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

The system was reversed.

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

Loop of length 14 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.aaa.caaca
	rule aaa-> aabcc at position 0
.aabcc.caaca
	rule cc-> cab at position 4
.aabccab.aaca
	rule cc-> cab at position 3
.aabcabab.aaca
	rule b->  at position 5
.aabcaab.aaca
	rule b->  at position 6
.aabcaa.aaca
	rule aaa-> aabcc at position 5
.aabcaaabcc.ca
	rule b->  at position 7
.aabcaaacc.ca
	rule cc-> cab at position 8
.aabcaaaccab.a
	rule cc-> cab at position 7
.aabcaaacabab.a
	rule b->  at position 9
.aabcaaacaab.a
	rule b->  at position 10
.aabcaaacaa.a
	rule aaa-> aabcc at position 8
.aabcaaacaabcc.
	rule b->  at position 10
.aabcaaacaacc.
	rule cc-> cab at position 10
.aabcaaacaacab.