/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 { a ↦ 0, b ↦ 1, c ↦ 2 },
it remains to prove termination of the 4-rule system
{ 0 ⟶ ,
  0 1 ⟶ 1 0 2 1 2 ,
  1 ⟶ 0 ,
  2 2 ⟶ }

Loop of length 11 starting with a string of length 4
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ab.cb
	rule ab-> bacbc at position 0
.bacbc.cb
	rule b-> a at position 0
.aacbc.cb
	rule b-> a at position 3
.aacac.cb
	rule cc->  at position 4
.aaca.b
	rule ab-> bacbc at position 3
.aacbacbc.
	rule b-> a at position 3
.aacaacbc.
	rule a->  at position 3
.aacacbc.
	rule a->  at position 3
.aaccbc.
	rule cc->  at position 2
.aabc.
	rule ab-> bacbc at position 1
.abacbcc.
	rule a->  at position 2
.abcbcc.