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For converting english statements to logical statements, I've understood that $\forall$ uses $\rightarrow$ and $\exists$ uses $\land$ (generally). I referred a particular question on this site, but I lost track of it (apologies).

Considering the same example used in the reference, "some real numbers are rational",

What is the difference between

$\exists x ( f(x) \rightarrow g(x) )$ and $\exists x ( f(x) \land g(x) )$ ?

where, $f(x) = $"$x$ is real" and $g(x) = $"$x$ is rational"

Also, where can $\exists$ be used with $\rightarrow$ and $\land$ ?

I understood that I could depict $f(x) \rightarrow g(x)$ as $\neg f(x) \vee g(x)$ and how this could prove why a conditional would work with $\forall$. But I could not figure out how and where could $\rightarrow$ and $\land$ be used with $\land$.

mahesh Rao
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    I disagree with your first paragraph. $\forall$ just means something is supposed to happen to all, while $\exists$ just means there is something that happens to at least one. Ther eis nothing aboutg such statemnets that will require, even “generally”, for all statements to be implications and existential statements to be conjunctions. – Arturo Magidin Aug 06 '19 at 06:07
  • In your case, $\exists x(f(x)\rightarrow g(x))$ would be true if there is an $x$ for which either $g(x)$ is true, or $f(x)$ is false. Thus, a non-real complex number would be a witness to the first statement, but not to the conjunction one. For an example of an existential statement with an implication, “There is at least one student who, if they come to class, then they will pass the course.” – Arturo Magidin Aug 06 '19 at 06:10
  • So in what scenario would $\rightarrow$ be preferred over $\wedge$ ? considering that an existential quantifier is in use. I'm not able to wrap my head around why we would use a $\rightarrow$ over $\wedge$ when an existential quantifier is in use. – mahesh Rao Aug 06 '19 at 08:02
  • This isn't a grammar rule, it's about accurately representing what you are saying. If you are saying that two things happen, it's a conjunction $\wedge$; if you are saying that at least one of two things happen, it's a disjunction $\vee$; if you are saying that if one thing happens, then a second thing will happen, then it's an implication $\rightarrow$. – Arturo Magidin Aug 06 '19 at 20:58

1 Answers1

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In logic, the symbols $\forall$ (for all) and $\exists$ (exists) are called quantifiers, whereas the symbols $\wedge$ and $\rightarrow$ (among others) are called connectors. Each quantifier may be used with any connector, and using one instead of the other results in a different proposition. Thus, I do not agree with your first claim:

I've understood that ∀ uses → and ∃ uses ∧ (generally)

Regarding the examples you ask about. Let $f(x)$ be the proposition "$x$ is real" and let $g(x)$ be the proposition "$x$ is rational."

$$\exists x: (f(x) \rightarrow g(x)) \qquad \textrm{(A)}$$

This may be read as "there is at least one $x$ such that, if $x$ is real, then $x$ is rational." This proposition is true, take for example $x=1$: $f(1)$ is true, $g(1)$ is true, then $f(1) \rightarrow g(1)$ is true. The truth of this implication derives from the definition of the logical connector $\rightarrow$ -you may check it with a truth table- and has little to do with mathematical properties of real and rational numbers.

The other proposition:

$$\exists x: (f(x) \wedge g(x)) \qquad \textrm{(B)}$$

is also true and you make take again $x=1$ to prove the existence. The truth table is different from the previous one, but yields the same truth value as before (in this case.)

These two cases may not be helpful to get the difference between both connectors. Let's build a different example. It can be easily shown that for any $f$ and $g$, if proposition (B) is true, then (A) is also true. So you could only get different truth values by getting (B) false and (A) true, for which is necessary and sufficient that $f(x)$ is false for all $x$.

Such an example may be achieved by considering $f(x) = $ " $x$ is a real solution to the equation $x^2 = -1$" and $g(x)$ may be whatever you want, e.g. $g(x) = $ "$1 \neq 0$" of $g(x) = $ "$1 = 0$", or $g(x) = $ "Thick as a Brick is the best album ever." In any of these cases (A) is true and (B) is false.

Javi
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  • So in what scenario would $\rightarrow$ be preferred over $\wedge$ ? considering that an existential quantifier is in use. I'm not able to wrap my head around why we would use a $\rightarrow$ over $\wedge$ when an existential quantifier is in use. – mahesh Rao Aug 06 '19 at 08:01
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    It is not a matter of "preference", they have different meanings. The example I wrote on my answer shows this, since they yield different truth values. The one you use depends on what you want say: do you want to state "there is some $x$ for which the truth of $f(x)$ implies the truth of $g(x)$"? Then use $\rightarrow$. Do you want to say "there is some $x$ for which $f(x)$ and $g(x)$ are both true"? Then go fo $\wedge$ – Javi Aug 06 '19 at 20:24