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"If there were no computers with antivirus every computer would work fine".

Use: O(x) = x is a computer, A(x) = Computer x has an antivirus, F(x) = x works fine.

My take is $$∀x((O(x)∧¬A(x))→F(x)).$$ What I'm trying to say here is that if every computer doesn't have an antivirus, then it must work fine no matter what. But my professor tells me it's not like that.

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    $\begingroup$ Maybe A(x) = Computer x has an antivirus $\endgroup$ Mar 27 at 8:38
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    $\begingroup$ Your formula means: "All computers without antivirus work fine". This is not the same statement as the one given. $\endgroup$
    – hff1
    Mar 27 at 8:43
  • $\begingroup$ You can rephrase (b) as "at least one number is a divisor of 3 and if x and y are divisors of 3 then x equals y" $\endgroup$
    – gandalf61
    Mar 27 at 9:40

1 Answer 1

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"If there were no computers with antivirus, then every computer would work fine".

Use: O(x) = x is a computer, A(x) = Computer x has an antivirus, F(x) = x works fine.

My take is: $$∀\color\red x\Big((O\color\red x∧¬A\color\red x)→F\color\red x\Big)$$

This says wrongly that every computer without an antivirus works fine. Here, you have specified a universal quantification outside the "If..., then..." sentence, whereas the given sentence is structured with all quantifications within the "If..., then...". (Side note: the use of tenses technically is misleading, and doesn't correspond to the given legend which is fully in present tense.)

The antecedent "there are no computers with antivirus" means $$\lnot\exists x\:(Ox\land Ax);$$ the consequent "every computer works fine" means $$\forall x\:(Ox\to Fx);$$ putting them together: $$\lnot\exists x\:(Ox\land Ax)\to\forall x\:(Ox\to Fx);$$ more clearly: $$\lnot\exists \color{violet}x\:(O\color{violet}x\land A\color{violet}x)\to\forall \color{brown}y\:(O\color{brown}y\to F\color{brown}y).$$

P.S. Whatever it is, do not make the common mistake of writing $$\forall x \,(Px\to Q)$$ as $$\forall x Px\to Q;$$ they are not logically equivalent!

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