/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 4-rule system
{ 0 ⟶ ,
  0 1 ⟶ 1 1 2 0 ,
  1 ⟶ ,
  2 2 1 ⟶ 0 }

Loop of length 20 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ab.bbbbbb
	rule ab-> bbca at position 0
.bbca.bbbbbb
	rule ab-> bbca at position 3
.bbcbbca.bbbbb
	rule b->  at position 3
.bbcbca.bbbbb
	rule b->  at position 3
.bbcca.bbbbb
	rule ab-> bbca at position 4
.bbccbbca.bbbb
	rule ccb-> a at position 2
.bbabca.bbbb
	rule ab-> bbca at position 5
.bbabcbbca.bbb
	rule b->  at position 5
.bbabcbca.bbb
	rule b->  at position 5
.bbabcca.bbb
	rule ab-> bbca at position 6
.bbabccbbca.bb
	rule ccb-> a at position 4
.bbababca.bb
	rule ab-> bbca at position 4
.bbabbbcaca.bb
	rule a->  at position 7
.bbabbbcca.bb
	rule ab-> bbca at position 8
.bbabbbccbbca.b
	rule ccb-> a at position 6
.bbabbbabca.b
	rule ab-> bbca at position 6
.bbabbbbbcaca.b
	rule a->  at position 9
.bbabbbbbcca.b
	rule ab-> bbca at position 10
.bbabbbbbccbbca.
	rule ccb-> a at position 8
.bbabbbbbabca.
	rule ab-> bbca at position 8
.bbabbbbbbbcaca.