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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

  1. $\left(\forall x \text{ fsa}\left(x\right) \right) \implies \left( \exists y \text{ pda}\left(y\right) \wedge \text{equivalent}\left(x,y\right)\right)$
  2. $\neg \forall y \left(\exists x \text{ fsa}\left(x\right)  \implies \text{pda}\left(y\right) \wedge \text{equivalent}\left(x,y\right)\right)$
  3. $\forall x \exists y \left(\text{fsa}\left(x\right) \wedge \text{pda}\left(y\right) \wedge \text{equivalent}\left(x,y\right)\right)$
  4. $\forall x \exists y \left(\text{fsa}\left(y\right) \wedge \text{pda}\left(x\right) \wedge \text{equivalent}\left(x,y\right)\right)$
asked in Mathematical Logic by Veteran (59.4k points)
edited by | 2.2k views
0
Can you please translate c into English
+7
It means everything is a FSA and there exist an equivalent PDA for each of them.
+1
Thank you.It helped me a lot
0
Is it the case that the chosen logical statement should be a tautology?

2 Answers

+38 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 exist an equivalent FSA.

The correct answer would be

$\forall x \left(\text{fsa}\left(x\right)\implies \left(\exists y \text{ pda}\left(y\right)\wedge \text{ equivalent}\left(x,y\right)\right)\right)$
answered by Veteran (342k points)
edited by
0
Yes,Thanks now i understood after further reading the binding of quantifiers

How '(' or a " , " after quantifier can change its binding scope to the predicate function at R.H.S
0
how c option is implying everything is fsa
0

 

It can also written as   ∀x∃y [fsa(x)⟹( pda(y)∧ equivalent(x,y))] .Is not it?

0

How both(given below) are equivalent to ∀x(fsa(x)⟹(∃y pda(y)∧ equivalent(x,y))).

  • ∀x(fsa(x)→  (∃y( pda(y) ∧ equivalent(x, y))))
  •  ∀x∃y(fsa(x)→  ( pda(y) ∧ equivalent(x, y)))

I did not getting.Plz verify me.

0
great solution
0
Is it possible to interpret above correct answer in following manner:

1. Every FSA has some equivalent PDA

2. If each x is a FSA then there exists Atleast one equivalent PDA for it.
+1

@Warrior , I think

we can take out "there exist y" out of the expression, but it should be after "for all x", it should not come before "for all x".

0
@rahul sharma  I too think that the scope of x will be finished by the time we reach RHS. And if this is the case then the option doesnt make any sense. So how can you convert such sentences to english.

@Arjun converted option A into english sentence. My doubt is if the scope of x is not defined while we are on RHS then how can we interpret it.
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What is the negation of option B?
0

Arjun sir is right because simple thing is here scope of x is different from both sides...

once we complete LHS then from RHS, x is free variable so we can take it any value..

Refer Kenneth. H Rosen for the same...

–2 votes
Let a and b be two proposition
a: Good Mobile phones.
b: Cheap Mobile Phones.

P and Q can be written in logic as
P: a-->~b
Q: b-->~a.

Truth Table
a   b   ~a  ~b   P   Q
T   T   F    F   F   F
T   F   F    T   T   T
F   T   T    F   T   T
F   F   T    T   T   T

it clearly shows P and Q are equivalent.
so option D is Correct
answered by Active (4.1k points)
+2


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