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SRS Standard pair #516969989
details
property
value
status
complete
benchmark
96438.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n078.star.cs.uiowa.edu
space
ICFP_2010
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
3.23590993881 seconds
cpu usage
11.072914958
max memory
1.873584128E9
stage attributes
key
value
output-size
270855
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/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 2 3 ⟶ 1 4 3 , 1 2 0 4 2 ⟶ 3 4 5 4 , 3 3 0 3 2 ⟶ 3 2 4 1 , 5 2 1 0 4 ⟶ 5 5 4 5 , 0 1 4 5 0 5 ⟶ 3 0 4 0 3 5 , 0 2 5 2 1 5 ⟶ 5 2 1 5 4 , 0 3 5 1 5 5 ⟶ 0 5 1 5 4 , 2 2 5 1 4 5 4 3 ⟶ 2 2 3 3 0 1 0 3 , 0 0 0 2 4 1 2 2 4 ⟶ 0 0 0 4 3 3 4 4 , 2 4 4 3 4 2 0 3 2 ⟶ 3 4 3 4 1 3 3 1 , 0 1 4 3 2 3 2 3 5 3 ⟶ 1 4 3 0 3 4 2 5 3 , 0 2 1 5 0 4 3 1 4 2 2 ⟶ 0 0 4 0 5 1 0 5 4 4 2 , 4 4 3 3 0 2 4 2 4 4 1 5 ⟶ 4 4 0 4 3 1 5 5 5 2 5 , 5 3 0 0 5 5 2 0 2 2 3 3 ⟶ 5 5 4 1 5 2 0 5 2 4 3 , 0 2 0 2 5 5 1 5 0 2 5 2 2 ⟶ 5 4 3 1 4 0 1 0 1 5 2 , 0 3 1 4 4 2 0 5 4 0 1 0 4 ⟶ 0 0 4 4 1 0 4 0 0 2 2 4 4 , 1 2 4 1 1 0 2 5 3 4 0 1 4 0 3 ⟶ 1 1 0 4 5 0 3 2 0 5 2 0 0 3 3 , 2 3 5 2 4 4 5 1 4 4 4 1 3 2 3 ⟶ 2 3 3 1 1 1 5 3 3 5 1 3 5 2 3 , 5 4 2 5 2 5 3 2 3 3 1 5 0 0 5 ⟶ 5 2 3 5 4 1 3 1 4 4 4 4 1 3 5 , 0 0 2 4 1 2 2 1 3 2 0 4 5 5 4 2 ⟶ 0 5 1 1 0 2 3 5 4 3 0 2 5 2 3 5 1 , 0 1 1 5 3 0 1 4 2 2 4 0 1 0 2 3 4 ⟶ 0 4 2 2 0 1 5 4 4 0 2 2 2 5 5 4 , 0 2 3 2 5 2 4 3 2 4 3 0 2 4 5 1 3 ⟶ 3 4 3 5 2 5 3 1 5 0 1 0 5 5 2 3 , 4 5 4 5 4 2 1 4 5 0 2 0 4 3 0 0 1 0 0 2 ⟶ 4 4 2 2 5 2 0 0 0 5 3 4 1 2 2 1 1 5 0 0 , 3 0 1 0 5 2 4 4 4 5 2 4 1 1 4 5 4 0 3 2 1 ⟶ 3 4 2 0 4 5 2 2 2 3 3 3 5 0 5 5 3 2 1 1 4 , 3 3 2 1 0 0 3 1 2 0 2 1 2 3 5 4 0 2 2 1 1 ⟶ 3 4 5 2 4 3 4 3 5 3 4 5 4 2 5 5 1 0 1 } The system was reversed. After renaming modulo the bijection { 3 ↦ 0, 2 ↦ 1, 1 ↦ 2, 0 ↦ 3, 4 ↦ 4, 5 ↦ 5 }, it remains to prove termination of the 25-rule system { 0 1 2 3 ⟶ 0 4 2 , 1 4 3 1 2 ⟶ 4 5 4 0 , 1 0 3 0 0 ⟶ 2 4 1 0 , 4 3 2 1 5 ⟶ 5 4 5 5 , 5 3 5 4 2 3 ⟶ 5 0 3 4 3 0 , 5 2 1 5 1 3 ⟶ 4 5 2 1 5 , 5 5 2 5 0 3 ⟶ 4 5 2 5 3 , 0 4 5 4 2 5 1 1 ⟶ 0 3 2 3 0 0 1 1 , 4 1 1 2 4 1 3 3 3 ⟶ 4 4 0 0 4 3 3 3 , 1 0 3 1 4 0 4 4 1 ⟶ 2 0 0 2 4 0 4 0 , 0 5 0 1 0 1 0 4 2 3 ⟶ 0 5 1 4 0 3 0 4 2 , 1 1 4 2 0 4 3 5 2 1 3 ⟶ 1 4 4 5 3 2 5 3 4 3 3 , 5 2 4 4 1 4 1 3 0 0 4 4 ⟶ 5 1 5 5 5 2 0 4 3 4 4 , 