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

Loop of length 6 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.bb.aabaab
	rule bb-> aba at position 0
.aba.aabaab
	rule aaa-> bb at position 2
.abbb.baab
	rule bb-> aba at position 3
.abbaba.aab
	rule aaa-> bb at position 5
.abbabbb.b
	rule bb-> aba at position 4
.abbaabab.b
	rule bb-> aba at position 7
.abbaabaaba.