/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/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 2-rule system
{ 0 0 1 ⟶ 1 2 1 0 0 0 ,
  0 2 ⟶ 1 }

Loop of length 12 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.aab.babcb
	rule aab-> bcbaaa at position 0
.bcbaaa.babcb
	rule aab-> bcbaaa at position 4
.bcbabcbaaa.abcb
	rule aab-> bcbaaa at position 9
.bcbabcbaabcbaaa.cb
	rule aab-> bcbaaa at position 7
.bcbabcbbcbaaacbaaa.cb
	rule ac-> b at position 12
.bcbabcbbcbaabbaaa.cb
	rule aab-> bcbaaa at position 10
.bcbabcbbcbbcbaaabaaa.cb
	rule aab-> bcbaaa at position 14
.bcbabcbbcbbcbabcbaaaaaa.cb
	rule ac-> b at position 22
.bcbabcbbcbbcbabcbaaaaab.b
	rule aab-> bcbaaa at position 20
.bcbabcbbcbbcbabcbaaabcbaaa.b
	rule aab-> bcbaaa at position 18
.bcbabcbbcbbcbabcbabcbaaacbaaa.b
	rule ac-> b at position 23
.bcbabcbbcbbcbabcbabcbaabbaaa.b
	rule aab-> bcbaaa at position 26
.bcbabcbbcbbcbabcbabcbaabbabcbaaa.