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	n151.star.cs.uiowa.edu
space	Transformed_CSR_innermost_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.46391296387 seconds
cpu usage	6.235140279
max memory	4.16006144E8

stage attributes

key	value
output-size	10557
starexec-result	YES

output

/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 41 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 77 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 2 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)) popout

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

job log

popout

actions

all output return to TRS Innermost