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

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

.abb.bbbbbb
	rule abb-> bbbaca at position 0
.bbbaca.bbbbbb
	rule abb-> bbbaca at position 5
.bbbacbbbaca.bbbb
	rule cb->  at position 4
.bbbabbaca.bbbb
	rule abb-> bbbaca at position 8
.bbbabbacbbbaca.bb
	rule cb->  at position 7
.bbbabbabbaca.bb
	rule abb-> bbbaca at position 6
.bbbabbbbbacaaca.bb
	rule a->  at position 11
.bbbabbbbbacaca.bb
	rule abb-> bbbaca at position 13
.bbbabbbbbacacbbbaca.
	rule cb->  at position 12
.bbbabbbbbacabbaca.
	rule abb-> bbbaca at position 11
.bbbabbbbbacbbbacaaca.
	rule cb->  at position 10
.bbbabbbbbabbacaaca.
	rule abb-> bbbaca at position 9
.bbbabbbbbbbbacaacaaca.