/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/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 0 1 1 ,
  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.aabababa
	rule aaaa-> aabb at position 0
.aabb.aabababa
	rule baab-> aaba at position 3
.aabaaba.ababa
	rule baab-> aaba at position 5
.aabaaaaba.aba
	rule aaaa-> aabb at position 3
.aabaabbba.aba
	rule baab-> aaba at position 7
.aabaabbaaba.a
	rule baab-> aaba at position 6
.aabaabaabaa.a
	rule baab-> aaba at position 5
.aabaaaabaaa.a
	rule aaaa-> aabb at position 3
.aabaabbbaaa.a
	rule baab-> aaba at position 2
.aaaababbaaa.a
	rule aaaa-> aabb at position 8
.aaaababbaabb.
	rule baab-> aaba at position 7
.aaaababaabab.
	rule baab-> aaba at position 6
.aaaabaaabaab.
	rule baab-> aaba at position 8
.aaaabaaaaaba.
	rule aaaa-> aabb at position 5
.aaaabaabbaba.
	rule baab-> aaba at position 4
.aaaaaabababa.