details

property	value
status	complete
benchmark	2.29.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n056.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.60898709297 seconds
cpu usage	3.620311101
max memory	2.30653952E8

stage attributes

key	value
output-size	2217
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 65 ms] (2) QTRS (3) RisEmptyProof [EQUIVALENT, 0 ms] (4) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: prime(0) -> false prime(s(0)) -> false prime(s(s(x))) -> prime1(s(s(x)), s(x)) prime1(x, 0) -> false prime1(x, s(0)) -> true prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y))) divp(x, y) -> =(rem(x, y), 0) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: prime/1(YES) 0/0) false/0) s/1(YES) prime1/2(YES,YES) true/0) and/2(YES,YES) not/1)YES( divp/2(YES,YES) =/2(YES,YES) rem/2(YES,YES) Quasi precedence: prime_1 > [false, prime1_2, true] > s_1 > [=_2, rem_2] prime_1 > [false, prime1_2, true] > and_2 > [=_2, rem_2] prime_1 > [false, prime1_2, true] > divp_2 > 0 > [=_2, rem_2] Status: prime_1: multiset status 0: multiset status false: multiset status s_1: [1] prime1_2: [1,2] true: multiset status and_2: [1,2] divp_2: multiset status =_2: [1,2] rem_2: [1,2] With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: prime(0) -> false prime(s(0)) -> false prime(s(s(x))) -> prime1(s(s(x)), s(x)) prime1(x, 0) -> false prime1(x, s(0)) -> true prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y))) divp(x, y) -> =(rem(x, y), 0) ---------------------------------------- (2) Obligation: Q restricted rewrite system: R is empty. Q is empty. ---------------------------------------- (3) RisEmptyProof (EQUIVALENT) The TRS R is empty. Hence, termination is trivially proven. ---------------------------------------- (4) YES popout

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