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Ok, so given the following automaton:

I need to find the language that accepts it (no need for formal prove, a short intuitive explanation is good enough).
I think the answer here is {$a^{11+6k}, k≥0$}.
I can explain why the shortest word that can be received here is $a^{11}$, and I can explain that this language contains the language {$a^{11+6k}, k≥0$}, but I don't know how to explain that this is exactly the language {$a^{11+6k}, k≥0$}.
How can you explain that it doesn't conatin more words, for example?

bof
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John
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  • with the rule that at the end it should arrive to the $O$ state with the stack empty, I also find $a^{11+6k}$ by counting the possible number of $a$ on the stack at each state. – reuns Dec 31 '15 at 21:28
  • Usage: languages don't accept PDAs; PDAs accept languages. – BrianO Dec 31 '15 at 21:36

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