details

property	value
status	complete
benchmark	13.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n085.star.cs.uiowa.edu
space	Zantema_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	10.9836909771 seconds
cpu usage	40.215543716
max memory	4.38849536E9

stage attributes

key	value
output-size	4910
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 47 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 13 ms] (4) QDP (5) MRRProof [EQUIVALENT, 59 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 134 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(b(b(x1))))) -> b(b(b(x1))) b(a(a(a(b(x1))))) -> a(a(a(b(a(a(a(x1))))))) a(a(a(x1))) -> a(a(x1)) b(b(x1)) -> a(b(a(x1))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a(x_1)) = 1 + x_1 POL(b(x_1)) = 3 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a(a(a(x1))) -> a(a(x1)) b(b(x1)) -> a(b(a(x1))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(b(b(x1))))) -> b(b(b(x1))) b(a(a(a(b(x1))))) -> a(a(a(b(a(a(a(x1))))))) Q is empty. ---------------------------------------- (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: A(a(a(b(b(x1))))) -> B(b(b(x1))) B(a(a(a(b(x1))))) -> A(a(a(b(a(a(a(x1))))))) B(a(a(a(b(x1))))) -> A(a(b(a(a(a(x1)))))) B(a(a(a(b(x1))))) -> A(b(a(a(a(x1))))) B(a(a(a(b(x1))))) -> B(a(a(a(x1)))) B(a(a(a(b(x1))))) -> A(a(a(x1))) B(a(a(a(b(x1))))) -> A(a(x1)) B(a(a(a(b(x1))))) -> A(x1) The TRS R consists of the following rules: a(a(a(b(b(x1))))) -> b(b(b(x1))) b(a(a(a(b(x1))))) -> a(a(a(b(a(a(a(x1))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. popout

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