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

The system was reversed.

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

Loop of length 24 starting with a string of length 6
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.cb.bbbb
	rule cb-> acac at position 0
.acac.bbbb
	rule cb-> acac at position 3
.acaacac.bbb
	rule ca-> bb at position 4
.acaabbc.bbb
	rule ab->  at position 3
.acabc.bbb
	rule ab->  at position 2
.acc.bbb
	rule cb-> acac at position 2
.acacac.bb
	rule ca-> bb at position 3
.acabbc.bb
	rule ab->  at position 2
.acbc.bb
	rule cb-> acac at position 1
.aacacc.bb
	rule cb-> acac at position 5
.aacacacac.b
	rule ca-> bb at position 4
.aacabbcac.b
	rule ab->  at position 3
.aacbcac.b
	rule cb-> acac at position 2
.aaacaccac.b
	rule ca-> bb at position 6
.aaacacbbc.b
	rule cb-> acac at position 5
.aaacaacacbc.b
	rule ca-> bb at position 6
.aaacaabbcbc.b
	rule ab->  at position 5
.aaacabcbc.b
	rule ab->  at position 4
.aaaccbc.b
	rule cb-> acac at position 4
.aaacacacc.b
	rule ca-> bb at position 5
.aaacabbcc.b
	rule ab->  at position 4
.aaacbcc.b
	rule cb-> acac at position 6
.aaacbcacac.
	rule ca-> bb at position 5
.aaacbbbcac.
	rule ca-> bb at position 7
.aaacbbbbbc.