/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, c ↦ 2 },
it remains to prove termination of the 4-rule system
{ 0 ⟶ ,
  0 1 ⟶ 1 1 2 0 ,
  1 ⟶ ,
  1 2 2 ⟶ 0 }

Loop of length 10 starting with a string of length 7
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ab.bbcbb
	rule ab-> bbca at position 0
.bbca.bbcbb
	rule ab-> bbca at position 3
.bbcbbca.bcbb
	rule b->  at position 3
.bbcbca.bcbb
	rule b->  at position 3
.bbcca.bcbb
	rule bcc-> a at position 1
.baa.bcbb
	rule ab-> bbca at position 2
.babbca.cbb
	rule a->  at position 5
.babbc.cbb
	rule bcc-> a at position 3
.baba.bb
	rule ab-> bbca at position 3
.babbbca.b
	rule ab-> bbca at position 6
.babbbcbbca.