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SRS Standard pair #487510650
details
property
value
status
complete
benchmark
random-27.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n178.star.cs.uiowa.edu
space
Waldmann_19
run statistics
property
value
solver
MultumNonMulta 3.16 29 June 2020 60G
configuration
default
runtime (wallclock)
1.67086100578 seconds
cpu usage
1.456729128
max memory
4.07953408E8
stage attributes
key
value
output-size
1882
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 { b->0, a->1 }, it remains to prove termination of the 3-rule system { 0 1 1 0 -> 1 1 1 0 , 0 1 1 1 -> 1 1 0 0 , 0 1 0 1 -> 0 0 1 0 } Applying the dependency pairs transformation. After renaming modulo { (0,true)->0, (1,false)->1, (0,false)->2 }, it remains to prove termination of the 9-rule system { 0 1 1 2 -> 0 , 0 1 1 1 -> 0 2 , 0 1 1 1 -> 0 , 0 1 2 1 -> 0 2 1 2 , 0 1 2 1 -> 0 1 2 , 0 1 2 1 -> 0 , 2 1 1 2 ->= 1 1 1 2 , 2 1 1 1 ->= 1 1 2 2 , 2 1 2 1 ->= 2 2 1 2 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 2: 0 is interpreted by / \ | 1 0 | | 0 1 | \ / 1 is interpreted by / \ | 1 1 | | 0 1 | \ / 2 is interpreted by / \ | 1 1 | | 0 1 | \ / After renaming modulo { 0->0, 1->1, 2->2 }, it remains to prove termination of the 4-rule system { 0 1 2 1 -> 0 2 1 2 , 2 1 1 2 ->= 1 1 1 2 , 2 1 1 1 ->= 1 1 2 2 , 2 1 2 1 ->= 2 2 1 2 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 5: 0 is interpreted by / \ | 1 0 1 0 0 | | 0 1 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 1 0 0 0 | | 0 0 0 1 0 | | 0 0 0 0 0 | | 0 1 0 0 0 | \ / 2 is interpreted by / \ | 1 0 0 0 0 | | 0 1 0 0 0 | | 0 0 0 0 0 | | 0 0 0 0 1 | | 0 1 0 0 0 | \ / After renaming modulo { 2->0, 1->1 }, it remains to prove termination of the 3-rule system { 0 1 1 0 ->= 1 1 1 0 , 0 1 1 1 ->= 1 1 0 0 , 0 1 0 1 ->= 0 0 1 0 } The system is trivially terminating.
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