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SRS Standard pair #516971303
details
property
value
status
complete
benchmark
86816.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n004.star.cs.uiowa.edu
space
ICFP_2010
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
12.7008149624 seconds
cpu usage
49.37422632
max memory
4.197306368E9
stage attributes
key
value
output-size
294624
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 30-rule system { 0 1 2 ⟶ 3 2 2 , 4 2 3 2 1 ⟶ 0 5 3 4 , 2 2 5 1 1 2 ⟶ 4 1 4 5 3 , 3 2 5 0 0 2 ⟶ 4 3 1 5 1 2 , 2 3 3 1 3 1 4 2 3 ⟶ 5 2 5 5 0 2 3 , 2 4 3 0 5 2 4 4 1 2 4 2 ⟶ 4 1 5 0 4 0 0 5 5 0 2 , 2 2 5 4 0 2 0 0 4 1 3 3 1 ⟶ 2 2 4 4 0 5 4 4 5 4 5 3 , 4 0 3 5 5 5 4 5 0 4 1 3 0 ⟶ 4 0 3 3 5 2 4 4 4 0 0 3 4 4 , 4 3 2 2 5 5 2 0 0 4 3 2 4 ⟶ 1 2 4 4 1 1 3 5 1 2 4 4 , 5 2 0 0 0 3 3 5 1 4 0 0 5 ⟶ 3 1 4 1 3 1 5 1 0 3 1 2 2 , 4 0 2 5 4 0 1 2 1 0 2 5 1 1 ⟶ 4 0 1 4 2 3 1 3 3 2 4 4 4 , 3 1 0 1 3 4 0 4 1 1 3 0 4 1 1 ⟶ 3 1 2 0 4 0 5 1 4 1 3 5 0 5 4 , 0 1 0 5 5 1 0 0 3 0 1 0 1 4 3 1 ⟶ 3 4 5 4 3 5 1 5 1 4 3 0 1 3 5 0 , 2 4 5 3 0 3 0 2 2 0 5 5 5 1 1 0 ⟶ 3 3 1 1 4 4 1 4 4 2 2 4 1 5 3 , 5 2 1 2 1 4 3 2 4 3 3 1 5 1 3 3 ⟶ 5 5 3 3 3 3 3 5 2 4 3 4 4 0 3 , 2 0 1 2 5 5 0 3 5 2 2 3 0 3 1 0 1 ⟶ 5 3 2 4 0 1 2 0 2 1 5 3 4 1 0 2 2 , 0 2 0 3 3 4 0 5 2 5 5 5 1 5 2 0 1 0 ⟶ 3 2 2 2 1 0 5 3 5 0 1 2 0 2 1 4 4 0 , 4 3 0 0 3 0 4 3 5 3 2 1 4 5 5 3 3 1 ⟶ 4 3 3 5 5 0 3 5 3 5 0 0 5 4 2 3 0 , 4 4 0 4 2 0 5 4 1 1 5 0 3 2 4 2 0 3 ⟶ 4 3 0 2 4 1 0 4 3 1 1 1 4 0 2 5 3 3 , 5 5 3 5 2 4 0 4 2 1 3 0 5 3 1 2 1 2 ⟶ 5 0 1 2 3 5 1 2 3 0 5 0 3 0 3 3 4 2 , 1 4 5 2 0 2 2 2 4 1 4 3 1 3 0 2 0 5 2 ⟶ 1 4 2 3 5 0 2 0 2 0 0 4 0 4 4 1 2 1 , 2 0 0 4 0 2 3 0 2 5 5 3 4 5 4 0 4 0 2 ⟶ 1 4 2 2 2 2 0 2 4 1 0 1 4 1 0 4 2 1 , 2 1 5 4 4 1 4 0 3 5 3 0 1 5 5 5 3 2 1 ⟶ 2 2 1 4 2 2 1 5 3 2 4 4 5 4 2 1 0 1 2 0 , 3 3 2 0 4 4 0 4 2 3 1 5 3 2 0 0 2 1 1 ⟶ 5 5 3 1 1 1 0 5 5 0 4 1 1 1 5 0 , 5 3 4 3 3 