is it none?

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$\left \{ \left ( b^{n}ab^{n}a\right )^{m} | n,m \geq 0 \right \}$

$\left \{ \left ( b^{n}ab^{n}a\right )^{m} | n,m \geq 0 \right \} \cup \left \{ b^{n} | n\geq 0 \right \}$

$\left \{ \left ( b^{n}ab^{n}\right )^{m}a | n,m \geq 0 \right \}$

$NONE$

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The above PDA accepts by emptying the stack.So you push N b's onto the stack and then on seeing an a move to q1.In q1 pop each time for input b and if N b's are there then only z0 would be left on the stack.So on seeing an a move to z0.If the string is completely parsed then on seeing epsilon stack would be emptied and string would be accepted.Else the process could repeat again.So the language is of the form {(b^n a b^n a)^m|n,m>=0].

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Yes it is accepted by the above PDA.On seeing a go to q1 and then on seeing a move to q0 and on seeing epsilon stack is emptied and hence it is accepted.

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@Hemanth_13 @Rutvik Reshamwala

cannot 'aa' accepted with these two transition

$\delta \left ( q_{0},a,X \right )=\left \{ q_{1},X \right \}$

$\delta \left ( q_{1},a,Z_{0} \right )=\left \{ q_{0},Z_{0} \right \}$

Why cannot it accepted?

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I havenot got u

Plz tell more

X is pushing in stack

So, while popping with 'a' , why it is not on top on stack?

Plz tell more

X is pushing in stack

So, while popping with 'a' , why it is not on top on stack?

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