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SRS Standard pair #516969113
details
property
value
status
complete
benchmark
uni-3.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n175.star.cs.uiowa.edu
space
Waldmann_06_SRS
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
0.581143856049 seconds
cpu usage
1.015947704
max memory
4.36334592E8
stage attributes
key
value
output-size
1319
starexec-result
NO
output
/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO After renaming modulo the bijection { a ↦ 0, b ↦ 1, c ↦ 2 }, it remains to prove termination of the 3-rule system { 0 0 0 ⟶ 1 1 2 0 0 , 1 1 ⟶ 2 0 1 , 2 ⟶ } The system was reversed. After renaming modulo the bijection { 0 ↦ 0, 2 ↦ 1, 1 ↦ 2 }, it remains to prove termination of the 3-rule system { 0 0 0 ⟶ 0 0 1 2 2 , 2 2 ⟶ 2 0 1 , 1 ⟶ } Loop of length 14 starting with a string of length 8 using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }: .aaa.caaca rule aaa-> aabcc at position 0 .aabcc.caaca rule cc-> cab at position 4 .aabccab.aaca rule cc-> cab at position 3 .aabcabab.aaca rule b-> at position 5 .aabcaab.aaca rule b-> at position 6 .aabcaa.aaca rule aaa-> aabcc at position 5 .aabcaaabcc.ca rule b-> at position 7 .aabcaaacc.ca rule cc-> cab at position 8 .aabcaaaccab.a rule cc-> cab at position 7 .aabcaaacabab.a rule b-> at position 9 .aabcaaacaab.a rule b-> at position 10 .aabcaaacaa.a rule aaa-> aabcc at position 8 .aabcaaacaabcc. rule b-> at position 10 .aabcaaacaacc. rule cc-> cab at position 10 .aabcaaacaacab.
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