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

Loop of length 24 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ab.bbbbbb
	rule ab-> ca at position 0
.ca.bbbbbb
	rule ab-> ca at position 1
.cca.bbbbb
	rule ab-> ca at position 2
.ccca.bbbb
	rule ab-> ca at position 3
.cccca.bbb
	rule ab-> ca at position 4
.ccccca.bb
	rule ab->  at position 5
.ccccc.b
	rule cb-> bbba at position 4
.ccccbbba.
	rule cb-> bbba at position 3
.cccbbbabba.
	rule cb-> bbba at position 2
.ccbbbabbabba.
	rule cb-> bbba at position 1
.cbbbabbabbabba.
	rule cb-> bbba at position 0
.bbbabbabbabbabba.
	rule ab-> ca at position 6
.bbbabbcababbabba.
	rule ab-> ca at position 7
.bbbabbccaabbabba.
	rule ab-> ca at position 9
.bbbabbccacababba.
	rule ab->  at position 10
.bbbabbccacabba.
	rule ab->  at position 10
.bbbabbccacba.
	rule cb-> bbba at position 9
.bbbabbccabbbaa.
	rule ab-> ca at position 8
.bbbabbcccabbaa.
	rule ab->  at position 9
.bbbabbcccbaa.
	rule cb-> bbba at position 8
.bbbabbccbbbaaa.
	rule cb-> bbba at position 7
.bbbabbcbbbabbaaa.
	rule cb-> bbba at position 6
.bbbabbbbbabbabbaaa.
	rule ab->  at position 9
.bbbabbbbbbabbaaa.
	rule ab->  at position 10
.bbbabbbbbbbaaa.