/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/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 0 0 ⟶ 1 ,
  1 2 ⟶ 0 2 2 1 ,
  2 ⟶ }

Loop of length 10 starting with a string of length 7
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.aaa.cccc
	rule aaa-> b at position 0
.b.cccc
	rule bc-> accb at position 0
.accb.ccc
	rule c->  at position 1
.acb.ccc
	rule c->  at position 1
.ab.ccc
	rule bc-> accb at position 1
.aaccb.cc
	rule c->  at position 2
.aacb.cc
	rule c->  at position 2
.aab.cc
	rule bc-> accb at position 2
.aaaccb.c
	rule bc-> accb at position 5
.aaaccaccb.
	rule a->  at position 5
.aaaccccb.