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

Loop of length 13 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ab.cbbcbc
	rule ab-> bca at position 0
.bca.cbbcbc
	rule cac-> baa at position 1
.bbaa.bbcbc
	rule ab-> bca at position 3
.bbabca.bcbc
	rule ab-> bca at position 2
.bbbcaca.bcbc
	rule cac-> baa at position 3
.bbbbaaa.bcbc
	rule ab-> bca at position 6
.bbbbaabca.cbc
	rule ab-> bca at position 5
.bbbbabcaca.cbc
	rule a->  at position 7
.bbbbabcca.cbc
	rule cac-> baa at position 7
.bbbbabcbaa.bc
	rule ab-> bca at position 9
.bbbbabcbabca.c
	rule ab-> bca at position 8
.bbbbabcbbcaca.c
	rule cac-> baa at position 11
.bbbbabcbbcabaa.
	rule ab-> bca at position 10
.bbbbabcbbcbcaaa.