/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 3-rule system
{ 0 ⟶ ,
  0 ⟶ 1 ,
  1 1 2 ⟶ 0 2 2 2 0 1 }

The system was reversed.

After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2 },
it remains to prove termination of the 3-rule system
{ 0 ⟶ ,
  0 ⟶ 1 ,
  2 1 1 ⟶ 1 0 2 2 2 0 }

Loop of length 11 starting with a string of length 5
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.cbb.bb
	rule cbb-> baccca at position 0
.baccca.bb
	rule a-> b at position 5
.bacccb.bb
	rule cbb-> baccca at position 4
.baccbaccca.b
	rule a-> b at position 5
.baccbbccca.b
	rule a-> b at position 9
.baccbbcccb.b
	rule cbb-> baccca at position 8
.baccbbccbaccca.
	rule a-> b at position 9
.baccbbccbbccca.
	rule cbb-> baccca at position 7
.baccbbcbacccaccca.
	rule a-> b at position 8
.baccbbcbbcccaccca.
	rule cbb-> baccca at position 6
.baccbbbacccacccaccca.
	rule a-> b at position 7
.baccbbbbcccacccaccca.