details

property	value
status	complete
benchmark	Ex9_BLR02_FR.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n172.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	5.01346683502 seconds
cpu usage	15.942502932
max memory	8.63215616E8

stage attributes

key	value
output-size	25115
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 1 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 117 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 88 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 81 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 78 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPOrderProof [EQUIVALENT, 68 ms] (16) QDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPSizeChangeProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: filter(cons(X, Y), 0, M) -> cons(0, n__filter(activate(Y), M, M)) filter(cons(X, Y), s(N), M) -> cons(X, n__filter(activate(Y), N, M)) sieve(cons(0, Y)) -> cons(0, n__sieve(activate(Y))) sieve(cons(s(N), Y)) -> cons(s(N), n__sieve(n__filter(activate(Y), N, N))) nats(N) -> cons(N, n__nats(n__s(N))) zprimes -> sieve(nats(s(s(0)))) filter(X1, X2, X3) -> n__filter(X1, X2, X3) sieve(X) -> n__sieve(X) nats(X) -> n__nats(X) s(X) -> n__s(X) activate(n__filter(X1, X2, X3)) -> filter(activate(X1), activate(X2), activate(X3)) activate(n__sieve(X)) -> sieve(activate(X)) activate(n__nats(X)) -> nats(activate(X)) activate(n__s(X)) -> s(activate(X)) 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: FILTER(cons(X, Y), 0, M) -> ACTIVATE(Y) FILTER(cons(X, Y), s(N), M) -> ACTIVATE(Y) SIEVE(cons(0, Y)) -> ACTIVATE(Y) SIEVE(cons(s(N), Y)) -> ACTIVATE(Y) ZPRIMES -> SIEVE(nats(s(s(0)))) ZPRIMES -> NATS(s(s(0))) ZPRIMES -> S(s(0)) ZPRIMES -> S(0) ACTIVATE(n__filter(X1, X2, X3)) -> FILTER(activate(X1), activate(X2), activate(X3)) ACTIVATE(n__filter(X1, X2, X3)) -> ACTIVATE(X1) ACTIVATE(n__filter(X1, X2, X3)) -> ACTIVATE(X2) ACTIVATE(n__filter(X1, X2, X3)) -> ACTIVATE(X3) ACTIVATE(n__sieve(X)) -> SIEVE(activate(X)) ACTIVATE(n__sieve(X)) -> ACTIVATE(X) ACTIVATE(n__nats(X)) -> NATS(activate(X)) ACTIVATE(n__nats(X)) -> ACTIVATE(X) ACTIVATE(n__s(X)) -> S(activate(X)) ACTIVATE(n__s(X)) -> ACTIVATE(X) The TRS R consists of the following rules: filter(cons(X, Y), 0, M) -> cons(0, n__filter(activate(Y), M, M)) filter(cons(X, Y), s(N), M) -> cons(X, n__filter(activate(Y), N, M)) sieve(cons(0, Y)) -> cons(0, n__sieve(activate(Y))) sieve(cons(s(N), Y)) -> cons(s(N), n__sieve(n__filter(activate(Y), N, N))) nats(N) -> cons(N, n__nats(n__s(N))) zprimes -> sieve(nats(s(s(0)))) filter(X1, X2, X3) -> n__filter(X1, X2, X3) popout

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

job log

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actions

all output return to TRS Standard