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in 7e what is the ans and in 7g if specific condition that is not equal to 6 is to be constructed then how can we consider middle situation?
7e-- L= {w :(na(w) – nb(w)) mod 3 > 0}.
7g--L= {w: |w| mod 3 = 0, |w| ≠6}.

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for 1st one

and for second one

ie,we want to accept the string whose length 3,9,12,15,18,21,24,27,,,,,,,,,,,,,,,,,,,,

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Can anyone help me in this exercise:Show that the language L = {an : n is a multiple of three, but not a multiple of 5} is regular with DFA.i make DFA but i don’t sure ...