/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 0 ,
  0 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 0 2 1 ,
  2 2 0 ⟶ 0 1 }

Loop of length 24 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ba.aaaaaa
	rule ba-> aacb at position 0
.aacb.aaaaaa
	rule ba-> aacb at position 3
.aacaacb.aaaaa
	rule a->  at position 3
.aacacb.aaaaa
	rule a->  at position 3
.aaccb.aaaaa
	rule ba-> aacb at position 4
.aaccaacb.aaaa
	rule cca-> ab at position 2
.aaabacb.aaaa
	rule ba-> aacb at position 6
.aaabacaacb.aaa
	rule a->  at position 6
.aaabacacb.aaa
	rule a->  at position 6
.aaabaccb.aaa
	rule ba-> aacb at position 7
.aaabaccaacb.aa
	rule cca-> ab at position 5
.aaabaabacb.aa
	rule ba-> aacb at position 6
.aaabaaaacbcb.aa
	rule ba-> aacb at position 11
.aaabaaaacbcaacb.a
	rule a->  at position 11
.aaabaaaacbcacb.a
	rule a->  at position 11
.aaabaaaacbccb.a
	rule ba-> aacb at position 12
.aaabaaaacbccaacb.
	rule cca-> ab at position 10
.aaabaaaacbabacb.
	rule ba-> aacb at position 9
.aaabaaaacaacbbacb.
	rule a->  at position 9
.aaabaaaacacbbacb.
	rule a->  at position 9
.aaabaaaaccbbacb.
	rule ba-> aacb at position 11
.aaabaaaaccbaacbcb.
	rule ba-> aacb at position 10
.aaabaaaaccaacbacbcb.
	rule cca-> ab at position 8
.aaabaaaaabacbacbcb.
	rule ba-> aacb at position 9
.aaabaaaaaaacbcbacbcb.