/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 0 0 2 0 0 ,
  2 2 ⟶ 1 }

Loop of length 7 starting with a string of length 3
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ab.c
	rule ab-> baacaa at position 0
.baacaa.c
	rule a->  at position 4
.baaca.c
	rule a->  at position 4
.baac.c
	rule cc-> b at position 3
.baab.
	rule ab-> baacaa at position 2
.babaacaa.
	rule a->  at position 3
.babacaa.
	rule a->  at position 3
.babcaa.