0 0 1 1 3 1 5 5 3 3 0 5 ⟶ 0 4 1 5 3 1 5 2 4 5 5 , 1 1 5 1 3 5 2 5 5 1 3 1 3 ⟶ 1 5 2 3 2 3 4 2 0 4 5 , 4 3 2 3 4 5 3 1 4 4 2 0 3 ⟶ 4 4 1 1 3 3 4 3 2 4 4 3 3 , 0 3 4 2 3 4 0 5 1 3 2 2 4 1 2 ⟶ 0 0 3 3 1 5 3 1 0 3 5 4 3 2 2 , 0 1 0 2 4 4 4 2 5 4 4 1 5 0 1 ⟶ 0 1 5 0 2 5 0 0 5 2 2 2 0 0 1 , 5 3 3 5 2 0 0 1 0 5 1 5 1 4 5 ⟶ 5 0 2 4 4 4 4 2 0 2 4 5 0 1 5 , 1 4 5 5 4 3 1 0 2 1 1 2 4 1 3 3 ⟶ 2 5 0 1 5 1 3 0 4 5 0 1 3 2 2 5 3 , 4 0 1 3 2 3 4 1 1 4 2 3 0 5 2 2 3 ⟶ 4 5 5 1 1 1 3 4 4 5 2 3 1 1 4 3 , 0 2 5 4 1 3 0 4 1 0 4 1 5 1 0 1 3 ⟶ 0 1 5 5 3 2 3 5 2 0 5 1 5 0 4 0 , 1 3 3 2 3 3 0 4 3 1 3 5 4 2 1 4 5 4 5 4 ⟶ 3 3 5 2 2 1 1 2 4 0 5 3 3 3 1 5 1 1 4 4 , 2 1 0 3 4 5 4 2 2 4 1 5 4 4 4 1 5 3 2 3 0 ⟶ 4 2 2 1 0 5 5 3 5 0 0 0 1 1 1 5 4 3 1 4 0 , 2 2 1 1 3 4 5 0 1 2 1 3 1 2 0 3 3 2 1 0 0 ⟶ 2 3 2 5 5 1 4 5 4 0 5 0 4 0 4 1 5 4 0 } Applying sparse tiling TRFC(2) [Geser/Hofbauer/Waldmann, FSCD 2019]. After renaming modulo the bijection { (0,0) ↦ 0, (0,1) ↦ 1, (1,2) ↦ 2, (2,3) ↦ 3, (3,0) ↦ 4, (0,4) ↦ 5, (4,2) ↦ 6, (2,0) ↦ 7, (3,1) ↦ 8, (2,1) ↦ 9, (3,2) ↦ 10, (2,2) ↦ 11, (3,3) ↦ 12, (3,4) ↦ 13, (2,4) ↦ 14, (3,5) ↦ 15, (2,5) ↦ 16, (3,7) ↦ 17, (2,7) ↦ 18, (1,0) ↦ 19, (4,0) ↦ 20, (5,0) ↦ 21, (6,0) ↦ 22, (1,4) ↦ 23, (4,3) ↦ 24, (4,5) ↦ 25, (5,4) ↦ 26, (0,2) ↦ 27, (0,3) ↦ 28, (0,5) ↦ 29, (0,7) ↦ 30, (1,1) ↦ 31, (4,1) ↦ 32, (4,4) ↦ 33, (5,1) ↦ 34, (6,1) ↦ 35, (6,4) ↦ 36, (5,2) ↦ 37, (6,2) ↦ 38, (1,5) ↦ 39, (5,5) ↦ 40, (5,3) ↦ 41, (5,7) ↦ 42, (6,5) ↦ 43, (1,3) ↦ 44, (1,7) ↦ 45, (4,7) ↦ 46, (6,3) ↦ 47 }, it remains to prove termination of the 1225-rule system { 0 1 2 3 4 ⟶ 0 5 6 7 , 0 1 2 3 8 ⟶ 0 5 6 9 , 0 1 2 3 10 ⟶ 0 5 6 11 , 0 1 2 3 12 ⟶ 0 5 6 3 , 0 1 2 3 13 ⟶ 0 5 6 14 , 0 1 2 3 15 ⟶ 0 5 6 16 , 0 1 2 3 17 ⟶ 0 5 6 18 , 19 1 2 3 4 ⟶ 19 5 6 7 , 19 1 2 3 8 ⟶ 19 5 6 9 , 19 1 2 3 10 ⟶ 19 5 6 11 , 19 1 2 3 12 ⟶ 19 5 6 3 , 19 1 2 3 13 ⟶ 19 5 6 14 , 19 1 2 3 15 ⟶ 19 5 6 16 , 19 1 2 3 17 ⟶ 19 5 6 18 , 7 1 2 3 4 ⟶ 7 5 6 7 , 7 1 2 3 8 ⟶ 7 5 6 9 , 7 1 2 3 10 ⟶ 7 5 6 11 , 7 1 2 3 12 ⟶ 7 5 6 3 , 7 1 2 3 13 ⟶ 7 5 6 14 , 7 1 2 3 15 ⟶ 7 5 6 16 , 7 1 2 3 17 ⟶ 7 5 6 18 , 4 1 2 3 4 ⟶ 4 5 6 7 , 4 1 2 3 8 ⟶ 4 5 6 9 , 4 1 2 3 10 ⟶ 4 5 6 11 , 4 1 2 3 12 ⟶ 4 5 6 3 , 4 1 2 3 13 ⟶ 4 5 6 14 , 4 1 2 3 15 ⟶ 4 5 6 16 , 4 1 2 3 17 ⟶ 4 5 6 18 , 20 1 2 3 4 ⟶ 20 5 6 7 , 20 1 2 3 8 ⟶ 20 5 6 9 ,
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