/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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NO

After renaming modulo the bijection { a ↦ 0, b ↦ 1 },
it remains to prove termination of the 2-rule system
{ 0 0 1 1 ⟶ 1 1 1 0 0 0 ,
  0 ⟶ }

Loop of length 13 starting with a string of length 15
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.aabb.abbbbbbabbb
	rule aabb-> bbbaaa at position 0
.bbbaaa.abbbbbbabbb
	rule aabb-> bbbaaa at position 5
.bbbaabbbaaa.bbbbabbb
	rule aabb-> bbbaaa at position 3
.bbbbbbaaabaaa.bbbbabbb
	rule a->  at position 10
.bbbbbbaaabaa.bbbbabbb
	rule aabb-> bbbaaa at position 10
.bbbbbbaaabbbbaaa.bbabbb
	rule aabb-> bbbaaa at position 7
.bbbbbbabbbaaabbaaa.bbabbb
	rule aabb-> bbbaaa at position 16
.bbbbbbabbbaaabbabbbaaa.abbb
	rule aabb-> bbbaaa at position 21
.bbbbbbabbbaaabbabbbaabbbaaa.b
	rule aabb-> bbbaaa at position 19
.bbbbbbabbbaaabbabbbbbbaaabaaa.b
	rule a->  at position 26
.bbbbbbabbbaaabbabbbbbbaaabaa.b
	rule a->  at position 26
.bbbbbbabbbaaabbabbbbbbaaaba.b
	rule a->  at position 26
.bbbbbbabbbaaabbabbbbbbaaab.b
	rule aabb-> bbbaaa at position 23
.bbbbbbabbbaaabbabbbbbbabbbaaa.