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When translating English phrases to mathematical statements using logical quantifiers, I find that I'm having trouble knowing the difference between $P(x)\wedge Q(x)$ and $P(x)\to Q(x)$.

For example:

Translate the following sentences using logical quantifiers:

  1. All rationals are real
  2. No rationals are real
  3. Some rationals are real
  4. Some reals are rational

My notes say the following:

  • $Q(x)$: $x$ is rational
  • $R(x)$: $x$ is real

    1. All rationals are real: $(\forall x)(Q(x)\to R(x))$
    2. No rationals re rel: $(\forall x)(Q(x)\to\neg R(x))$
    3. Some rationals are real: $(\exists x)(Q(x)\wedge R(x))$
    4. Some reals are rational: $(\exists x)(R(x)\wedge Q(x))$

So I have to wonder: Why, for number 1 and 2 is it not right to say $(\forall x)(Q(x)\wedge R(x))$? I've gotten many questions like this wrong because I said $A\to B$ when it was supposed to be $A\wedge B$.

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To say $(\forall x)(Q(x)\wedge R(x))$ is to say that all numbers in your universe are real and rational. – Alex Provost Apr 6 '13 at 23:38
Well - this example may not be the best, given that I already know that this can't be true. But I've seen examples where stuff was made up completely, like "All my poultry dance the Tango". I had said $(\forall x)(Poultry(x)\to Dance(x))$ and it was marked wrong. To this day I don't know if I was right and the marker of the assignment was wrong... but I didn't contest it. – agent154 Apr 6 '13 at 23:43
That example looks right to me (compare with your #1). – Alex Provost Apr 6 '13 at 23:45
...unless you needed to specify that the poultry was yours, in which case you'd need something like $$\forall x(\text{Poultry}(x) \wedge \text{Mine}(x) \to \text{DanceTango}(x))$$ – Clive Newstead Apr 6 '13 at 23:47
In the example you give, you'd be correct, unless as Clive noted, you were simply missing a conjunct: "x is mine". then you'd want "for all x[if(x is poultry and x is mine) then x dances the tango.] The primary connective under the "for all" quantifier is an implication. – amWhy Apr 6 '13 at 23:52
up vote 3 down vote accepted

Note that it is typically the case that $\forall$ is accompanied by $\rightarrow$, and $\exists$ by $\land$.

For all $x$, if $x$ is $P$ then $x$ is $Q$.

Exists some $x$ such that $x$ is $P$ and $x$ is $Q$.


Negate $\forall x(P(x) \rightarrow Q(x))$

  • $\lnot \forall x(P(x) \rightarrow Q(x)) \equiv \exists x\lnot(P(x) \rightarrow Q(x)) \equiv \exists x \lnot(\lnot P(x) \lor Q(x)) \equiv \exists x (P(x) \land \lnot Q(x))$

Negate $\exists x(P(x) \land Q(x))$

  • $\lnot \exists x(P(x) \land Q(x)) \equiv \forall x \lnot(P(x) \land Q(x)), \equiv \forall x (\lnot P(x) \lor \lnot Q(x)) \equiv \forall x(P(x) \rightarrow Q(x))$

Moving the negation inward, and see the form you arrive at in each case.

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Suppose that $P(x)$ is something like "$x$ is rational". When you say "for all rational $x$, $Q(x)$ holds", what you're saying is "for all $x$, if $x$ is rational then $Q(x)$". That is: $$\forall x(P(x) \to Q(x))$$ When you say "there exists a rational $x$ such that $Q(x)$", what you're saying is "there exists an $x$, which is rational and also $Q(x)$". That is: $$\exists x(P(x) \wedge Q(x))$$

So the general rule is:

  • $\forall$ uses $\to$
  • $\exists$ uses $\wedge$

There's a more fundamental reason for this. $A \to B$ is the same as $\neg A \vee B$, and $\exists$ is the same as $\neg \forall \neg$. ("There exists a red car" is the same as saying "it's not the case that all cars aren't red".) So... $$\begin{align} &\exists x (P(x) \wedge Q(x))\\ \text{is equivalent to}\ &\neg \forall x \neg (P(x) \wedge Q(x))\\ \text{is equivalent to}\ &\neg \forall x (\neg P(x) \vee \neg Q(x))\\ \text{is equivalent to}\ &\neg \forall x (P(x) \to \neg Q(x))\end{align}$$ So really the $\wedge$ in the $\exists$ case does, in some sense, 'come from' a $\to$. The converse also holds.

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