/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 2 ,
  2 2 ⟶ 1 0 2 0 0 }

Loop of length 13 starting with a string of length 6
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ab.ccbc
	rule ab-> bc at position 0
.bc.ccbc
	rule cc-> bacaa at position 1
.bbacaa.cbc
	rule a->  at position 4
.bbaca.cbc
	rule a->  at position 4
.bbac.cbc
	rule cc-> bacaa at position 3
.bbabacaa.bc
	rule ab-> bc at position 2
.bbbcacaa.bc
	rule a->  at position 4
.bbbccaa.bc
	rule cc-> bacaa at position 3
.bbbbacaaaa.bc
	rule ab-> bc at position 9
.bbbbacaaabc.c
	rule ab-> bc at position 8
.bbbbacaabcc.c
	rule ab-> bc at position 7
.bbbbacabccc.c
	rule cc-> bacaa at position 10
.bbbbacabccbacaa.
	rule a->  at position 11
.bbbbacabccbcaa.