/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 4-rule system
{ 0 ⟶ ,
  0 1 ⟶ ,
  1 1 ⟶ 2 ,
  2 0 ⟶ 0 0 1 2 }

Loop of length 14 starting with a string of length 7
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.bb.aaaaa
	rule bb-> c at position 0
.c.aaaaa
	rule ca-> aabc at position 0
.aabc.aaaa
	rule ca-> aabc at position 3
.aabaabc.aaa
	rule a->  at position 3
.aababc.aaa
	rule a->  at position 3
.aabbc.aaa
	rule ca-> aabc at position 4
.aabbaabc.aa
	rule ca-> aabc at position 7
.aabbaabaabc.a
	rule a->  at position 7
.aabbaababc.a
	rule a->  at position 7
.aabbaabbc.a
	rule bb-> c at position 6
.aabbaacc.a
	rule ca-> aabc at position 7
.aabbaacaabc.
	rule ca-> aabc at position 6
.aabbaaaabcabc.
	rule ab->  at position 7
.aabbaaacabc.
	rule ca-> aabc at position 7
.aabbaaaaabcbc.