/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 4-rule system
{ 0 ⟶ ,
  0 ⟶ 1 ,
  0 1 2 ⟶ 2 2 1 0 0 ,
  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 ⟶ ,
  0 ⟶ 1 ,
  2 1 0 ⟶ 0 0 1 2 2 ,
  2 ⟶ }

Loop of length 11 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.cba.babaa
	rule cba-> aabcc at position 0
.aabcc.babaa
	rule cba-> aabcc at position 4
.aabcaabcc.baa
	rule a-> b at position 4
.aabcbabcc.baa
	rule cba-> aabcc at position 8
.aabcbabcaabcc.a
	rule a-> b at position 8
.aabcbabcbabcc.a
	rule cba-> aabcc at position 7
.aabcbabaabccbcc.a
	rule a->  at position 7
.aabcbababccbcc.a
	rule c->  at position 9
.aabcbababcbcc.a
	rule c->  at position 11
.aabcbababcbc.a
	rule c->  at position 11
.aabcbababcb.a
	rule cba-> aabcc at position 9
.aabcbababaabcc.