details

property	value
status	complete
benchmark	x03.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n092.star.cs.uiowa.edu
space	Secret_07_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.82777190208 seconds
cpu usage	11.219313372
max memory	9.19375872E8

stage attributes

key	value
output-size	4636
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 44 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 20 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 56 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 73 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(b(c(x1))) -> b(x1) c(b(b(x1))) -> a(x1) c(x1) -> b(x1) a(a(x1)) -> c(b(a(c(x1)))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a(x_1)) = 3 + x_1 POL(b(x_1)) = 1 + x_1 POL(c(x_1)) = 1 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a(b(c(x1))) -> b(x1) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: c(b(b(x1))) -> a(x1) c(x1) -> b(x1) a(a(x1)) -> c(b(a(c(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: C(b(b(x1))) -> A(x1) A(a(x1)) -> C(b(a(c(x1)))) A(a(x1)) -> A(c(x1)) A(a(x1)) -> C(x1) The TRS R consists of the following rules: c(b(b(x1))) -> a(x1) c(x1) -> b(x1) a(a(x1)) -> c(b(a(c(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. popout

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all output return to SRS Stand 10685