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

The system was reversed.

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

Loop of length 16 starting with a string of length 9
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.cba.bababa
	rule cba-> aaaacc at position 0
.aaaacc.bababa
	rule cba-> aaaacc at position 5
.aaaacaaaacc.baba
	rule aa-> b at position 5
.aaaacbaacc.baba
	rule cba-> aaaacc at position 9
.aaaacbaacaaaacc.ba
	rule aa-> b at position 9
.aaaacbaacbaacc.ba
	rule cba-> aaaacc at position 8
.aaaacbaaaaaaccacc.ba
	rule aa-> b at position 7
.aaaacbabaaaccacc.ba
	rule aa-> b at position 9
.aaaacbababccacc.ba
	rule cba-> aaaacc at position 14
.aaaacbababccacaaaacc.
	rule aa-> b at position 14
.aaaacbababccacbaacc.
	rule cba-> aaaacc at position 13
.aaaacbababccaaaaaccacc.
	rule aa-> b at position 12
.aaaacbababccbaaaccacc.
	rule cba-> aaaacc at position 11
.aaaacbababcaaaaccaaccacc.
	rule aa-> b at position 11
.aaaacbababcbaaccaaccacc.
	rule cba-> aaaacc at position 10
.aaaacbababaaaaccaccaaccacc.
	rule aa-> b at position 11
.aaaacbabababaccaccaaccacc.