SRS Standard pair #516969875

details

property	value
status	complete
benchmark	212480.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n168.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	MnM 3.18b
configuration	default
runtime (wallclock)	3.4023630619 seconds
cpu usage	12.014349015
max memory	1.988333568E9

stage attributes

key	value
output-size	114918
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES

After renaming modulo the bijection { 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 ⟶ 1 2 1 2 0 ,
  0 3 1 ⟶ 1 3 2 2 0 ,
  0 3 1 ⟶ 3 2 1 2 0 ,
  0 3 1 ⟶ 1 3 3 3 2 0 ,
  0 4 1 ⟶ 2 1 2 0 4 ,
  0 0 4 5 ⟶ 0 0 2 5 4 ,
  0 1 4 1 ⟶ 0 1 2 2 4 1 ,
  0 1 4 5 ⟶ 4 0 1 2 5 4 ,
  0 1 5 1 ⟶ 1 2 2 5 0 1 ,
  0 1 5 3 ⟶ 0 5 3 2 1 ,
  0 2 4 1 ⟶ 1 3 3 2 0 4 ,
  0 2 4 1 ⟶ 4 2 1 2 0 4 ,
  0 2 4 5 ⟶ 0 2 2 5 0 4 ,
  0 3 1 5 ⟶ 0 1 2 5 3 ,
  0 3 1 5 ⟶ 1 2 5 3 0 4 ,
  0 3 5 1 ⟶ 1 2 5 3 0 ,
  0 3 5 1 ⟶ 0 5 2 1 2 3 ,
  0 3 5 5 ⟶ 0 3 2 5 5 ,
  0 4 0 1 ⟶ 2 0 4 4 0 1 ,
  0 4 1 5 ⟶ 1 2 5 0 4 ,
  0 4 3 5 ⟶ 0 4 3 2 5 4 ,
  0 4 5 1 ⟶ 2 5 4 4 0 1 ,
  3 0 1 5 ⟶ 3 1 4 0 5 4 ,
  3 0 3 1 ⟶ 1 3 3 2 0 ,
  3 0 3 5 ⟶ 3 2 5 0 2 3 ,
  3 3 0 1 ⟶ 0 1 3 2 2 3 ,
  3 4 5 1 ⟶ 3 2 5 4 2 1 ,
  4 1 3 5 ⟶ 1 2 5 3 4 4 ,
  4 1 5 1 ⟶ 4 4 5 1 2 1 ,
  4 4 1 5 ⟶ 4 1 2 5 4 ,
  0 1 4 5 5 ⟶ 0 5 1 4 2 5 ,
  0 2 1 4 5 ⟶ 0 0 1 2 5 4 ,
  0 2 1 5 5 ⟶ 0 1 2 2 5 5 ,
  0 4 2 4 1 ⟶ 1 3 2 0 4 4 ,
  0 4 5 4 3 ⟶ 2 5 0 4 4 3 ,
  0 5 1 5 1 ⟶ 0 5 1 1 2 5 ,
  0 5 2 1 5 ⟶ 1 2 5 5 0 4 ,
  0 5 2 4 1 ⟶ 4 5 2 1 2 0 ,
  3 0 1 4 1 ⟶ 0 4 4 1 3 1 ,
  3 0 1 4 1 ⟶ 4 3 2 0 1 1 ,
  3 0 3 5 5 ⟶ 3 3 2 5 0 5 ,
  3 0 5 3 1 ⟶ 1 0 3 3 2 5 ,
  4 0 1 4 1 ⟶ 4 4 0 1 3 1 ,
  4 0 1 5 1 ⟶ 0 1 2 5 4 1 ,
  4 0 2 4 5 ⟶ 4 0 2 5 0 4 ,
  4 1 1 5 1 ⟶ 1 1 2 5 4 1 ,
  4 5 1 4 1 ⟶ 4 4 1 2 1 5 ,
  4 5 2 3 1 ⟶ 4 3 1 2 2 5 ,
  4 5 4 3 1 ⟶ 4 1 2 5 3 4 ,
  4 5 5 3 1 ⟶ 1 3 2 5 5 4 }

The system was reversed.

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

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