3 2 4 2 0 3 1 1 2 4 2 5 3 2 ⟶ 0 3 2 2 4 3 5 2 4 0 4 5 3 5 5 4 0 2 4 , 5 3 5 3 2 5 0 2 0 3 0 2 1 1 5 5 4 0 0 ⟶ 5 5 2 3 5 2 1 1 4 0 4 2 1 4 3 5 1 1 0 , 3 5 2 5 4 4 0 1 5 5 5 0 5 1 0 0 0 4 5 1 ⟶ 5 3 1 0 5 1 4 4 2 1 4 3 0 5 4 0 0 5 0 3 , 4 4 2 0 5 1 4 4 0 1 0 4 5 0 3 4 0 5 4 1 ⟶ 4 4 4 5 5 2 4 1 1 4 4 5 5 2 4 4 1 4 4 , 4 3 4 5 2 5 2 2 1 4 0 0 4 3 0 4 5 4 1 4 3 ⟶ 4 0 1 0 0 1 5 2 5 4 5 5 4 5 4 1 1 5 5 , 4 5 2 3 0 2 3 4 3 1 3 3 4 0 2 4 3 0 4 1 1 ⟶ 4 4 5 4 2 4 2 4 1 3 4 4 2 0 1 4 1 2 2 1 } The system was reversed. After renaming modulo the bijection { 2 ↦ 0, 1 ↦ 1, 0 ↦ 2, 3 ↦ 3, 4 ↦ 4, 5 ↦ 5 }, it remains to prove termination of the 30-rule system { 0 1 2 ⟶ 0 0 3 , 1 0 3 0 4 ⟶ 4 3 5 2 , 0 1 1 5 0 0 ⟶ 3 5 4 1 4 , 0 2 2 5 0 3 ⟶ 0 1 5 1 3 4 , 3 0 4 1 3 1 3 3 0 ⟶ 3 0 2 5 5 0 5 , 0 4 0 1 4 4 0 5 2 3 4 0 ⟶ 0 2 5 5 2 2 4 2 5 1 4 , 1 3 3 1 4 2 2 0 2 4 5 0 0 ⟶ 3 5 4 5 4 4 5 2 4 4 0 0 , 2 3 1 4 2 5 4 5 5 5 3 2 4 ⟶ 4 4 3 2 2 4 4 4 0 5 3 3 2 4 , 4 0 3 4 2 2 0 5 5 0 0 3 4 ⟶ 4 4 0 1 5 3 1 1 4 4 0 1 , 5 2 2 4 1 5 3 3 2 2 2 0 5 ⟶ 0 0 1 3 2 1 5 1 3 1 4 1 3 , 1 1 5 0 2 1 0 1 2 4 5 0 2 4 ⟶ 4 4 4 0 3 3 1 3 0 4 1 2 4 , 1 1 4 2 3 1 1 4 2 4 3 1 2 1 3 ⟶ 4 5 2 5 3 1 4 1 5 2 4 2 0 1 3 , 1 3 4 1 2 1 2 3 2 2 1 5 5 2 1 2 ⟶ 2 5 3 1 2 3 4 1 5 1 5 3 4 5 4 3 , 2 1 1 5 5 5 2 0 0 2 3 2 3 5 4 0 ⟶ 3 5 1 4 0 0 4 4 1 4 4 1 1 3 3 , 3 3 1 5 1 3 3 4 0 3 4 1 0 1 0 5 ⟶ 3 2 4 4 3 4 0 5 3 3 3 3 3 5 5 , 1 2 1 3 2 3 0 0 5 3 2 5 5 0 1 2 0 ⟶ 0 0 2 1 4 3 5 1 0 2 0 1 2 4 0 3 5 , 2 1 2 0 5 1 5 5 5 0 5 2 4 3 3 2 0 2 ⟶ 2 4 4 1 0 2 0 1 2 5 3 5 2 1 0 0 0 3 , 1 3 3 5 5 4 1 0 3 5 3 4 2 3 2 2 3 4 ⟶ 2 3 0 4 5 2 2 5 3 5 3 2 5 5 3 3 4 , 3 2 0 4 0 3 2 5 1 1 4 5 2 0 4 2 4 4 ⟶ 3 3 5 0 2 4 1 1 1 3 4 2 1 4 0 2 3 4 , 0 1 0 1 3 5 2 3 1 0 4 2 4 0 5 3 5 5 ⟶ 0 4 3 3 2 3 2 5 2 3 0 1 5 3 0 1 2 5 , 0 5 2 0 2 3 1 3 4 1 4 0 0 0 2 0 5 4 1 ⟶ 1 0 1 4 4 2 4 2 2 0 2 0 2 5 3 0 4 1 , 0 2 4 2 4 5 4 3 5 5 0 2 3 0 2 4 2 2 0 ⟶ 1 0 4 2 1 4 1 2 1 4 0 2 