/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 ,
  1 1 2 ⟶ 2 2 0 0 0 ,
  2 ⟶ }

Loop of length 10 starting with a string of length 6
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.a.bcbcc
	rule a-> b at position 0
.b.bcbcc
	rule bbc-> ccaaa at position 0
.ccaaa.bcc
	rule a-> b at position 3
.ccaba.bcc
	rule a-> b at position 4
.ccabb.bcc
	rule bbc-> ccaaa at position 4
.ccabccaaa.c
	rule c->  at position 4
.ccabcaaa.c
	rule a-> b at position 5
.ccabcbaa.c
	rule a-> b at position 6
.ccabcbba.c
	rule a-> b at position 7
.ccabcbbb.c
	rule bbc-> ccaaa at position 6
.ccabcbccaaa.