SRS Standard pair #516971255

details

property	value
status	complete
benchmark	96302.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n138.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	MnM 3.18b
configuration	default
runtime (wallclock)	104.699012995 seconds
cpu usage	413.730553455
max memory	2.3676096512E10

stage attributes

key	value
output-size	331028
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 25-rule system
{ 0 1 2 1 ⟶ 0 3 1 ,
  3 4 2 0 ⟶ 3 1 4 ,
  1 2 1 2 5 ⟶ 3 4 5 5 ,
  1 2 3 1 2 1 ⟶ 1 5 0 4 3 ,
  3 3 5 2 4 3 ⟶ 0 3 4 5 3 ,
  5 2 3 1 3 4 ⟶ 5 0 1 4 3 ,
  0 0 3 4 0 2 0 ⟶ 4 3 4 3 0 2 0 ,
  0 2 5 3 5 0 4 ⟶ 0 2 3 3 5 4 1 ,
  1 2 0 4 2 1 1 5 ⟶ 2 3 2 0 5 2 4 ,
  2 3 4 5 5 2 2 0 ⟶ 5 4 5 3 0 0 2 ,
  4 0 1 1 5 3 4 1 5 ⟶ 0 1 4 4 2 4 5 3 ,
  4 0 3 1 5 4 0 2 4 5 ⟶ 5 5 0 2 1 5 3 0 2 4 ,
  3 0 2 0 2 4 2 1 1 0 1 ⟶ 3 0 4 5 3 1 5 5 4 1 ,
  1 4 3 2 0 3 0 5 5 0 4 2 ⟶ 1 0 4 2 0 0 4 2 4 3 3 2 ,
  3 3 2 0 1 0 3 5 5 5 2 5 ⟶ 5 3 0 4 2 4 0 4 2 5 1 2 ,
  2 3 2 5 1 5 3 0 5 1 4 5 5 ⟶ 1 4 2 1 3 0 1 5 4 1 5 4 ,
  2 5 3 3 2 1 4 5 4 0 3 2 5 ⟶ 5 0 3 1 0 5 5 3 5 0 1 ,
  4 3 5 1 0 0 1 1 2 1 1 2 1 3 ⟶ 4 2 1 0 0 0 0 4 2 2 2 0 0 ,
  1 5 3 5 0 1 3 2 2 0 4 0 4 4 5 0 ⟶ 2 1 1 2 1 2 4 2 3 3 4 2 5 5 0 0 0 0 ,
  5 2 3 0 5 4 1 0 4 1 4 3 3 3 5 5 2 ⟶ 5 2 3 4 5 4 0 5 3 5 3 4 4 5 0 2 ,
  4 0 2 2 1 1 1 0 1 5 0 2 0 4 1 1 4 4 ⟶ 4 4 1 3 4 1 5 2 1 1 4 1 5 1 1 2 1 4 ,
  5 0 5 0 1 3 1 3 2 4 1 4 4 0 4 5 0 1 ⟶ 5 3 3 4 2 0 2 2 0 1 0 5 2 2 0 0 3 0 1 ,
  0 1 5 3 3 4 1 0 1 4 3 4 0 5 5 2 3 5 5 ⟶ 3 1 2 3 4 5 2 4 0 4 5 2 2 3 1 0 2 2 3 ,
  2 2 5 2 3 0 3 1 4 4 3 4 4 0 3 0 1 0 0 ⟶ 3 1 1 5 4 3 0 0 1 0 2 4 4 4 1 5 0 0 ,
  1 5 0 0 5 4 4 3 4 3 2 2 2 2 3 4 4 1 3 0 2 ⟶ 2 5 3 0 2 1 5 4 2 3 4 1 1 5 1 1 5 0 4 4 2 }

The system was reversed.

After renaming modulo the bijection { 1 ↦ 0, 2 ↦ 1, 0 ↦ 2, 3 ↦ 3, 4 ↦ 4, 5 ↦ 5 },
it remains to prove termination of the 25-rule system
{ 0 1 0 2 ⟶ 0 3 2 ,
  2 1 4 3 ⟶ 4 0 3 ,
  5 1 0 1 0 ⟶ 5 5 4 3 ,
  0 1 0 3 1 0 ⟶ 3 4 2 5 0 ,
  3 4 1 5 3 3 ⟶ 3 5 4 3 2 ,
  4 3 0 3 1 5 ⟶ 3 4 0 2 5 ,
  2 1 2 4 3 2 2 ⟶ 2 1 2 3 4 3 4 ,
  4 2 5 3 5 1 2 ⟶ 0 4 5 3 3 1 2 ,
  5 0 0 1 4 2 1 0 ⟶ 4 1 5 2 1 3 1 ,
  2 1 1 5 5 4 3 1 ⟶ 1 2 2 3 5 4 5 ,
  5 0 4 3 5 0 0 2 4 ⟶ 3 5 4 1 4 4 0 2 ,
  5 4 1 2 4 5 0 3 2 4 ⟶ 4 1 2 3 5 0 1 2 5 5 ,
  0 2 0 0 1 4 1 2 1 2 3 ⟶ 0 4 5 5 0 3 5 4 2 3 ,
  1 4 2 5 5 2 3 2 1 3 4 0 ⟶ 1 3 3 4 1 4 2 2 1 4 2 0 ,
  5 1 5 5 5 3 2 0 2 1 3 3 ⟶ 1 0 5 1 4 2 4 1 4 2 3 5 ,
  5 5 4 0 5 2 3 5 0 5 1 3 1 ⟶ 4 5 0 4 5 0 2 3 0 1 4 0 ,
  5 1 3 2 4 5 4 0 1 3 3 5 1 ⟶ 0 2 5 3 5 5 2 0 3 2 5 ,
  3 0 1 0 0 1 0 0 2 2 0 5 3 4 ⟶ 2 2 1 1 1 4 2 2 2 2 0 1 4 ,
  2 5 4 4 2 4 2 1 1 3 0 2 5 3 5 0 ⟶ 2 2 2 2 5 5 1 4 3 3 1 4 1 0 1 0 0 1 ,
  1 5 5 3 3 3 4 0 4 2 0 4 5 2 3 1 5 ⟶ 1 2 5 4 4 3 5 3 5 2 4 5 4 3 1 5 ,
  4 4 0 0 4 2 1 2 5 0 2 0 0 0 1 1 2 4 ⟶ 4 0 1 0 0 5 0 4 0 0 1 5 0 4 3 0 4 4 ,
  0 2 5 4 2 4 4 0 4 1 3 0 3 0 2 5 2 5 ⟶ 0 2 3 2 2 1 1 5 2 0 2 1 1 2 1 4 3 3 5 ,
  5 5 3 1 5 5 2 4 3 4 0 2 0 4 3 3 5 0 2 ⟶ 3 1 1 2 0 3 1 1 5 4 2 4 1 5 4 3 1 0 3 ,
  2 2 0 2 3 2 4 4 3 4 4 0 3 2 3 1 5 1 1 ⟶ 2 2 5 0 4 4 4 1 2 0 2 2 3 4 5 0 0 3 ,
  1 2 3 0 4 4 3 1 1 1 1 3 4 3 4 4 5 2 2 5 0 ⟶ 1 4 4 2 5 0 0 5 0 0 4 3 1 4 5 0 1 2 3 5 1 }

