Modelling properties as "inference" (implications) or "and" (conjunctions) in logic

Question

In translating sentences to propositional logic, I am often found puzzled if I should model a property as a conjuction between sentences, or as an inference.

e.g. (I use symbol "V" as "For all, for any" and symbol "E" as "there is one, some")

Consider this:

Nobody who knows Peter is a friend of Ann's.

I have:

~Ex Kxp // There is not someone who knows Peter 
Ex Fxa // There is someone who is friend of Ann

How to put them together ?

Shall I:

~Ex (Kxp -> Fxa)

or

~Ex (Kxp && Fxa)

Example 2:

None of Ann's friends are friends with each other.

Ex Fxa  // Someone x among Ann's friends
Ey Fya  // Someone y among Ann's friends
Exy Fxy // There are some x, y guys who are friend with each other

Shall I go for:

~Exy( Fxa && Fya && Fxy)

or

~Exy( (Fxa V Fya) -> Fxy)

Example 3

None of Ann's friends know any friend of Bob.


Ex Fxa  // Someone x friend of Ann
Ey Fyb  // Someone y friend of Bob
Exy Kxy  // Someone x and some y who know each other

Shall I go for:

~Exy((Fxa V Fyb) -> Kxy)

or

~Exy (Kxy -> (Fxa && Fxb))

or

~Exy (Kxy && Fxa && Fxb)

Could you help understand when natural language relationships like : "is ", "can be", "knows " etc.

should be translated with conjuctions && (which I interepret as properties) and as inferences ->(which I interpret as relationships... but i also think properties could be relationships) ?

Could you comment the meaning in natural languages of the answers above, to understand the logical differences ?

Any suggestion to go through translations is appreciated!

(And this last one should be:
Vx // any suggestion x
Vy // any translation y

Ax // x is appreciated 
Vxy ((x -> y) -> Ax) // If suggestions go through translations, suggestions are appreciated ? )

Answer 1

Nobody who knows Peter is a friend of Ann's.

$¬∃x\; (Kxp → Fxa)\quad$ or $\quad¬∃x\; (Kxp ∧ Fxa)$

Suggestion 2 is correct since it literally says that no one knows Peter and is Ann's friend.

I don't like double negation, so when I read the given sentence my mind instinctively translates it to "if someone knows Peter, then they are not Ann's friend", that is, $∀x\; (Kxp → ¬Fxa);$ this is indeed equivalent to Suggestion 2.

Suggestion 1 is a stronger claim than Suggestion 2 because by stating that no person $x$ satisfies $Kxp → Fxa,$ and consequently that $Kxp$ is never false, it informs us that everyone knows Peter.

None of Ann's friends are friends with each other.

$¬∃x,y\; ( Fxa ∧ Fya ∧ Fxy)$

or

$¬∃x,y\; ( (Fxa ∨ Fya) → Fxy)$

Similarly, Suggestion 1 is correct. Suggestion 2 is stronger than Suggestion 1 and says that everone is Ann's friend.

None of Ann's friends know any friend of Bob.

$¬∃x,y\; ((Fxa ∨ Fyb) → Kxy)$

or

$¬∃x,y\; (Kxy → (Fxa ∧ Fxb))$

or

$¬∃x,y\; (Kxy ∧ Fxa ∧ Fxb)$

Suggestion 3 is correct; however, my mind instinctively interprets the given sentence as $∀x\; (Fxa → ∀y\; (Fxb → ¬Kxy)).$ Here is a step-by-step sequence of equivalences:

2. $∀x,y\; (Fxa → (Fxb → ¬Kxy))$

3. $∀x,y\; ((Fxa ∧ Fxb) → ¬Kxy)$

4. $∀x,y\; (¬Fxa ∨ ¬Fxb ∨ ¬Kxy)$

5. $¬∃x,y\; (Fxa ∧ Fxb ∧ Kxy).$

Answer 2

You would encode "Someone knows Peter and is a friend of Ann" as follows:

$\exists x (Kxp \land Fxa)$

But you want the negation of this:

$\neg \exists x (Kxp \land Fxa)~~~~$ or equivalently $~~~~\forall x (Kxp \to \neg Fxa) $

Note: You will rarely if ever have an existential quantifier on an implication. Such a construct, though not necessarily wrong, is likely to be an error in such translation exercises.