/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 3-rule system
{ 0 ⟶ ,
  0 1 ⟶ 2 0 ,
  2 2 ⟶ 1 1 2 1 0 }

The system was reversed.

After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2 },
it remains to prove termination of the 3-rule system
{ 0 ⟶ ,
  1 0 ⟶ 0 2 ,
  2 2 ⟶ 0 1 2 1 1 }

Loop of length 17 starting with a string of length 6
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.cc.acac
	rule cc-> abcbb at position 0
.abcbb.acac
	rule ba-> ac at position 4
.abcbac.cac
	rule a->  at position 4
.abcbc.cac
	rule cc-> abcbb at position 4
.abcbabcbb.ac
	rule ba-> ac at position 3
.abcacbcbb.ac
	rule a->  at position 3
.abccbcbb.ac
	rule ba-> ac at position 7
.abccbcbac.c
	rule ba-> ac at position 6
.abccbcacc.c
	rule a->  at position 6
.abccbccc.c
	rule cc-> abcbb at position 5
.abccbabcbbc.c
	rule ba-> ac at position 4
.abccacbcbbc.c
	rule cc-> abcbb at position 10
.abccacbcbbabcbb.
	rule ba-> ac at position 9
.abccacbcbacbcbb.
	rule ba-> ac at position 8
.abccacbcaccbcbb.
	rule a->  at position 8
.abccacbcccbcbb.
	rule cc-> abcbb at position 7
.abccacbabcbbcbcbb.
	rule ba-> ac at position 6
.abccacacbcbbcbcbb.