/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 { 0 ↦ 0, 1 ↦ 1 },
it remains to prove termination of the 2-rule system
{ 0 0 0 0 ⟶ 0 1 1 0 ,
  1 0 0 1 ⟶ 0 0 1 0 }

Loop of length 15 starting with a string of length 12
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.aaaa.abababaa
	rule aaaa-> abba at position 0
.abba.abababaa
	rule baab-> aaba at position 2
.abaaba.ababaa
	rule baab-> aaba at position 4
.abaaaaba.abaa
	rule baab-> aaba at position 6
.abaaaaaaba.aa
	rule aaaa-> abba at position 3
.abaabbaaba.aa
	rule baab-> aaba at position 5
.abaabaabaa.aa
	rule baab-> aaba at position 4
.abaaaabaaa.aa
	rule aaaa-> abba at position 8
.abaaaabaabba.
	rule baab-> aaba at position 6
.abaaaaaababa.
	rule aaaa-> abba at position 3
.abaabbaababa.
	rule baab-> aaba at position 1
.aaababaababa.
	rule baab-> aaba at position 5
.aaabaaabaaba.
	rule baab-> aaba at position 7
.aaabaaaaabaa.
	rule aaaa-> abba at position 5
.aaabaabbabaa.
	rule baab-> aaba at position 3
.aaaaabababaa.