5. Let CONTAINPDA DFA L(M1) C (M2)}. Show that CONTAIN PDA DFA is decidable. {{M1, M2)...

Question

Question

5. Let CONTAINPDA DFA L(M1) C (M2)}. Show that CONTAIN PDA DFA is decidable. {{M1, M2) M1 is a PDA and M2 is a DFA such that

Answer 1

Answer #1

WE KNOW THAT THE DFA LANGUAGES ARE DECIDABLE.

THE PROOF OF THIS IS AS FOLLOWS :-

$ApEA = {{D, w} D¥@@iaatha, accepts input w} ble _- Turing Machine details: Check input is in correct format. (Transition func$

AS WE CAN SEE A DFA MACHINE EITHER ACCEPTS A STRING OR REJECT IT.

THUS, DFA LANGUAGES ARE DECIDABLES.

IN THE GIVEN QUESTIONS,

L(M₂)IS A DFA WHICH IS DECIDABLE.

AS WE CAN SEE IN THE QUESTION THAT

L(M₁) IS A SUBSET OF THE L(M₂).

THE L(M₁) WILL ALSO BE DECIDABLE AS L(M₂) IS DECIDABLE.

NOW, CONTAIN_{PDA_DFA} LANGUAGE HAS BOTH L(M1) AND L(M₂) WHICH ARE BOTH DECIDABLES, SO THE CONTAIN_{PDA_DFA} IS ALSO DECIDABLE.

Answer 2

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