/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 4-rule system
{ 0 ⟶ ,
  0 ⟶ 1 ,
  0 1 2 ⟶ 2 1 0 2 0 ,
  2 ⟶ }

The system was reversed.

After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2 },
it remains to prove termination of the 4-rule system
{ 0 ⟶ ,
  0 ⟶ 1 ,
  2 1 0 ⟶ 0 2 0 1 2 ,
  2 ⟶ }

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

.cba.baba
	rule cba-> acabc at position 0
.acabc.baba
	rule a->  at position 2
.acbc.baba
	rule cba-> acabc at position 3
.acbacabc.ba
	rule a->  at position 5
.acbacbc.ba
	rule cba-> acabc at position 6
.acbacbacabc.
	rule cba-> acabc at position 4
.acbaacabccabc.
	rule a-> b at position 4
.acbabcabccabc.
	rule c->  at position 5
.acbababccabc.
	rule c->  at position 7
.acbababcabc.
	rule c->  at position 7
.acbabababc.