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

The system was reversed.

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

Loop of length 19 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.bca.aacaa
	rule bca-> acbc at position 0
.acbc.aacaa
	rule c-> ab at position 1
.aabbc.aacaa
	rule bca-> acbc at position 3
.aabacbc.acaa
	rule a->  at position 3
.aabcbc.acaa
	rule bca-> acbc at position 4
.aabcacbc.caa
	rule c-> ab at position 5
.aabcaabbc.caa
	rule c-> ab at position 8
.aabcaabbab.caa
	rule a->  at position 8
.aabcaabbb.caa
	rule bca-> acbc at position 8
.aabcaabbacbc.a
	rule a->  at position 8
.aabcaabbcbc.a
	rule bca-> acbc at position 9
.aabcaabbcacbc.
	rule bca-> acbc at position 7
.aabcaabacbccbc.
	rule a->  at position 7
.aabcaabcbccbc.
	rule c-> ab at position 10
.aabcaabcbcabbc.
	rule bca-> acbc at position 8
.aabcaabcacbcbbc.
	rule bca-> acbc at position 6
.aabcaaacbccbcbbc.
	rule c-> ab at position 10
.aabcaaacbcabbcbbc.
	rule bca-> acbc at position 8
.aabcaaacacbcbbcbbc.
	rule c-> ab at position 9
.aabcaaacaabbcbbcbbc.