/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 0 ⟶ 1 1 2 ,
  1 ⟶ ,
  2 1 ⟶ 1 0 2 }

Loop of length 15 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.aa.bbbbbb
	rule aa-> bbc at position 0
.bbc.bbbbbb
	rule cb-> bac at position 2
.bbbac.bbbbb
	rule cb-> bac at position 4
.bbbabac.bbbb
	rule b->  at position 4
.bbbaac.bbbb
	rule cb-> bac at position 5
.bbbaabac.bbb
	rule cb-> bac at position 7
.bbbaababac.bb
	rule b->  at position 7
.bbbaabaac.bb
	rule aa-> bbc at position 6
.bbbaabbbcc.bb
	rule cb-> bac at position 9
.bbbaabbbcbac.b
	rule cb-> bac at position 8
.bbbaabbbbacac.b
	rule a->  at position 11
.bbbaabbbbacc.b
	rule cb-> bac at position 11
.bbbaabbbbacbac.
	rule cb-> bac at position 10
.bbbaabbbbabacac.
	rule b->  at position 10
.bbbaabbbbaacac.
	rule aa-> bbc at position 9
.bbbaabbbbbbccac.