/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 0 1 ⟶ 2 2 2 ,
  2 ⟶ 1 0 ,
  2 1 ⟶ }

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 ⟶ 2 2 2 ,
  2 ⟶ 0 1 ,
  1 2 ⟶ }

Loop of length 16 starting with a string of length 5
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.baa.aa
	rule baa-> ccc at position 0
.ccc.aa
	rule c-> ab at position 0
.abcc.aa
	rule c-> ab at position 2
.ababc.aa
	rule c-> ab at position 4
.ababab.aa
	rule baa-> ccc at position 5
.ababaccc.
	rule c-> ab at position 5
.ababaabcc.
	rule baa-> ccc at position 3
.abacccbcc.
	rule c-> ab at position 3
.abaabccbcc.
	rule c-> ab at position 5
.abaababcbcc.
	rule bc->  at position 6
.abaababcc.
	rule bc->  at position 6
.abaabac.
	rule c-> ab at position 6
.abaabaab.
	rule baa-> ccc at position 4
.abaacccb.
	rule c-> ab at position 4
.abaaabccb.
	rule bc->  at position 5
.abaaacb.
	rule c-> ab at position 5
.abaaaabb.