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

Loop of length 9 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.aaaa.aaaa
	rule aaaa-> baaa at position 0
.baaa.aaaa
	rule aaaa-> baaa at position 1
.bbaaa.aaa
	rule bb-> aab at position 0
.aabaaa.aaa
	rule aaaa-> baaa at position 3
.aabbaaa.aa
	rule bb-> aab at position 2
.aaaabaaa.aa
	rule aaaa-> baaa at position 5
.aaaabbaaa.a
	rule bb-> aab at position 4
.aaaaaabaaa.a
	rule aaaa-> baaa at position 7
.aaaaaabbaaa.
	rule bb-> aab at position 6
.aaaaaaaabaaa.