details

property	value
status	complete
benchmark	abacababab-ababababacabac.srs.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n018.star.cs.uiowa.edu
space	Wenzel_16

run statistics

property	value
solver	MultumNonMulta 3.16 29 June 2020 60G
configuration	default
runtime (wallclock)	6.92276597023 seconds
cpu usage	26.024777704
max memory	3.821756416E9

stage attributes

key	value
output-size	2747
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES After renaming modulo { a->0, b->1, c->2 }, it remains to prove termination of the 1-rule system { 0 1 0 2 0 1 0 1 0 1 -> 0 1 0 1 0 1 0 1 0 2 0 1 0 2 } The system was reversed. After renaming modulo { 1->0, 0->1, 2->2 }, it remains to prove termination of the 1-rule system { 0 1 0 1 0 1 2 1 0 1 -> 2 1 0 1 2 1 0 1 0 1 0 1 0 1 } Applying the dependency pairs transformation. After renaming modulo { (0,true)->0, (1,false)->1, (0,false)->2, (2,false)->3 }, it remains to prove termination of the 6-rule system { 0 1 2 1 2 1 3 1 2 1 -> 0 1 3 1 2 1 2 1 2 1 2 1 , 0 1 2 1 2 1 3 1 2 1 -> 0 1 2 1 2 1 2 1 , 0 1 2 1 2 1 3 1 2 1 -> 0 1 2 1 2 1 , 0 1 2 1 2 1 3 1 2 1 -> 0 1 2 1 , 0 1 2 1 2 1 3 1 2 1 -> 0 1 , 2 1 2 1 2 1 3 1 2 1 ->= 3 1 2 1 3 1 2 1 2 1 2 1 2 1 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 11: 0 is interpreted by / \ | 1 0 1 0 0 0 0 0 0 0 0 | | 0 1 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | \ / 1 is interpreted by / \ | 1 0 0 0 0 0 0 0 0 0 0 | | 0 1 0 0 0 0 0 0 0 0 0 | | 0 0 0 2 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 1 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 1 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 1 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 1 0 0 0 1 0 0 0 0 0 | \ / 2 is interpreted by / \ | 1 0 0 0 0 0 0 0 0 0 0 | | 0 1 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 1 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 1 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 1 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 1 0 0 0 0 0 0 0 1 | | 0 0 0 0 0 0 0 0 0 0 0 | \ / 3 is interpreted by / \ | 1 0 0 0 0 0 0 0 0 0 0 | | 0 1 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 1 0 0 0 1 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 0 0 | \ / After renaming modulo { 2->0, 1->1, 3->2 }, it remains to prove termination of the 1-rule system { 0 1 0 1 0 1 2 1 0 1 ->= 2 1 0 1 2 1 0 1 0 1 0 1 0 1 } The system is trivially terminating. popout

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all output return to SRS Standard