details

property	value
status	complete
benchmark	6.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n190.star.cs.uiowa.edu
space	Beerendonk_07

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.01392 seconds
cpu usage	4.38854
user time	4.1705
system time	0.218033
max virtual memory	1.841096E7
max residence set size	304100.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPOrderProof [EQUIVALENT, 18 ms] (20) QDP (21) PisEmptyProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: cond(true, x) -> cond(odd(x), p(x)) odd(0) -> false odd(s(0)) -> true odd(s(s(x))) -> odd(x) p(0) -> 0 p(s(x)) -> x Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is odd(0) -> false odd(s(0)) -> true odd(s(s(x))) -> odd(x) p(0) -> 0 p(s(x)) -> x The TRS R 2 is cond(true, x) -> cond(odd(x), p(x)) The signature Sigma is {cond_2} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: cond(true, x) -> cond(odd(x), p(x)) odd(0) -> false odd(s(0)) -> true odd(s(s(x))) -> odd(x) p(0) -> 0 p(s(x)) -> x The set Q consists of the following terms: cond(true, x0) odd(0) odd(s(0)) odd(s(s(x0))) p(0) p(s(x0)) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: COND(true, x) -> COND(odd(x), p(x)) COND(true, x) -> ODD(x) COND(true, x) -> P(x) popout

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all output return to TRS Standard