/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 { b ↦ 0, a ↦ 1 },
it remains to prove termination of the 3-rule system
{ 0 1 1 ⟶ 1 ,
  1 1 1 ⟶ 0 0 0 ,
  0 0 ⟶ 1 0 1 }

Loop of length 8 starting with a string of length 4
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.aa.bb
	rule aa-> bab at position 0
.bab.bb
	rule bbb-> aaa at position 2
.baaaa.
	rule aa-> bab at position 1
.bbabaa.
	rule aa-> bab at position 4
.bbabbab.
	rule abb-> b at position 2
.bbbab.
	rule bbb-> aaa at position 0
.aaaab.
	rule aa-> bab at position 2
.aababb.
	rule abb-> b at position 3
.aabb.