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

Loop of length 21 starting with a string of length 13
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.cbbb.bbbbbbbbb
	rule cbbb-> aacc at position 0
.aacc.bbbbbbbbb
	rule cbbb-> aacc at position 3
.aacaacc.bbbbbb
	rule a-> bb at position 3
.aacbbacc.bbbbbb
	rule a-> bb at position 5
.aacbbbbcc.bbbbbb
	rule cbbb-> aacc at position 8
.aacbbbbcaacc.bbb
	rule a-> bb at position 8
.aacbbbbcbbacc.bbb
	rule a-> bb at position 10
.aacbbbbcbbbbcc.bbb
	rule cbbb-> aacc at position 7
.aacbbbbaaccbcc.bbb
	rule a-> bb at position 7
.aacbbbbbbaccbcc.bbb
	rule a-> bb at position 9
.aacbbbbbbbbccbcc.bbb
	rule cbbb-> aacc at position 15
.aacbbbbbbbbccbcaacc.
	rule a-> bb at position 15
.aacbbbbbbbbccbcbbacc.
	rule a-> bb at position 17
.aacbbbbbbbbccbcbbbbcc.
	rule cbbb-> aacc at position 14
.aacbbbbbbbbccbaaccbcc.
	rule a-> bb at position 14
.aacbbbbbbbbccbbbaccbcc.
	rule cbbb-> aacc at position 12
.aacbbbbbbbbcaaccaccbcc.
	rule a-> bb at position 12
.aacbbbbbbbbcbbaccaccbcc.
	rule a-> bb at position 14
.aacbbbbbbbbcbbbbccaccbcc.
	rule cbbb-> aacc at position 11
.aacbbbbbbbbaaccbccaccbcc.
	rule a-> bb at position 11
.aacbbbbbbbbbbaccbccaccbcc.
	rule a-> bb at position 13
.aacbbbbbbbbbbbbccbccaccbcc.