/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


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 0 2 1 1 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-> aacbba at position 0
.aacbba.b
	rule b->  at position 3
.aacba.b
	rule b->  at position 3
.aaca.b
	rule ab-> aacbba at position 3
.aacaacbba.
	rule a->  at position 3
.aacacbba.
	rule a->  at position 3
.aaccbba.
	rule cc->  at position 2
.aabba.