@Arjun sir Does it means There exist atleast one PDA for every FSA.

Also can we Say option A and Option C are interpreting the same meaning

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Let $\text{fsa}$ and $\text{pda}$ be two predicates such that $\text{fsa}(x)$ means $x$ is a finite state automaton and $\text{pda}(y)$ means that $y$ is a pushdown automaton. Let $\text{equivalent}$ be another predicate such that $\text{equivalent} (a,b)$ means $a$ and $b$ are equivalent. Which of the following first order logic statements represent the following?

Each finite state automaton has an equivalent pushdown automaton

- $\left(\forall x \text{ fsa}\left(x\right) \right) \implies \left( \exists y \text{ pda}\left(y\right) \wedge \text{equivalent}\left(x,y\right)\right)$
- $\neg \forall y \left(\exists x \text{ fsa}\left(x\right) \implies \text{pda}\left(y\right) \wedge \text{equivalent}\left(x,y\right)\right)$
- $\forall x \exists y \left(\text{fsa}\left(x\right) \wedge \text{pda}\left(y\right) \wedge \text{equivalent}\left(x,y\right)\right)$
- $\forall x \exists y \left(\text{fsa}\left(y\right) \wedge \text{pda}\left(x\right) \wedge \text{equivalent}\left(x,y\right)\right)$

@Arjun sir Does it means There exist atleast one PDA for every FSA.

Also can we Say option A and Option C are interpreting the same meaning

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How can we have option (b) using variable y in it's english equivalent...while no other option uses these free variable for the same. It ain't the case that x and y are necessarily bound to FSA and PDA respectively.

We need to first define the domain for variable x and y .

Or as per the given defination of "FSA(x)" and PDA (y) , everything that is supplied to FSA becomes FSA and same for PDA. Then following that option (a) must be correct.

We need to first define the domain for variable x and y .

Or as per the given defination of "FSA(x)" and PDA (y) , everything that is supplied to FSA becomes FSA and same for PDA. Then following that option (a) must be correct.

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89 votes

Best answer

None of these.

A. If everything is a FSA, then there exists an equivalent PDA for everything.

B. It is not the case that for all y if there exist a FSA then it has an equivalent PDA.

C. Everything is a FSA and has an equivalent PDA.

D. Everything is a PDA and has an equivalent FSA.

The correct answer would be

$\forall x \left(\text{fsa}\left(x\right)\implies \exists y \left( \text{pda}\left(y\right)\wedge \text{ equivalent}\left(x,y\right)\right)\right)$

A. If everything is a FSA, then there exists an equivalent PDA for everything.

B. It is not the case that for all y if there exist a FSA then it has an equivalent PDA.

C. Everything is a FSA and has an equivalent PDA.

D. Everything is a PDA and has an equivalent FSA.

The correct answer would be

$\forall x \left(\text{fsa}\left(x\right)\implies \exists y \left( \text{pda}\left(y\right)\wedge \text{ equivalent}\left(x,y\right)\right)\right)$

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@Arjun @Deepak Poonia sir i think @Gaganjot _Kaur is correct

∀x(fsa(x)⟹∃y (pda(y)∧ equivalent(x,y)))

Sir I think this is the correct answer.

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