/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 3-rule system
{ 0 ⟶ ,
  0 0 1 ⟶ 2 0 ,
  0 2 ⟶ 2 1 0 0 }

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 2 ,
  2 0 ⟶ 0 0 1 2 }

Loop of length 16 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.baa.aaaaa
	rule baa-> ac at position 0
.ac.aaaaa
	rule ca-> aabc at position 1
.aaabc.aaaa
	rule ca-> aabc at position 4
.aaabaabc.aaa
	rule ca-> aabc at position 7
.aaabaabaabc.aa
	rule baa-> ac at position 6
.aaabaaacbc.aa
	rule ca-> aabc at position 9
.aaabaaacbaabc.a
	rule baa-> ac at position 8
.aaabaaacacbc.a
	rule ca-> aabc at position 7
.aaabaaaaabccbc.a
	rule ca-> aabc at position 13
.aaabaaaaabccbaabc.
	rule baa-> ac at position 12
.aaabaaaaabccacbc.
	rule ca-> aabc at position 11
.aaabaaaaabcaabccbc.
	rule ca-> aabc at position 10
.aaabaaaaabaabcabccbc.
	rule baa-> ac at position 9
.aaabaaaaaacbcabccbc.
	rule ca-> aabc at position 12
.aaabaaaaaacbaabcbccbc.
	rule baa-> ac at position 11
.aaabaaaaaacacbcbccbc.
	rule ca-> aabc at position 10
.aaabaaaaaaaabccbcbccbc.