details

property	value
status	complete
benchmark	z111.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n033.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.7264778614 seconds
cpu usage	11.59671722
max memory	1.72886016E9

stage attributes

key	value
output-size	4125
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 57 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 1 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) MRRProof [EQUIVALENT, 48 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(x1)) -> b(c(c(c(x1)))) b(c(x1)) -> d(d(d(d(x1)))) a(x1) -> d(c(d(x1))) b(b(x1)) -> c(c(c(x1))) c(c(x1)) -> d(d(d(x1))) c(d(d(x1))) -> a(x1) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a(x_1)) = 25 + x_1 POL(b(x_1)) = 17 + x_1 POL(c(x_1)) = 11 + x_1 POL(d(x_1)) = 7 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: b(b(x1)) -> c(c(c(x1))) c(c(x1)) -> d(d(d(x1))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(x1)) -> b(c(c(c(x1)))) b(c(x1)) -> d(d(d(d(x1)))) a(x1) -> d(c(d(x1))) c(d(d(x1))) -> 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(x1)) -> B(c(c(c(x1)))) A(a(x1)) -> C(c(c(x1))) A(a(x1)) -> C(c(x1)) A(a(x1)) -> C(x1) A(x1) -> C(d(x1)) C(d(d(x1))) -> A(x1) The TRS R consists of the following rules: a(a(x1)) -> b(c(c(c(x1)))) b(c(x1)) -> d(d(d(d(x1)))) a(x1) -> d(c(d(x1))) c(d(d(x1))) -> a(x1) Q is empty. popout

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