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

Loop of length 4 starting with a string of length 7
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.abbcb.bb
	rule abbcb-> bbcbaabbc at position 0
.bbcbaabbc.bb
	rule abbcb-> bbcbaabbc at position 5
.bbcbabbcbaabbc.b
	rule abbcb-> bbcbaabbc at position 10
.bbcbabbcbabbcbaabbc.
	rule abbcb-> bbcbaabbc at position 9
.bbcbabbcbbbcbaabbcaabbc.