/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 ⟶ 1 2 1 0 ,
  1 ⟶ 0 ,
  2 2 ⟶ }

Loop of length 20 starting with a string of length 4
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ab.bb
	rule ab-> bcba at position 0
.bcba.bb
	rule b-> a at position 0
.acba.bb
	rule b-> a at position 2
.acaa.bb
	rule ab-> bcba at position 3
.acabcba.b
	rule ab-> bcba at position 2
.acbcbacba.b
	rule b-> a at position 2
.acacbacba.b
	rule a->  at position 2
.accbacba.b
	rule cc->  at position 1
.abacba.b
	rule b-> a at position 4
.abacaa.b
	rule ab-> bcba at position 5
.abacabcba.
	rule ab-> bcba at position 4
.abacbcbacba.
	rule b-> a at position 4
.abacacbacba.
	rule a->  at position 4
.abaccbacba.
	rule cc->  at position 3
.ababacba.
	rule ab-> bcba at position 2
.abbcbaacba.
	rule b-> a at position 4
.abbcaaacba.
	rule a->  at position 4
.abbcaacba.
	rule a->  at position 4
.abbcacba.
	rule a->  at position 4
.abbccba.
	rule cc->  at position 3
.abbba.