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

Loop of length 10 starting with a string of length 7
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.aab.babb
	rule aab-> ccaaa at position 0
.ccaaa.babb
	rule aab-> ccaaa at position 3
.ccaccaaa.abb
	rule ac-> ba at position 2
.ccbacaaa.abb
	rule ac-> ba at position 3
.ccbbaaaa.abb
	rule aab-> ccaaa at position 7
.ccbbaaaccaaa.b
	rule ac-> ba at position 6
.ccbbaabacaaa.b
	rule ac-> ba at position 7
.ccbbaabbaaaa.b
	rule aab-> ccaaa at position 10
.ccbbaabbaaccaaa.
	rule ac-> ba at position 9
.ccbbaabbabacaaa.
	rule ac-> ba at position 10
.ccbbaabbabbaaaa.