/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 3-rule system
{ 0 ⟶ ,
  0 1 ⟶ 1 0 2 1 0 0 ,
  2 1 2 ⟶ }

Loop of length 14 starting with a string of length 4
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ab.bb
	rule ab-> bacbaa at position 0
.bacbaa.bb
	rule ab-> bacbaa at position 5
.bacbabacbaa.b
	rule ab-> bacbaa at position 4
.bacbbacbaaacbaa.b
	rule a->  at position 8
.bacbbacbaacbaa.b
	rule a->  at position 8
.bacbbacbacbaa.b
	rule a->  at position 8
.bacbbacbcbaa.b
	rule cbc->  at position 6
.bacbbabaa.b
	rule ab-> bacbaa at position 8
.bacbbababacbaa.
	rule ab-> bacbaa at position 7
.bacbbabbacbaaacbaa.
	rule a->  at position 8
.bacbbabbcbaaacbaa.
	rule a->  at position 10
.bacbbabbcbaacbaa.
	rule a->  at position 10
.bacbbabbcbacbaa.
	rule a->  at position 10
.bacbbabbcbcbaa.
	rule cbc->  at position 8
.bacbbabbbaa.