	6.34094
system time	0.270417
max virtual memory	1.849014E7
max residence set size	428084.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 36 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 111 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QReductionProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M)) a__filter(cons(X, Y), s(N), M) -> cons(mark(X), filter(Y, N, M)) a__sieve(cons(0, Y)) -> cons(0, sieve(Y)) a__sieve(cons(s(N), Y)) -> cons(s(mark(N)), sieve(filter(Y, N, N))) a__nats(N) -> cons(mark(N), nats(s(N))) a__zprimes -> a__sieve(a__nats(s(s(0)))) mark(filter(X1, X2, X3)) -> a__filter(mark(X1), mark(X2), mark(X3)) mark(sieve(X)) -> a__sieve(mark(X)) mark(nats(X)) -> a__nats(mark(X)) mark(zprimes) -> a__zprimes mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(s(X)) -> s(mark(X)) a__filter(X1, X2, X3) -> filter(X1, X2, X3) a__sieve(X) -> sieve(X) a__nats(X) -> nats(X) a__zprimes -> zprimes The set Q consists of the following terms: a__nats(x0) a__zprimes mark(filter(x0, x1, x2)) mark(sieve(x0)) mark(nats(x0)) mark(zprimes) mark(cons(x0, x1)) mark(0) mark(s(x0)) a__filter(x0, x1, x2) a__sieve(x0) ---------------------------------------- (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: A__FILTER(cons(X, Y), s(N), M) -> MARK(X) A__SIEVE(cons(s(N), Y)) -> MARK(N) A__NATS(N) -> MARK(N) A__ZPRIMES -> A__SIEVE(a__nats(s(s(0)))) A__ZPRIMES -> A__NATS(s(s(0))) MARK(filter(X1, X2, X3)) -> A__FILTER(mark(X1), mark(X2), mark(X3)) MARK(filter(X1, X2, X3)) -> MARK(X1) MARK(filter(X1, X2, X3)) -> MARK(X2) MARK(filter(X1, X2, X3)) -> MARK(X3) MARK(sieve(X)) -> A__SIEVE(mark(X)) MARK(sieve(X)) -> MARK(X) MARK(nats(X)) -> A__NATS(mark(X)) MARK(nats(X)) -> MARK(X) MARK(zprimes) -> A__ZPRIMES MARK(cons(X1, X2)) -> MARK(X1) MARK(s(X)) -> MARK(X) The TRS R consists of the following rules: a__filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M)) a__filter(cons(X, Y), s(N), M) -> cons(mark(X), filter(Y, N, M)) a__sieve(cons(0, Y)) -> cons(0, sieve(Y)) a__sieve(cons(s(N), Y)) -> cons(s(mark(N)), sieve(filter(Y, N, N))) a__nats(N) -> cons(mark(N), nats(s(N))) a__zprimes -> a__sieve(a__nats(s(s(0)))) mark(filter(X1, X2, X3)) -> a__filter(mark(X1), mark(X2), mark(X3)) mark(sieve(X)) -> a__sieve(mark(X)) popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to TRS Innermost