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Hello people I have a simple question. I have this formula from which I need to remove all the implications. Here it is.

$\forall x ( [ Roman(x) \wedge know ( x, Marcus )] \rightarrow [ hate (x, Caesar ) \vee ( \forall y \exists z hate(y,z) \rightarrow thinkcrazy(x,y)) ])$

And here is the answer.

$\forall x \neg [Roman(x) \wedge know (x, Marcus)] \vee [ hate(x,Caesar) \vee (\forall y \neg \exists z hate(y,z) \vee thinkcrazy(x,y))]$

Then thing I can't understand is, why isn't the second universal quantificator negated (the one for y)? When do you do a negation on the quantifier ?

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You have to remove the implication connective from :

$∀x([Roman(x) \land know(x,Marcus)] \rightarrow [hate(x,Caesar) \lor (∀y∃zhate(y,z) \rightarrow thinkcrazy(x,y))])$

Starting from the rightmost one, you have correctly used the equivalence between $p \rightarrow q$ and $\lnot p \lor q$.

But we have a problem with some missing parentheses; if we assume that the subformula is :

$∀y(∃zhate(y,z) \rightarrow thinkcrazy(x,y))$

then we can rewrite it as :

$∀y(\lnot ∃zhate(y,z) \lor thinkcrazy(x,y)$.

But this is an assumption [please, note that the "standard" convention is that the quantifiers apply to as little as possible]; if instead we "read" the subformula with a "narrow" scope for the universal quantifier, we must have instead :

$\lnot ∀y∃zhate(y,z) \lor thinkcrazy(x,y))$.

Going on with the first interpretation, we proceed in the same way, for the leftmost $\rightarrow$, obtaining :

$\forall x(\lnot [Roman(x) \land know(x,Marcus)] \lor [hate(x,Caesar) \lor (∀y(\lnot ∃zhate(y,z) \lor thinkcrazy(x,y))])$


Conclusion

You have to check for the correct formula.

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