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	n178.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.51267 seconds
cpu usage	6.26219
user time	5.99201
system time	0.270178
max virtual memory	1.8494236E7
max residence set size	400328.0

stage attributes

key	value
starexec-result	YES

output

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 1 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 121 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) 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 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: 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)) 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 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) QDPOrderProof (EQUIVALENT) popout

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all output return to TRS Standard