/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 ⟶ 1 ,
  0 ⟶ 1 1 2 ,
  2 0 1 ⟶ 0 2 0 }

Loop of length 12 starting with a string of length 7
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.a.abbbbb
	rule a-> bbc at position 0
.bbc.abbbbb
	rule cab-> aca at position 2
.bbaca.bbbb
	rule cab-> aca at position 3
.bbaaca.bbb
	rule cab-> aca at position 4
.bbaaaca.bb
	rule a-> bbc at position 4
.bbaabbcca.bb
	rule cab-> aca at position 7
.bbaabbcaca.b
	rule cab-> aca at position 8
.bbaabbcaaca.
	rule a-> bbc at position 8
.bbaabbcabbcca.
	rule cab-> aca at position 6
.bbaabbacabcca.
	rule a-> b at position 6
.bbaabbbcabcca.
	rule cab-> aca at position 7
.bbaabbbacacca.
	rule a-> bbc at position 7
.bbaabbbbbccacca.