0 0 0 0 4 1 , 1 0 3 5 5 5 1 2 3 5 3 2 4 1 4 4 5 1 0 ⟶ 2 0 1 2 1 0 4 5 4 4 0 3 5 1 0 0 4 1 0 0 , 1 1 0 2 2 0 3 5 1 3 0 4 2 4 4 2 0 3 3 ⟶ 2 5 1 1 1 4 2 5 5 2 1 1 1 3 5 5 , 0 3 5 0 4 0 1 1 3 2 0 4 0 3 3 3 4 3 5 ⟶ 4 0 2 4 5 5 3 5 4 2 4 0 5 3 4 0 0 3 2 , 2 2 4 5 5 1 1 0 2 3 2 0 2 5 0 3 5 3 5 ⟶ 2 1 1 5 3 4 1 0 4 2 4 1 1 0 5 3 0 5 5 , 1 5 4 2 2 2 1 5 2 5 5 5 1 2 4 4 5 0 5 3 ⟶ 3 2 5 2 2 4 5 2 3 4 1 0 4 4 1 5 2 1 3 5 , 1 4 5 2 4 3 2 5 4 2 1 2 4 4 1 5 2 0 4 4 ⟶ 4 4 1 4 4 0 5 5 4 4 1 1 4 0 5 5 4 4 4 , 3 4 1 4 5 4 2 3 4 2 2 4 1 0 0 5 0 5 4 3 4 ⟶ 5 5 1 1 4 5 4 5 5 4 5 0 5 1 2 2 1 2 4 , 1 1 4 2 3 4 0 2 4 3 3 1 3 4 3 0 2 3 0 5 4 ⟶ 1 0 0 1 4 1 2 0 4 4 3 1 4 0 4 0 4 5 4 4 } Applying sparse tiling TRFC(2) [Geser/Hofbauer/Waldmann, FSCD 2019]. After renaming modulo the bijection { (0,0) ↦ 0, (0,1) ↦ 1, (1,2) ↦ 2, (2,0) ↦ 3, (0,3) ↦ 4, (3,0) ↦ 5, (2,1) ↦ 6, (3,1) ↦ 7, (2,2) ↦ 8, (3,2) ↦ 9, (2,3) ↦ 10, (3,3) ↦ 11, (2,4) ↦ 12, (3,4) ↦ 13, (2,5) ↦ 14, (3,5) ↦ 15, (2,7) ↦ 16, (3,7) ↦ 17, (1,0) ↦ 18, (4,0) ↦ 19, (5,0) ↦ 20, (6,0) ↦ 21, (0,4) ↦ 22, (4,3) ↦ 23, (5,2) ↦ 24, (4,1) ↦ 25, (4,2) ↦ 26, (4,4) ↦ 27, (4,5) ↦ 28, (4,7) ↦ 29, (1,1) ↦ 30, (1,4) ↦ 31, (5,1) ↦ 32, (5,4) ↦ 33, (6,1) ↦ 34, (6,4) ↦ 35, (1,5) ↦ 36, (0,2) ↦ 37, (0,5) ↦ 38, (0,7) ↦ 39, (1,3) ↦ 40, (5,3) ↦ 41, (6,3) ↦ 42, (5,5) ↦ 43, (5,7) ↦ 44, (6,2) ↦ 45, (1,7) ↦ 46, (6,5) ↦ 47 }, it remains to prove termination of the 1470-rule system { 0 1 2 3 ⟶ 0 0 4 5 , 0 1 2 6 ⟶ 0 0 4 7 , 0 1 2 8 ⟶ 0 0 4 9 , 0 1 2 10 ⟶ 0 0 4 11 , 0 1 2 12 ⟶ 0 0 4 13 , 0 1 2 14 ⟶ 0 0 4 15 , 0 1 2 16 ⟶ 0 0 4 17 , 18 1 2 3 ⟶ 18 0 4 5 , 18 1 2 6 ⟶ 18 0 4 7 , 18 1 2 8 ⟶ 18 0 4 9 , 18 1 2 10 ⟶ 18 0 4 11 , 18 1 2 12 ⟶ 18 0 4 13 , 18 1 2 14 ⟶ 18 0 4 15 , 18 1 2 16 ⟶ 18 0 4 17 , 3 1 2 3 ⟶ 3 0 4 5 , 3 1 2 6 ⟶ 3 0 4 7 , 3 1 2 8 ⟶ 3 0 4 9 , 3 1 2 10 ⟶ 3 0 4 11 , 3 1 2 12 ⟶ 3 0 4 13 , 3 1 2 14 ⟶ 3 0 4 15 ,
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