/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


NO

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

Loop of length 24 starting with a string of length 11
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.a.bacbaccacc
	rule a-> b at position 0
.b.bacbaccacc
	rule bbac-> ccaaaa at position 0
.ccaaaa.baccacc
	rule a-> b at position 2
.ccbaaa.baccacc
	rule a-> b at position 3
.ccbbaa.baccacc
	rule a-> b at position 5
.ccbbab.baccacc
	rule bbac-> ccaaaa at position 5
.ccbbaccaaaa.cacc
	rule bbac-> ccaaaa at position 2
.ccccaaaacaaaa.cacc
	rule a-> b at position 5
.ccccabaacaaaa.cacc
	rule a-> b at position 6
.ccccabbacaaaa.cacc
	rule bbac-> ccaaaa at position 5
.ccccaccaaaaaaaa.cacc
	rule a-> b at position 9
.ccccaccaabaaaaa.cacc
	rule a-> b at position 10
.ccccaccaabbaaaa.cacc
	rule a-> b at position 12
.ccccaccaabbabaa.cacc
	rule a-> b at position 13
.ccccaccaabbabba.cacc
	rule bbac-> ccaaaa at position 12
.ccccaccaabbaccaaaa.acc
	rule bbac-> ccaaaa at position 9
.ccccaccaaccaaaacaaaa.acc
	rule a-> b at position 13
.ccccaccaaccaabacaaaa.acc
	rule a-> b at position 16
.ccccaccaaccaabacbaaa.acc
	rule a-> b at position 18
.ccccaccaaccaabacbaba.acc
	rule a-> b at position 19
.ccccaccaaccaabacbabb.acc
	rule bbac-> ccaaaa at position 18
.ccccaccaaccaabacbaccaaaa.c
	rule a-> b at position 21
.ccccaccaaccaabacbaccabaa.c
	rule a-> b at position 22
.ccccaccaaccaabacbaccabba.c
	rule bbac-> ccaaaa at position 21
.ccccaccaaccaabacbaccaccaaaa.