Applying sparse tiling TRFC(2) [Geser/Hofbauer/Waldmann, FSCD 2019].

After renaming modulo the bijection { (0,0) ↦ 0, (0,1) ↦ 1, (1,0) ↦ 2, (0,2) ↦ 3, (2,0) ↦ 4, (0,3) ↦ 5, (3,2) ↦ 6, (2,1) ↦ 7, (2,2) ↦ 8, (2,3) ↦ 9, (2,4) ↦ 10, (2,5) ↦ 11, (2,7) ↦ 12, (3,0) ↦ 13, (4,0) ↦ 14, (5,0) ↦ 15, (6,0) ↦ 16, (1,4) ↦ 17, (4,3) ↦ 18, (0,4) ↦ 19, (3,1) ↦ 20, (3,3) ↦ 21, (3,4) ↦ 22, (3,5) ↦ 23, (3,7) ↦ 24, (1,2) ↦ 25, (4,2) ↦ 26, (4,4) ↦ 27, (5,2) ↦ 28, (5,4) ↦ 29, (6,2) ↦ 30, (6,4) ↦ 31, (0,5) ↦ 32, (5,1) ↦ 33, (5,5) ↦ 34, (0,7) ↦ 35, (1,5) ↦ 36, (4,5) ↦ 37, (6,5) ↦ 38, (1,3) ↦ 39, (5,3) ↦ 40, (6,3) ↦ 41, (4,1) ↦ 42, (5,7) ↦ 43, (4,7) ↦ 44, (1,1) ↦ 45, (1,7) ↦ 46, (6,1) ↦ 47 },
it remains to prove termination of the 1225-rule system
{ 0 1 2 3 4 ⟶ 0 5 6 4 ,
  0 1 2 3 7 ⟶ 0 5 6 7 ,
  0 1 2 3 8 ⟶ 0 5 6 8 ,
  0 1 2 3 9 ⟶ 0 5 6 9 ,
  0 1 2 3 10 ⟶ 0 5 6 10 ,
  0 1 2 3 11 ⟶ 0 5 6 11 ,
  0 1 2 3 12 ⟶ 0 5 6 12 ,
  2 1 2 3 4 ⟶ 2 5 6 4 ,
  2 1 2 3 7 ⟶ 2 5 6 7 ,
  2 1 2 3 8 ⟶ 2 5 6 8 ,
  2 1 2 3 9 ⟶ 2 5 6 9 ,
  2 1 2 3 10 ⟶ 2 5 6 10 ,
  2 1 2 3 11 ⟶ 2 5 6 11 ,
  2 1 2 3 12 ⟶ 2 5 6 12 ,
  4 1 2 3 4 ⟶ 4 5 6 4 ,
  4 1 2 3 7 ⟶ 4 5 6 7 ,
  4 1 2 3 8 ⟶ 4 5 6 8 ,
  4 1 2 3 9 ⟶ 4 5 6 9 ,
  4 1 2 3 10 ⟶ 4 5 6 10 ,
  4 1 2 3 11 ⟶ 4 5 6 11 ,
  4 1 2 3 12 ⟶ 4 5 6 12 ,
  13 1 2 3 4 ⟶ 13 5 6 4 ,
  13 1 2 3 7 ⟶ 13 5 6 7 ,
  13 1 2 3 8 ⟶ 13 5 6 8 ,
  13 1 2 3 9 ⟶ 13 5 6 9 ,
  13 1 2 3 10 ⟶ 13 5 6 10 ,
  13 1 2 3 11 ⟶ 13 5 6 11 ,
  13 1 2 3 12 ⟶ 13 5 6 12 ,
  14 1 2 3 4 ⟶ 14 5 6 4 ,
  14 1 2 3 7 ⟶ 14 5 6 7 ,

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