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SRS Standard pair #516971909
details
property
value
status
complete
benchmark
96239.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n055.star.cs.uiowa.edu
space
ICFP_2010
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
12.6100969315 seconds
cpu usage
48.707147371
max memory
4.428156928E9
stage attributes
key
value
output-size
427459
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- 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 1 2 ⟶ 3 4 4 2 , 2 1 0 2 5 5 ⟶ 4 0 5 2 5 , 4 2 0 3 1 2 ⟶ 5 5 0 5 2 5 , 0 1 5 0 4 4 5 ⟶ 3 0 0 4 5 0 , 0 4 1 2 2 4 5 ⟶ 0 0 1 5 4 5 , 3 5 4 0 0 4 0 ⟶ 0 0 0 5 4 0 , 2 3 4 3 1 0 2 4 ⟶ 0 4 5 3 1 4 , 2 5 2 4 4 3 1 0 0 ⟶ 4 2 5 5 4 5 5 0 , 5 4 1 5 5 5 3 3 0 ⟶ 5 3 5 0 4 3 5 4 0 , 4 0 5 1 2 1 3 3 0 5 3 ⟶ 1 2 1 4 4 3 2 5 0 2 3 , 4 5 2 5 0 0 2 2 0 4 1 ⟶ 4 4 4 3 3 1 3 5 5 1 1 , 1 2 3 1 4 1 3 2 0 5 1 4 ⟶ 1 0 4 4 5 1 2 4 3 3 0 1 , 1 1 5 1 4 3 4 1 1 5 1 2 5 ⟶ 5 0 0 3 3 0 5 5 5 4 2 3 , 1 4 0 0 1 3 4 3 3 0 3 0 4 5 ⟶ 5 5 3 3 5 0 4 5 4 1 5 0 0 , 2 1 0 3 2 0 2 0 3 3 2 3 4 0 5 ⟶ 2 2 3 3 5 5 5 3 1 4 0 5 0 2 2 , 0 5 5 4 0 5 3 1 5 4 4 2 2 5 3 4 ⟶ 0 4 2 1 2 0 2 2 4 3 3 2 2 4 4 3 3 4 , 2 4 4 5 0 1 5 0 3 5 1 1 5 2 0 4 4 ⟶ 2 5 2 4 3 5 4 2 4 3 4 1 3 4 5 5 4 1 2 4 , 1 2 2 2 5 0 0 3 3 5 2 4 5 1 1 0 4 3 ⟶ 1 1 3 0 0 0 3 1 5 1 0 2 3 0 0 , 0 2 1 4 1 2 5 0 0 1 3 1 5 5 5 5 1 3 4 ⟶ 3 5 5 1 3 2 2 2 5 2 5 5 5 2 0 4 1 1 4 4 , 1 3 1 0 2 2 4 4 1 3 4 2 0 5 2 1 3 5 4 ⟶ 1 4 2 0 2 0 5 5 2 2 0 5 4 3 4 4 4 , 1 3 2 0 0 1 1 5 5 3 1 3 0 2 3 4 5 5 5 ⟶ 3 0 0 5 2 5 2 2 1 4 2 2 0 0 2 5 5 3 0 2 , 2 2 2 5 1 3 5 5 0 4 0 4 5 1 1 2 0 2 5 ⟶ 3 2 4 4 2 3 3 1 1 2 0 3 3 1 1 4 5 , 2 5 3 4 5 1 1 5 5 1 2 1 3 