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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. $\forall x \exists y \left(\text{fsa}\left(y\right) \wedge \text{pda}\left(x\right) \wedge \text{equivalent}\left(x,y\right)\right)$
in Mathematical Logic by
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0
Can you please translate c into English
+19
It means everything is a FSA and there exist an equivalent PDA for each of them.
+2
Thank you.It helped me a lot
0
Is it the case that the chosen logical statement should be a tautology?
0

@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 

0

Meanings of A and C are also different. Option C says that " everything  is a fsa and there exist an equivalent pda for each of them". (As arjun sir already describes the meaning of C). And option A is if then statement.

3 Answers

+62 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  

$\forall x \left(\text{fsa}\left(x\right)\implies \left(\exists y \text{ pda}\left(y\right)\wedge \text{ equivalent}\left(x,y\right)\right)\right)$
by
edited by
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.
0
What is the negation of option B?
+1

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

Arjun Sir in your corrected answer, if we don't use parenthesis after "∃y"  than won't its scope be limited to pda(y) only?

0
Yes for every x fsa(x) mean everything is fsa here
+3
@Arjun Sir, I also think that $\exists y$ should be outside the parenthesis. Please correct me If I am wrong.
0
Thq soo much sir....
0

@ankitgupta.1729 yes you are right..

+1
In option A the RHS is having 'x' as free variable?
0
@khushtak        After Implies(→) symbol , is that the negation symbol in both 1 and 2?

If yes then what will be there english conversion considering the negation?
0

@Arjun sir, as @ankitgupta.1729 mentioned in the comments, shouldn't∃y be outside the parenthesis as the scope extends to equivalent predicate as well. Or is it implicit?

0
I think that year every one got marks for this question
0
Hi Arjun sir,From option C and D,how do you take out the meaning as "Everything ".
+1

@Arjun Sir , Shouldn't the correct answer be 

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

According to me , parenthesis is after ∃y

+1 vote
This depicts even iit professors can commit mistakes :p
by
–3 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
by
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