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

Loop of length 25 starting with a string of length 11
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.aaaa.abbbbab
	rule aaaa-> abbb at position 0
.abbb.abbbbab
	rule babb-> aaba at position 3
.abbaaba.bbab
	rule babb-> aaba at position 5
.abbaaaaba.ab
	rule aaaa-> abbb at position 3
.abbabbbba.ab
	rule babb-> aaba at position 2
.abaababba.ab
	rule babb-> aaba at position 4
.abaaaabaa.ab
	rule aaaa-> abbb at position 2
.ababbbbaa.ab
	rule babb-> aaba at position 1
.aaababbaa.ab
	rule babb-> aaba at position 3
.aaaaabaaa.ab
	rule aaaa-> abbb at position 1
.aabbbbaaa.ab
	rule aaaa-> abbb at position 6
.aabbbbabbb.b
	rule babb-> aaba at position 5
.aabbbaabab.b
	rule babb-> aaba at position 7
.aabbbaaaaba.
	rule aaaa-> abbb at position 5
.aabbbabbbba.
	rule babb-> aaba at position 4
.aabbaababba.
	rule babb-> aaba at position 6
.aabbaaaabaa.
	rule aaaa-> abbb at position 4
.aabbabbbbaa.
	rule babb-> aaba at position 3
.aabaababbaa.
	rule babb-> aaba at position 5
.aabaaaabaaa.
	rule aaaa-> abbb at position 3
.aababbbbaaa.
	rule babb-> aaba at position 2
.aaaababbaaa.
	rule babb-> aaba at position 4
.aaaaaabaaaa.
	rule aaaa-> abbb at position 7
.aaaaaababbb.
	rule babb-> aaba at position 6
.aaaaaaaabab.
	rule aaaa-> abbb at position 4
.aaaaabbbbab.