/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 ⟶ 1 2 ,
  0 1 1 ⟶ 2 0 0 0 ,
  2 2 ⟶ }

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 ,
  2 2 0 ⟶ 0 0 0 1 ,
  1 1 ⟶ }

Loop of length 17 starting with a string of length 9
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.a.cabcabca
	rule a-> bc at position 0
.bc.cabcabca
	rule cca-> aaab at position 1
.baaab.bcabca
	rule a-> bc at position 2
.babcab.bcabca
	rule a-> bc at position 4
.babcbcb.bcabca
	rule bb->  at position 6
.babcbc.cabca
	rule cca-> aaab at position 5
.babcbaaab.bca
	rule a-> bc at position 5
.babcbbcaab.bca
	rule bb->  at position 4
.babccaab.bca
	rule cca-> aaab at position 3
.babaaabab.bca
	rule a-> bc at position 3
.babbcaabab.bca
	rule bb->  at position 2
.bacaabab.bca
	rule a-> bc at position 4
.bacabcbab.bca
	rule a-> bc at position 7
.bacabcbbcb.bca
	rule bb->  at position 6
.bacabccb.bca
	rule bb->  at position 7
.bacabcc.ca
	rule cca-> aaab at position 6
.bacabcaaab.
	rule a-> bc at position 7
.bacabcabcab.