3 3 5 4 5 1 ⟶ 5 4 5 4 5 1 1 0 4 2 5 4 1 2 0 1 3 1 , 3 2 1 3 3 5 5 5 4 2 4 4 5 1 5 1 3 0 4 ⟶ 0 4 4 3 0 1 1 3 0 4 2 2 3 3 0 4 0 1 , 2 5 5 2 2 0 1 2 0 3 3 5 1 3 2 2 0 2 3 4 3 ⟶ 4 4 0 4 1 1 4 5 2 5 1 0 1 4 4 1 1 3 3 } The system was reversed. After renaming modulo the bijection { 2 ↦ 0, 1 ↦ 1, 0 ↦ 2, 4 ↦ 3, 3 ↦ 4, 5 ↦ 5 }, it remains to prove termination of the 25-rule system { 0 1 1 2 ⟶ 0 3 3 4 , 5 5 0 2 1 0 ⟶ 5 0 5 2 3 , 0 1 4 2 0 3 ⟶ 5 0 5 2 5 5 , 5 3 3 2 5 1 2 ⟶ 2 5 3 2 2 4 , 5 3 0 0 1 3 2 ⟶ 5 3 5 1 2 2 , 2 3 2 2 3 5 4 ⟶ 2 3 5 2 2 2 , 3 0 2 1 4 3 4 0 ⟶ 3 1 4 5 3 2 , 2 2 1 4 3 3 0 5 0 ⟶ 2 5 5 3 5 5 0 3 , 2 4 4 5 5 5 1 3 5 ⟶ 2 3 5 4 3 2 5 4 5 , 4 5 2 4 4 1 0 1 5 2 3 ⟶ 4 0 2 5 0 4 3 3 1 0 1 , 1 3 2 0 0 2 2 5 0 5 3 ⟶ 1 1 5 5 4 1 4 4 3 3 3 , 3 1 5 2 0 4 1 3 1 4 0 1 ⟶ 1 2 4 4 3 0 1 5 3 3 2 1 , 5 0 1 5 1 1 3 4 3 1 5 1 1 ⟶ 4 0 3 5 5 5 2 4 4 2 2 5 , 5 3 2 4 2 4 4 3 4 1 2 2 3 1 ⟶ 2 2 5 1 3 5 3 2 5 4 4 5 5 , 5 2 3 4 0 4 4 2 0 2 0 4 2 1 0 ⟶ 0 0 2 5 2 3 1 4 5 5 5 4 4 0 0 , 3 4 5 0 0 3 3 5 1 4 5 2 3 5 5 2 ⟶ 3 4 4 3 3 0 0 4 4 3 0 0 2 0 1 0 3 2 , 3 3 2 0 5 1 1 5 4 2 5 1 2 5 3 3 0 ⟶ 3 0 1 3 5 5 3 4 1 3 4 3 0 3 5 4 3 0 5 0 , 4 3 2 1 1 5 3 0 5 4 4 2 2 5 0 0 0 1 ⟶ 2 2 4 0 2 1 5 1 4 2 2 2 4 1 1 , 3 4 1 5 5 5 5 1 4 1 2 2 5 0 1 3 1 0 2 ⟶ 3 3 1 1 3 2 0 5 5 5 0 5 0 0 0 4 1 5 5 4 , 3 5 4 1 0 5 2 0 3 4 1 3 3 0 0 2 1 4 1 ⟶ 3 3 3 4 3 5 2 0 0 5 5 2 0 2 0 3 1 , 5 5 5 3 4 0 2 4 1 4 5 5 1 1 2 2 0 4 1 ⟶ 0 2 4 5 5 0 2 2 0 0 3 1 0 0 5 0 5 2 2 4 , 5 0 2 0 1 1 5 3 2 3 2 5 5 4 1 5 0 0 0 ⟶ 5 3 1 1 4 4 2 0 1 1 4 4 0 3 3 0 4 , 1 5 3 5 4 4 4 1 0 1 5 5 1 1 5 3 4 5 0 ⟶ 1 4 1 2 0 1 3 5 0 3 2 1 1 5 3 5 3 5 , 3 2 4 1 5 1 5 3 3 0 3 5 5 5 4 4 1 0 4 ⟶ 1 2 3 2 4 4 0 0 3 2 4 1 1 2 4 3 3 2 , 4 3 4 0 2 0 0 4 1 5 4 4 2 0 1 2 0 0 5 5 0 ⟶ 4 4 1 1 3 3 1 2 1 5 0 5 3 1 1 3 2 3 3 } Applying sparse tiling TRFC(2) [Geser/Hofbauer/Waldmann, FSCD 2019]. After renaming modulo the bijection { (0,0) ↦ 0, (0,1) ↦ 1, (1,1) ↦ 2, (1,2) ↦ 3, (2,0) ↦ 4, (0,3) ↦ 5, (3,3) ↦ 6, (3,4) ↦ 7, (4,0) ↦ 8, (2,1) ↦ 9, (4,1) ↦ 10, (2,2) ↦ 11, (4,2) ↦ 12, (2,3) ↦ 13, (4,3) ↦ 14, (2,4) ↦ 15, (4,4) ↦ 16, (2,5) ↦ 17, (4,5) ↦ 18, (2,7) ↦ 19, (4,7) ↦ 20, (1,0) ↦ 21, (3,0) ↦ 22, (5,0) ↦ 23, (6,0) ↦ 24, (0,5) ↦ 25, (5,5) ↦ 26, (0,2) ↦ 27, (5,2) ↦ 28, (3,1) ↦ 29, (3,2) ↦ 30, (0,4) ↦ 31, (3,5) ↦ 32, (0,7) ↦ 33, (3,7) ↦ 34, (1,5) ↦ 35, (6,5) ↦ 36, (1,4) ↦ 37, (5,1) ↦ 38, (5,3) ↦ 39, (5,4) ↦ 40, (5,7) ↦ 41, (6,2) ↦ 42, (1,3) ↦ 43, (6,3) ↦ 44, (1,7) ↦ 45, (6,4) ↦ 46, (6,1) ↦ 47 }, it remains to prove termination of the 1225-rule system { 0 1 2 3 4 ⟶ 0 5 6 7 8 , 0 1 2 3 9 ⟶ 0 5 6 7 10 , 0 1 2 3 11 ⟶ 0 5 6 7 12 , 0 1 2 3 13 ⟶ 0 5 6 7 14 , 0 1 2 3 15 ⟶ 0 5 6 7 16 , 0 1 2 3 17 ⟶ 0 5 6 7 18 , 0 1 2 3 19 ⟶ 0 5 6 7 20 , 21 1 2 3 4 ⟶ 21 5 6 7 8 , 21 1 2 3 9 ⟶ 21 5 6 7 10 , 21 1 2 3 11 ⟶ 21 5 6 7 12 , 21 1 2 3 13 ⟶ 21 5 6 7 14 , 21 1 2 3 15 ⟶ 21 5 6 7 16 , 21 1 2 3 17 ⟶ 21 5 6 7 18 , 21 1 2 3 19 ⟶ 21 5 6 7 20 , 4 1 2 3 4 ⟶ 4 5 6 7 8 , 4 1 2 3 9 ⟶ 4 5 6 7 10 , 4 1 2 3 11 ⟶ 4 5 6 7 12 , 4 1 2 3 13 ⟶ 4 5 6 7 14 , 4 1 2 3 15 ⟶ 4 5 6 7 16 , 4 1 2 3 17 ⟶ 4 5 6 7 18 , 4 1 2 3 19 ⟶ 4 5 6 7 20 , 22 1 2 3 4 ⟶ 22 5 6 7 8 , 22 1 2 3 9 ⟶ 22 5 6 7 10 , 22 1 2 3 11 ⟶ 22 5 6 7 12 , 22 1 2 3 13 ⟶ 22 5 6 7 14 , 22 1 2 3 15 ⟶ 22 5 6 7 16 , 22 1 2 3 17 ⟶ 22 5 6 7 18 , 22 1 2 3 19 ⟶ 22 5 6 7 20 , 8 1 2 3 4 ⟶ 8 5 6 7 8 , 8 1 2 3 9 ⟶ 8 5 6 7 10 ,
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