/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 ⟶ 0 2 1 1 0 0 ,
  1 ⟶ ,
  2 2 ⟶ }

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

.ab.b
	rule ab-> acbbaa at position 0
.acbbaa.b
	rule b->  at position 2
.acbaa.b
	rule b->  at position 2
.acaa.b
	rule a->  at position 2
.aca.b
	rule ab-> acbbaa at position 2
.acacbbaa.
	rule a->  at position 2
.accbbaa.
	rule cc->  at position 1
.abbaa.