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

Loop of length 29 starting with a string of length 11
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ba.acbcbcbcb
	rule ba-> acb at position 0
.acb.acbcbcbcb
	rule ba-> acb at position 2
.acacb.cbcbcbcb
	rule a->  at position 2
.accb.cbcbcbcb
	rule ccb-> abc at position 1
.aabc.cbcbcbcb
	rule ccb-> abc at position 3
.aababc.cbcbcb
	rule ccb-> abc at position 5
.aabababc.cbcb
	rule ba-> acb at position 4
.aabaacbbc.cbcb
	rule ccb-> abc at position 8
.aabaacbbabc.cb
	rule ba-> acb at position 7
.aabaacbacbbc.cb
	rule ba-> acb at position 6
.aabaacacbcbbc.cb
	rule a->  at position 6
.aabaaccbcbbc.cb
	rule ccb-> abc at position 11
.aabaaccbcbbabc.
	rule ba-> acb at position 10
.aabaaccbcbacbbc.
	rule ba-> acb at position 9
.aabaaccbcacbcbbc.
	rule a->  at position 9
.aabaaccbccbcbbc.
	rule ccb-> abc at position 8
.aabaaccbabccbbc.
	rule ba-> acb at position 7
.aabaaccacbbccbbc.
	rule a->  at position 7
.aabaacccbbccbbc.
	rule ccb-> abc at position 10
.aabaacccbbabcbc.
	rule ba-> acb at position 9
.aabaacccbacbbcbc.
	rule ba-> acb at position 8
.aabaacccacbcbbcbc.
	rule a->  at position 8
.aabaaccccbcbbcbc.
	rule ccb-> abc at position 7
.aabaaccabccbbcbc.
	rule a->  at position 7
.aabaaccbccbbcbc.
	rule ccb-> abc at position 8
.aabaaccbabcbcbc.
	rule ba-> acb at position 7
.aabaaccacbbcbcbc.
	rule a->  at position 7
.aabaacccbbcbcbc.
	rule ccb-> abc at position 6
.aabaacabcbcbcbc.
	rule a->  at position 6
.aabaacbcbcbcbc.