SRS Standard pair #516970367

details
property	value
status	complete
benchmark	231043.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n083.star.cs.uiowa.edu
space	ICFP_2010
run statistics
property	value
solver	MnM 3.18b
configuration	default
runtime (wallclock)	1.02049398422 seconds
cpu usage	2.404223676
max memory	6.99953152E8
stage attributes
key	value
output-size	5121
starexec-result	YES
output
/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES

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

Applying sparse untiling TRFCU(2) [Geser/Hofbauer/Waldmann, FSCD 2019].

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

The system was filtered by the following matrix interpretation
of type E_J with J = {1,...,2} and dimension 2:

0 ↦ ⎛     ⎞
    ⎜ 1 2 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

1 ↦ ⎛     ⎞
    ⎜ 1 0 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

2 ↦ ⎛     ⎞
    ⎜ 1 3 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

After renaming modulo the bijection {  },
it remains to prove termination of the 0-rule system
{ }

The system is trivially terminating.
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output may be truncated. 'popout' for the full output.
job log
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actions all output return to SRS Standard