/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/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 5-rule system
{ 0 ⟶ ,
  0 0 ⟶ 1 2 ,
  1 ⟶ ,
  2 ⟶ ,
  2 1 ⟶ 1 0 2 }

The system was reversed.

After renaming modulo the bijection { 0 ↦ 0, 2 ↦ 1, 1 ↦ 2 },
it remains to prove termination of the 5-rule system
{ 0 ⟶ ,
  0 0 ⟶ 1 2 ,
  2 ⟶ ,
  1 ⟶ ,
  2 1 ⟶ 1 0 2 }

Loop of length 16 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.aa.bbbbbb
	rule aa-> bc at position 0
.bc.bbbbbb
	rule cb-> bac at position 1
.bbac.bbbbb
	rule cb-> bac at position 3
.bbabac.bbbb
	rule b->  at position 3
.bbaac.bbbb
	rule cb-> bac at position 4
.bbaabac.bbb
	rule cb-> bac at position 6
.bbaababac.bb
	rule b->  at position 6
.bbaabaac.bb
	rule aa-> bc at position 5
.bbaabbcc.bb
	rule cb-> bac at position 7
.bbaabbcbac.b
	rule cb-> bac at position 6
.bbaabbbacac.b
	rule c->  at position 8
.bbaabbbaac.b
	rule aa-> bc at position 7
.bbaabbbbcc.b
	rule cb-> bac at position 9
.bbaabbbbcbac.
	rule cb-> bac at position 8
.bbaabbbbbacac.
	rule c->  at position 10
.bbaabbbbbaac.
	rule aa-> bc at position 9
.bbaabbbbbbcc.