/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 0 ⟶ 1 0 1 2 ,
  2 ⟶ ,
  2 1 ⟶ 0 2 }

Loop of length 18 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.cb.bbbbab
	rule cb-> ac at position 0
.ac.bbbbab
	rule cb-> ac at position 1
.aac.bbbab
	rule aa-> babc at position 0
.babcc.bbbab
	rule cb-> ac at position 4
.babcac.bbab
	rule cb-> ac at position 5
.babcaac.bab
	rule aa-> babc at position 4
.babcbabcc.bab
	rule a->  at position 5
.babcbbcc.bab
	rule cb-> ac at position 7
.babcbbcac.ab
	rule c->  at position 8
.babcbbca.ab
	rule aa-> babc at position 7
.babcbbcbabc.b
	rule cb-> ac at position 6
.babcbbacabc.b
	rule c->  at position 7
.babcbbaabc.b
	rule aa-> babc at position 6
.babcbbbabcbc.b
	rule a->  at position 7
.babcbbbbcbc.b
	rule cb-> ac at position 8
.babcbbbbacc.b
	rule c->  at position 9
.babcbbbbac.b
	rule cb-> ac at position 9
.babcbbbbaac.
	rule aa-> babc at position 8
.babcbbbbbabcc.