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

Loop of length 4 starting with a string of length 13
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.abcbc.bcbcbcbc
	rule abcbc-> cbcbcbaba at position 0
.cbcbcbaba.bcbcbcbc
	rule abcbc-> cbcbcbaba at position 8
.cbcbcbabcbcbcbaba.bcbc
	rule abcbc-> cbcbcbaba at position 16
.cbcbcbabcbcbcbabcbcbcbaba.
	rule abcbc-> cbcbcbaba at position 14
.cbcbcbabcbcbcbcbcbcbababcbaba.