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m →Channel system: close the angle bracket |
→Channel system: It is mentioned that the alphabet can be empty, depending on the author. Here let's define $\epsilon$ is always in $A$, so that the notation is simpler |
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* <math>Q=\{q_1,\dots,q_m\}</math> a finite set of '''control states''',
* <math>q_0\in Q</math> an '''initial state''',
* <math>A</math> a finite alphabet (for the sake of notation simplicity, let <math>\epsilon \in A</math>),
* <math>C=\{c_1,\dots,c_n\}</math> a finite set of '''channels''',
* <math>\Sigma=\{a_1,\dots,a_p\}</math> a finite alphabet of '''messages''',
* <math>\Delta\subseteq Q\times C\times\{?,!\}\times\Sigma^*\times
Depending of the author, a '''channel system''' may have no initial state and may have an empty alphabet.<ref name="Nonprimitive recursive">{{cite journal |last1=Schnoebelen |first1=Ph |title=Verifying Lossy Channel Systems has Nonprimitive Recursive Complexity |journal =Information Processing Letters |date=15 September 2002 |volume=83 |issue=5 |pages=251–261 |doi=10.1016/S0020-0190(01)00337-4}}</ref>
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