/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/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 4-rule system
{ 0 ⟶ ,
  0 1 ⟶ 1 0 2 1 0 ,
  1 ⟶ ,
  2 2 ⟶ }

The system was reversed.

After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2 },
it remains to prove termination of the 4-rule system
{ 0 ⟶ ,
  1 0 ⟶ 0 1 2 0 1 ,
  1 ⟶ ,
  2 2 ⟶ }

Loop of length 12 starting with a string of length 6
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ba.aaca
	rule ba-> abcab at position 0
.abcab.aaca
	rule a->  at position 3
.abcb.aaca
	rule ba-> abcab at position 3
.abcabcab.aca
	rule a->  at position 3
.abcbcab.aca
	rule b->  at position 3
.abccab.aca
	rule cc->  at position 2
.abab.aca
	rule ba-> abcab at position 3
.abaabcab.ca
	rule a->  at position 6
.abaabcb.ca
	rule b->  at position 6
.abaabc.ca
	rule cc->  at position 5
.abaab.a
	rule ba-> abcab at position 4
.abaaabcab.
	rule b->  at position 5
.abaaacab.