/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 2 0 0 ,
  1 ⟶ 0 ,
  2 2 ⟶ 1 }

Loop of length 12 starting with a string of length 4
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ab.bb
	rule ab-> bcaa at position 0
.bcaa.bb
	rule ab-> bcaa at position 3
.bcabcaa.b
	rule ab-> bcaa at position 2
.bcbcaacaa.b
	rule b-> a at position 2
.bcacaacaa.b
	rule a->  at position 4
.bcacacaa.b
	rule a->  at position 4
.bcaccaa.b
	rule cc-> b at position 3
.bcabaa.b
	rule ab-> bcaa at position 5
.bcababcaa.
	rule ab-> bcaa at position 4
.bcabbcaacaa.
	rule a->  at position 6
.bcabbcacaa.
	rule a->  at position 6
.bcabbccaa.
	rule cc-> b at position 5
.bcabbbaa.