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

Loop of length 4 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.abcb.cbcb
	rule abcb-> bcbcaab at position 0
.bcbcaab.cbcb
	rule abcb-> bcbcaab at position 5
.bcbcabcbcaab.cb
	rule abcb-> bcbcaab at position 10
.bcbcabcbcabcbcaab.
	rule abcb-> bcbcaab at position 9
.bcbcabcbcbcbcaabcaab.