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

The system was reversed.

After renaming modulo the bijection { 1 ↦ 0, 0 ↦ 1 },
it remains to prove termination of the 1-rule system
{ 0 1 1 1 1 ⟶ 1 1 1 1 0 0 1 }

Loop of length 16 starting with a string of length 17
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.abbbb.bbbbbbbbbbbb
	rule abbbb-> bbbbaab at position 0
.bbbbaab.bbbbbbbbbbbb
	rule abbbb-> bbbbaab at position 5
.bbbbabbbbaab.bbbbbbbbb
	rule abbbb-> bbbbaab at position 10
.bbbbabbbbabbbbaab.bbbbbb
	rule abbbb-> bbbbaab at position 9
.bbbbabbbbbbbbaabaab.bbbbbb
	rule abbbb-> bbbbaab at position 17
.bbbbabbbbbbbbaababbbbaab.bbb
	rule abbbb-> bbbbaab at position 16
.bbbbabbbbbbbbaabbbbbaabaab.bbb
	rule abbbb-> bbbbaab at position 14
.bbbbabbbbbbbbabbbbaabbaabaab.bbb
	rule abbbb-> bbbbaab at position 13
.bbbbabbbbbbbbbbbbaabaabbaabaab.bbb
	rule abbbb-> bbbbaab at position 28
.bbbbabbbbbbbbbbbbaabaabbaababbbbaab.
	rule abbbb-> bbbbaab at position 27
.bbbbabbbbbbbbbbbbaabaabbaabbbbbaabaab.
	rule abbbb-> bbbbaab at position 25
.bbbbabbbbbbbbbbbbaabaabbabbbbaabbaabaab.
	rule abbbb-> bbbbaab at position 24
.bbbbabbbbbbbbbbbbaabaabbbbbbaabaabbaabaab.
	rule abbbb-> bbbbaab at position 21
.bbbbabbbbbbbbbbbbaababbbbaabbbaabaabbaabaab.
	rule abbbb-> bbbbaab at position 20
.bbbbabbbbbbbbbbbbaabbbbbaabaabbbaabaabbaabaab.
	rule abbbb-> bbbbaab at position 18
.bbbbabbbbbbbbbbbbabbbbaabbaabaabbbaabaabbaabaab.
	rule abbbb-> bbbbaab at position 17
.bbbbabbbbbbbbbbbbbbbbaabaabbaabaabbbaabaabbaabaab.