/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 ⟶ 2 ,
  2 2 ⟶ 1 2 1 0 0 }

Loop of length 25 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.cc.bcbcbc
	rule cc-> bcbaa at position 0
.bcbaa.bcbcbc
	rule ab-> c at position 4
.bcbac.cbcbc
	rule cc-> bcbaa at position 4
.bcbabcbaa.bcbc
	rule ab-> c at position 3
.bcbccbaa.bcbc
	rule cc-> bcbaa at position 3
.bcbbcbaabaa.bcbc
	rule ab-> c at position 7
.bcbbcbacaa.bcbc
	rule ab-> c at position 9
.bcbbcbacac.cbc
	rule cc-> bcbaa at position 9
.bcbbcbacabcbaa.bc
	rule ab-> c at position 8
.bcbbcbacccbaa.bc
	rule cc-> bcbaa at position 7
.bcbbcbabcbaacbaa.bc
	rule ab-> c at position 6
.bcbbcbccbaacbaa.bc
	rule cc-> bcbaa at position 6
.bcbbcbbcbaabaacbaa.bc
	rule ab-> c at position 10
.bcbbcbbcbacaacbaa.bc
	rule a->  at position 11
.bcbbcbbcbacacbaa.bc
	rule a->  at position 11
.bcbbcbbcbaccbaa.bc
	rule cc-> bcbaa at position 10
.bcbbcbbcbabcbaabaa.bc
	rule ab-> c at position 9
.bcbbcbbcbccbaabaa.bc
	rule a->  at position 12
.bcbbcbbcbccbabaa.bc
	rule ab-> c at position 12
.bcbbcbbcbccbcaa.bc
	rule ab-> c at position 14
.bcbbcbbcbccbcac.c
	rule cc-> bcbaa at position 14
.bcbbcbbcbccbcabcbaa.
	rule ab-> c at position 13
.bcbbcbbcbccbcccbaa.
	rule cc-> bcbaa at position 13
.bcbbcbbcbccbcbcbaabaa.
	rule a->  at position 16
.bcbbcbbcbccbcbcbabaa.
	rule ab-> c at position 16
.bcbbcbbcbccbcbcbcaa.