SRS Standard pair #487087265

details

property	value
status	complete
benchmark	212094.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n139.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	MultumNonMulta 20 June 2020 20G sparse
configuration	default
runtime (wallclock)	2.75365 seconds
cpu usage	9.45883
user time	8.49479
system time	0.964044
max virtual memory	2.6220536E7
max residence set size	808524.0

stage attributes

key	value
starexec-result	YES

output

YES

After renaming modulo { 0->0, 1->1, 2->2, 3->3, 4->4, 5->5 }, 
it remains to prove termination of the 50-rule system
{ 0 1 1 2 -> 1 3 0 1 2 ,
  0 1 1 2 -> 1 3 1 0 2 ,
  0 1 1 2 -> 2 3 1 3 0 1 ,
  0 1 2 0 -> 1 0 0 2 2 ,
  0 1 2 4 -> 0 2 3 4 1 ,
  0 1 2 4 -> 1 0 4 2 3 ,
  0 1 2 4 -> 1 2 3 0 4 ,
  0 1 2 4 -> 2 1 4 0 3 ,
  0 1 2 4 -> 1 2 3 4 0 5 ,
  0 1 2 4 -> 2 1 1 0 3 4 ,
  0 1 2 4 -> 4 1 2 2 3 0 ,
  0 1 2 4 -> 4 5 0 2 3 1 ,
  0 4 2 0 -> 0 4 2 3 0 ,
  0 4 2 0 -> 2 3 0 4 0 0 ,
  0 4 2 0 -> 3 4 0 2 3 0 ,
  0 4 2 0 -> 4 2 3 0 4 0 ,
  0 4 2 0 -> 4 3 0 0 5 2 ,
  0 4 2 4 -> 4 0 2 3 1 4 ,
  0 4 2 4 -> 4 0 4 2 0 3 ,
  5 1 2 0 -> 2 1 3 5 0 ,
  5 1 2 0 -> 3 1 0 2 5 ,
  5 1 2 0 -> 2 3 0 0 1 5 ,
  5 1 2 4 -> 3 2 5 1 4 ,
  5 1 2 4 -> 5 2 1 3 4 ,
  5 1 2 4 -> 5 5 2 1 4 ,
  5 1 2 4 -> 0 3 4 5 2 1 ,
  5 1 4 2 -> 3 4 5 2 1 ,
  5 1 4 2 -> 2 1 3 4 5 4 ,
  5 1 4 2 -> 3 3 4 2 1 5 ,
  5 4 1 4 -> 3 4 4 1 5 4 ,
  5 4 1 4 -> 4 1 3 5 0 4 ,
  5 4 1 4 -> 5 1 3 4 5 4 ,
  5 4 1 4 -> 5 1 5 3 4 4 ,
  5 4 2 0 -> 5 4 2 3 0 ,
  5 4 2 0 -> 0 1 2 3 4 5 ,
  5 4 2 0 -> 4 5 3 3 0 2 ,
  0 1 2 0 4 -> 0 2 4 1 3 0 ,
  0 1 2 0 4 -> 2 0 3 4 0 1 ,
  0 1 2 0 4 -> 4 0 2 3 1 0 ,
  0 1 4 2 2 -> 1 2 3 4 0 2 ,
  5 0 1 2 4 -> 3 0 2 1 4 5 ,
  5 1 2 2 4 -> 5 2 2 1 4 5 ,
  5 1 2 4 0 -> 5 0 2 1 3 4 ,
  5 1 2 4 0 -> 5 0 3 2 1 4 ,
  5 1 3 1 4 -> 1 5 1 3 4 0 ,
  5 1 4 1 2 -> 2 1 5 3 4 1 ,
  5 4 1 1 4 -> 1 1 3 4 5 4 ,
  5 4 1 4 0 -> 3 4 0 4 5 1 ,
  5 4 3 2 0 -> 5 4 0 3 2 3 ,
  5 4 5 2 0 -> 5 0 5 4 4 2 }


The system was reversed.

After renaming modulo { 2->0, 1->1, 0->2, 3->3, 4->4, 5->5 }, 
it remains to prove termination of the 50-rule system
{ 0 1 1 2 -> 0 1 2 3 1 ,
  0 1 1 2 -> 0 2 1 3 1 ,
  0 1 1 2 -> 1 2 3 1 3 0 ,
  2 0 1 2 -> 0 0 2 2 1 ,
  4 0 1 2 -> 1 4 3 0 2 ,
  4 0 1 2 -> 3 0 4 2 1 ,
  4 0 1 2 -> 4 2 3 0 1 ,
  4 0 1 2 -> 3 2 4 1 0 ,
  4 0 1 2 -> 5 2 4 3 0 1 ,
  4 0 1 2 -> 4 3 2 1 1 0 ,
  4 0 1 2 -> 2 3 0 0 1 4 ,
  4 0 1 2 -> 1 3 0 2 5 4 ,
  2 0 4 2 -> 2 3 0 4 2 ,
  2 0 4 2 -> 2 2 4 2 3 0 ,
  2 0 4 2 -> 2 3 0 2 4 3 ,
  2 0 4 2 -> 2 4 2 3 0 4 ,
  2 0 4 2 -> 0 5 2 2 3 4 ,
  4 0 4 2 -> 4 1 3 0 2 4 ,
  4 0 4 2 -> 3 2 0 4 2 4 ,
  2 0 1 5 -> 2 5 3 1 0 ,
  2 0 1 5 -> 5 0 2 1 3 ,
  2 0 1 5 -> 5 1 2 2 3 0 ,
  4 0 1 5 -> 4 1 5 0 3 ,
  4 0 1 5 -> 4 3 1 0 5 ,
  4 0 1 5 -> 4 1 0 5 5 ,
  4 0 1 5 -> 1 0 5 4 3 2 ,
  0 4 1 5 -> 1 0 5 4 3 ,
  0 4 1 5 -> 4 5 4 3 1 0 ,
  0 4 1 5 -> 5 1 0 4 3 3 ,
  4 1 4 5 -> 4 5 1 4 4 3 ,
  4 1 4 5 -> 4 2 5 3 1 4 ,
  4 1 4 5 -> 4 5 4 3 1 5 ,
  4 1 4 5 -> 4 4 3 5 1 5 ,
  2 0 4 5 -> 2 3 0 4 5 ,
  2 0 4 5 -> 5 4 3 0 1 2 ,
  2 0 4 5 -> 0 2 3 3 5 4 ,
  4 2 0 1 2 -> 2 3 1 4 0 2 ,
  4 2 0 1 2 -> 1 2 4 3 2 0 ,
  4 2 0 1 2 -> 2 1 3 0 2 4 ,
  0 0 4 1 2 -> 0 2 4 3 0 1 ,

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