/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 },
it remains to prove termination of the 2-rule system
{ 0 1 0 ⟶ 1 1 1 ,
  1 0 1 ⟶ 0 1 0 }

Loop of length 6 starting with a string of length 6
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.aba.abb
	rule aba-> bbb at position 0
.bbb.abb
	rule bab-> aba at position 2
.bbaba.b
	rule bab-> aba at position 3
.bbaaba.
	rule aba-> bbb at position 3
.bbabbb.
	rule bab-> aba at position 1
.bababb.
	rule bab-> aba at position 0
.abaabb.