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

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

.ababba.abbaa
	rule ababba-> aababa at position 0
.aababa.abbaa
	rule abaa-> aabbaba at position 3
.aabaabbaba.bbaa
	rule abaa-> aabbaba at position 1
.aaabbababbaba.bbaa
	rule ababba-> aababa at position 5
.aaabbaabababa.bbaa
	rule ababba-> aababa at position 10
.aaabbaababaababa.a
	rule abaa-> aabbaba at position 8
.aaabbaabaabbabababa.a
	rule abaa-> aabbaba at position 6
.aaabbaaabbababbabababa.a
	rule abaa-> aabbaba at position 19
.aaabbaaabbababbababaabbaba.
	rule abaa-> aabbaba at position 17
.aaabbaaabbababbabaabbababbaba.
	rule abaa-> aabbaba at position 15
.aaabbaaabbababbaabbababbababbaba.
	rule ababba-> aababa at position 19
.aaabbaaabbababbaabbaababababbaba.