/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 1 1 ⟶ 1 1 0 0 ,
  0 0 ⟶ 1 }

Loop of length 6 starting with a string of length 7
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.abb.bbbb
	rule abb-> bbaa at position 0
.bbaa.bbbb
	rule abb-> bbaa at position 3
.bbabbaa.bb
	rule abb-> bbaa at position 6
.bbabbabbaa.
	rule abb-> bbaa at position 5
.bbabbbbaaaa.
	rule aa-> b at position 7
.bbabbbbbaa.
	rule aa-> b at position 8
.bbabbbbbb.