/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


NO

After renaming modulo the bijection { a ↦ 0, c ↦ 1 },
it remains to prove termination of the 2-rule system
{ 0 0 0 0 ⟶ 0 1 0 1 1 ,
  1 1 1 ⟶ 0 0 0 }

Loop of length 7 starting with a string of length 13
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.aaaa.baaabaaba
	rule aaaa-> ababb at position 0
.ababb.baaabaaba
	rule bbb-> aaa at position 3
.abaaaa.aaabaaba
	rule aaaa-> ababb at position 5
.abaaaababb.baaba
	rule bbb-> aaa at position 8
.abaaaabaaaa.aaba
	rule aaaa-> ababb at position 9
.abaaaabaaababb.ba
	rule bbb-> aaa at position 12
.abaaaabaaabaaaa.a
	rule aaaa-> ababb at position 12
.abaaaabaaabaababb.