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

Loop of length 12 starting with a string of length 9
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.abc.bccccc
	rule abc-> ccaa at position 0
.ccaa.bccccc
	rule abc-> ccaa at position 3
.ccaccaa.cccc
	rule cc-> b at position 3
.ccabaa.cccc
	rule a-> b at position 5
.ccabab.cccc
	rule abc-> ccaa at position 4
.ccabccaa.ccc
	rule a-> b at position 7
.ccabccab.ccc
	rule abc-> ccaa at position 6
.ccabccccaa.cc
	rule cc-> b at position 5
.ccabcbcaa.cc
	rule a-> b at position 8
.ccabcbcab.cc
	rule abc-> ccaa at position 7
.ccabcbcccaa.c
	rule a-> b at position 10
.ccabcbcccab.c
	rule abc-> ccaa at position 9
.ccabcbcccccaa.