/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 3-rule system
{ 0 0 ⟶ 1 0 1 2 2 ,
  2 0 ⟶ ,
  2 1 ⟶ 0 }

Loop of length 22 starting with a string of length 10
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.aa.bababaab
	rule aa-> babcc at position 0
.babcc.bababaab
	rule cb-> a at position 4
.babca.ababaab
	rule aa-> babcc at position 4
.babcbabcc.babaab
	rule cb-> a at position 3
.babaabcc.babaab
	rule aa-> babcc at position 3
.babbabccbcc.babaab
	rule cb-> a at position 7
.babbabcacc.babaab
	rule cb-> a at position 9
.babbabcaca.abaab
	rule aa-> babcc at position 9
.babbabcacbabcc.baab
	rule cb-> a at position 8
.babbabcaaabcc.baab
	rule aa-> babcc at position 7
.babbabcbabccabcc.baab
	rule cb-> a at position 6
.babbabaabccabcc.baab
	rule ca->  at position 10
.babbabaabcbcc.baab
	rule cb-> a at position 9
.babbabaabacc.baab
	rule cb-> a at position 11
.babbabaabaca.aab
	rule aa-> babcc at position 11
.babbabaabacbabcc.ab
	rule cb-> a at position 10
.babbabaabaaabcc.ab
	rule aa-> babcc at position 10
.babbabaabababccbcc.ab
	rule cb-> a at position 14
.babbabaabababcacc.ab
	rule ca->  at position 16
.babbabaabababcac.b
	rule cb-> a at position 15
.babbabaabababcaa.
	rule aa-> babcc at position 14
.babbabaabababcbabcc.
	rule cb-> a at position 13
.babbabaabababaabcc.