/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 ⟶ 1 ,
  0 1 ⟶ 1 1 0 2 2 0 ,
  2 1 ⟶ }

Loop of length 10 starting with a string of length 4
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ab.bb
	rule ab-> bbacca at position 0
.bbacca.bb
	rule ab-> bbacca at position 5
.bbaccbbacca.b
	rule cb->  at position 4
.bbacbacca.b
	rule cb->  at position 3
.bbaacca.b
	rule ab-> bbacca at position 6
.bbaaccbbacca.
	rule cb->  at position 5
.bbaacbacca.
	rule cb->  at position 4
.bbaaacca.
	rule a-> b at position 4
.bbaabcca.
	rule ab-> bbacca at position 3
.bbabbaccacca.
	rule a-> b at position 5
.bbabbbccacca.