details

property	value
status	complete
benchmark	96563.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n036.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.18702197075 seconds
cpu usage	13.158969017
max memory	1.70463232E9

stage attributes

key	value
output-size	14093
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 200 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 94 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(2(3(x1)))) -> 0(2(2(3(x1)))) 1(1(4(4(5(x1))))) -> 5(1(0(0(5(x1))))) 2(3(3(2(5(x1))))) -> 3(5(4(5(5(x1))))) 5(1(1(5(3(x1))))) -> 5(1(4(5(3(x1))))) 0(2(5(1(1(4(x1)))))) -> 0(3(2(2(4(5(x1)))))) 1(2(1(4(0(3(x1)))))) -> 4(2(5(4(5(3(x1)))))) 1(5(2(0(4(3(x1)))))) -> 5(0(1(1(1(x1))))) 2(2(1(5(4(0(x1)))))) -> 2(3(5(4(0(3(x1)))))) 4(0(2(3(1(4(4(x1))))))) -> 4(1(3(0(5(0(5(x1))))))) 4(3(5(4(3(2(2(2(x1)))))))) -> 4(1(2(3(4(1(4(x1))))))) 0(2(2(2(3(4(3(3(3(x1))))))))) -> 0(5(4(0(5(3(5(5(3(x1))))))))) 4(5(3(4(1(1(0(5(2(x1))))))))) -> 4(4(3(1(3(2(0(5(2(x1))))))))) 3(3(4(1(0(4(5(1(5(2(x1)))))))))) -> 3(3(4(1(5(0(3(0(2(4(x1)))))))))) 2(4(5(4(0(5(5(1(2(1(4(2(x1)))))))))))) -> 4(1(0(1(1(3(3(0(1(4(2(x1))))))))))) 1(0(0(4(5(4(3(4(5(1(5(4(0(x1))))))))))))) -> 5(3(3(1(2(1(1(0(3(1(3(x1))))))))))) 1(2(2(1(2(3(2(4(4(0(0(3(2(x1))))))))))))) -> 4(4(4(3(2(0(1(3(1(2(2(4(5(x1))))))))))))) 4(1(5(4(5(3(3(3(1(2(0(2(0(x1))))))))))))) -> 4(1(4(3(4(4(0(1(3(1(1(5(x1)))))))))))) 2(2(3(3(0(2(0(2(5(0(5(5(5(1(0(3(1(x1))))))))))))))))) -> 5(5(5(2(0(2(5(0(0(4(3(2(1(4(3(3(1(x1))))))))))))))))) 1(5(1(1(3(0(3(3(3(3(1(1(5(1(5(2(5(2(x1)))))))))))))))))) -> 3(5(4(2(0(5(1(3(4(1(1(2(5(2(4(1(5(2(x1)))))))))))))))))) 2(4(3(3(4(2(3(5(4(0(4(0(2(3(0(3(3(0(x1)))))))))))))))))) -> 2(1(3(0(4(4(1(5(2(1(0(3(0(5(0(1(5(x1))))))))))))))))) 2(5(5(1(2(3(5(0(3(0(0(3(3(1(0(1(3(2(x1)))))))))))))))))) -> 2(1(5(0(0(1(5(2(2(2(3(1(0(5(0(4(3(3(x1)))))))))))))))))) 2(3(5(1(3(0(0(5(0(2(2(4(5(4(0(3(1(4(2(x1))))))))))))))))))) -> 5(2(0(2(0(3(2(3(5(2(2(4(1(3(5(5(1(2(x1)))))))))))))))))) 3(2(3(4(1(2(5(5(0(1(2(3(4(0(5(1(4(4(5(4(x1)))))))))))))))))))) -> 3(2(0(1(5(0(1(5(5(1(2(3(0(2(4(2(5(2(2(4(x1)))))))))))))))))))) 3(5(5(2(0(4(4(5(5(0(3(5(2(1(2(1(0(2(2(3(x1)))))))))))))))))))) -> 3(1(3(3(0(5(4(2(4(1(2(3(5(0(3(4(5(2(4(3(x1)))))))))))))))))))) 4(5(1(4(5(3(1(5(5(3(5(0(2(1(2(4(4(5(1(0(4(x1))))))))))))))))))))) -> 4(4(0(2(1(5(2(0(4(5(2(2(3(5(5(2(2(0(3(5(4(x1))))))))))))))))))))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 5 + x_1 POL(1(x_1)) = 6 + x_1 POL(2(x_1)) = 5 + x_1 POL(3(x_1)) = 5 + x_1 POL(4(x_1)) = 5 + x_1 POL(5(x_1)) = 5 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 0(1(2(3(x1)))) -> 0(2(2(3(x1)))) 1(1(4(4(5(x1))))) -> 5(1(0(0(5(x1))))) 5(1(1(5(3(x1))))) -> 5(1(4(5(3(x1))))) 0(2(5(1(1(4(x1)))))) -> 0(3(2(2(4(5(x1)))))) 1(2(1(4(0(3(x1)))))) -> 4(2(5(4(5(3(x1)))))) 1(5(2(0(4(3(x1)))))) -> 5(0(1(1(1(x1))))) 2(2(1(5(4(0(x1)))))) -> 2(3(5(4(0(3(x1)))))) 4(3(5(4(3(2(2(2(x1)))))))) -> 4(1(2(3(4(1(4(x1))))))) 4(5(3(4(1(1(0(5(2(x1))))))))) -> 4(4(3(1(3(2(0(5(2(x1))))))))) 3(3(4(1(0(4(5(1(5(2(x1)))))))))) -> 3(3(4(1(5(0(3(0(2(4(x1)))))))))) 2(4(5(4(0(5(5(1(2(1(4(2(x1)))))))))))) -> 4(1(0(1(1(3(3(0(1(4(2(x1))))))))))) 1(0(0(4(5(4(3(4(5(1(5(4(0(x1))))))))))))) -> 5(3(3(1(2(1(1(0(3(1(3(x1))))))))))) 4(1(5(4(5(3(3(3(1(2(0(2(0(x1))))))))))))) -> 4(1(4(3(4(4(0(1(3(1(1(5(x1)))))))))))) 1(5(1(1(3(0(3(3(3(3(1(1(5(1(5(2(5(2(x1)))))))))))))))))) -> 3(5(4(2(0(5(1(3(4(1(1(2(5(2(4(1(5(2(x1)))))))))))))))))) 2(4(3(3(4(2(3(5(4(0(4(0(2(3(0(3(3(0(x1)))))))))))))))))) -> 2(1(3(0(4(4(1(5(2(1(0(3(0(5(0(1(5(x1))))))))))))))))) 2(3(5(1(3(0(0(5(0(2(2(4(5(4(0(3(1(4(2(x1))))))))))))))))))) -> 5(2(0(2(0(3(2(3(5(2(2(4(1(3(5(5(1(2(x1)))))))))))))))))) 4(5(1(4(5(3(1(5(5(3(5(0(2(1(2(4(4(5(1(0(4(x1))))))))))))))))))))) -> 4(4(0(2(1(5(2(0(4(5(2(2(3(5(5(2(2(0(3(5(4(x1))))))))))))))))))))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 2(3(3(2(5(x1))))) -> 3(5(4(5(5(x1))))) popout

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all output return to SRS Stand 10685