/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 ,
  0 2 2 ⟶ 1 }

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 0 2 1 ,
  2 2 0 ⟶ 1 }

Loop of length 14 starting with a string of length 6
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ba.aaaa
	rule ba-> aaacb at position 0
.aaacb.aaaa
	rule ba-> aaacb at position 4
.aaacaaacb.aaa
	rule a->  at position 4
.aaacaacb.aaa
	rule a->  at position 4
.aaacacb.aaa
	rule a->  at position 4
.aaaccb.aaa
	rule ba-> aaacb at position 5
.aaaccaaacb.aa
	rule cca-> b at position 3
.aaabaacb.aa
	rule ba-> aaacb at position 7
.aaabaacaaacb.a
	rule a->  at position 7
.aaabaacaacb.a
	rule a->  at position 7
.aaabaacacb.a
	rule a->  at position 7
.aaabaaccb.a
	rule ba-> aaacb at position 8
.aaabaaccaaacb.
	rule cca-> b at position 6
.aaabaabaacb.
	rule ba-> aaacb at position 6
.aaabaaaaacbacb.