SRS Standard pair #487088339

details

property	value
status	complete
benchmark	214091.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)	2.23435 seconds
cpu usage	7.52105
user time	6.65905
system time	0.861997
max virtual memory	5.1236412E7
max residence set size	700604.0

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 2 -> 0 2 1 0 ,
  0 1 2 -> 1 0 2 3 ,
  0 1 2 -> 0 2 4 1 5 ,
  0 1 2 -> 0 3 2 1 0 ,
  0 1 2 -> 1 0 3 2 3 ,
  0 1 2 -> 0 1 3 4 2 3 ,
  0 5 2 -> 0 2 4 5 3 ,
  0 5 2 -> 5 4 2 3 0 4 ,
  2 0 1 -> 3 0 2 1 ,
  2 0 1 -> 0 2 1 1 4 ,
  2 0 1 -> 0 3 2 4 1 ,
  2 0 1 -> 3 0 2 1 4 ,
  2 0 1 -> 0 2 2 3 4 1 ,
  2 0 1 -> 0 3 2 3 1 1 ,
  2 0 1 -> 4 0 4 2 1 4 ,
  2 5 1 -> 0 2 1 5 1 ,
  2 5 1 -> 1 4 5 4 2 ,
  2 5 1 -> 5 0 2 1 4 ,
  2 5 1 -> 5 2 1 4 1 ,
  2 5 1 -> 1 5 0 2 4 1 ,
  2 5 1 -> 5 2 1 1 1 1 ,
  0 1 2 1 -> 3 1 4 0 2 1 ,
  0 1 3 1 -> 5 0 3 1 1 ,
  0 1 3 1 -> 1 0 3 4 2 1 ,
  0 1 5 1 -> 5 0 3 1 1 ,
  0 2 1 2 -> 0 2 2 1 5 ,
  0 2 5 1 -> 1 1 5 0 2 1 ,
  0 5 3 1 -> 0 1 4 4 3 5 ,
  0 5 5 2 -> 5 4 2 3 5 0 ,
  2 0 1 2 -> 0 2 3 2 1 1 ,
  2 0 1 2 -> 4 0 2 1 1 2 ,
  2 0 4 1 -> 3 0 2 4 1 ,
  2 0 5 1 -> 5 4 2 1 0 ,
  2 2 5 1 -> 3 2 2 4 5 1 ,
  2 4 0 1 -> 1 0 2 4 4 ,
  2 4 0 1 -> 3 0 0 2 4 1 ,
  2 4 0 1 -> 5 4 0 2 1 1 ,
  2 5 2 1 -> 1 5 2 2 3 1 ,
  2 5 4 1 -> 4 5 2 1 4 4 ,
  2 5 5 1 -> 1 5 4 2 4 5 ,
  2 5 5 2 -> 5 5 2 3 2 ,
  0 1 3 0 1 -> 0 3 1 0 1 1 ,
  0 2 4 3 1 -> 1 3 4 2 3 0 ,
  0 2 4 3 1 -> 4 0 3 2 1 0 ,
  0 2 5 3 1 -> 5 0 2 3 5 1 ,
  2 0 5 4 1 -> 0 4 5 3 2 1 ,
  2 2 0 1 2 -> 2 4 0 2 2 1 ,
  2 4 0 5 1 -> 1 4 5 0 4 2 ,
  2 4 2 3 1 -> 4 2 2 3 3 1 ,
  2 5 2 0 1 -> 0 2 4 1 5 2 }


The system was reversed.

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

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