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I am working on predicates and quantifier and got confused.

Can someone explain this?

“All lions are fierce.”

“Some lions do not drink coffee.”

“Some fierce creatures do not drink coffee.”

Let P(x), Q(x), and R(x) be the statements “x is a lion,” “x is fierce,” and “x drinks coffee,” respectively

Here is the answers for those statements

$\forall$x(P (x) → Q(x)).

∃x(P (x) ∧ ¬R(x)).

∃x(Q(x) ∧ ¬R(x)).

In this problem, Can someone explain why using "∧" in 2nd and 3rd statements ?

I do not really understand when i need to use → and ∧. What is the differences ?

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    Typing "\forall" (between two \$) leads to $\forall$ – 2017-02-20
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    The short answer is that $\implies$ and $\wedge" evaluate to different truth values for the same inputs/arguments. You may read the detailed discussion in my answer below. – 2017-02-20

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When you write "all lions are fierce" you are saying something like this:

"I look around and see all sorts of 'objects.' Whenever the object I see is a lion, I notice it's fierce. That is, every object that is a lion is fierce. That is, for every object, if it is a lion, it is fierce."

The point here is that the variable $x$ ranges over all elements of the universe of discourse. Since you want to say something only of the lions, you qualify your claim: "If it is a lion, then..."

Now let's look at the drinking habits of lions. You look around and observe that some lions don't drink coffee. What does it mean? That you observe at least one lion that doesn't drink coffee. That is, there is an 'object' that is a lion and doesn't drink coffee. Once again, $x$ may be any object around you, but you want to make sure it designates a non-coffee-drinking lion. That's why you use the $\wedge$.

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    I am still confused. so u mean, if i wanna say some specific features of some lions in all objects, i use ∧. And, i use → for the feature every lions has? – 2017-02-20
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    Almost, but not quite. I use 'object' to refer to things, animals, people,... all the objects that I want to talk about. I use $\wedge$ to say that one of these 'objects' is both a lion-kind-of-object *and* a non-coffee-drinking-kind-of-object. For features that every lion has, yes, I use $\rightarrow$. All lions like ice cream: $\forall x (P(x) \rightarrow I(x))$, where, of course, $I(x)$ is true if object $x$ likes ice cream. – 2017-02-20
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It's simpler to understand with sets and a proper choice of names. $P,Q,R$ are not very meaningful, while these below ease understanding.

$L=$ the set of lions

$F=$ the set of fierce creatures

$C=$ the set of coffee drinkers

So your propositions meaning are

  • $\forall x(P(x)\Rightarrow Q(x))$

    $x\in L\Rightarrow x\in F\qquad$ if x is a lion then it is a fierce creature

    $L\subset F\qquad$ or all the lions are fierces creatures


  • $\exists x(P(x)\land\lnot R(x))$

    $\exists x\in L$ and $x\notin C\qquad$ there is at least one lion that do not drink coffee

    $L\cap \overline C\neq \varnothing\qquad$ or some lions do not drink coffee

    $L\cap C\neq L\qquad$ or the lions that drink coffee do not represent all the lions


  • $\exists x(Q(x)\land\lnot R(x))$

    $\exists x\in F$ and $x\notin C\qquad$ there is at least one fierce creature that do not drink coffee

    $F\cap \overline C\neq \varnothing\qquad$or some fierce creatures do not drink coffee

    $F\cap C\neq F\qquad$or the fierce creatures that drink coffee do not represent all the fierce creatures


Note that in natural language $\exists x(X\land\lnot Y)$ or equivalently $X\cap \overline Y\neq \varnothing$ means : "there exist some $X$ that are not $Y$".

Clearly propositionnal logic is often quite hard to read compared to set vocabulary.

Edit: Meanwhile, reading a wiki page on quantifiers, I found this german one

wiki: Quantor

Let suppose we have :

$\Omega$ : Is everything let say all the animals. Quantifiers are implicitely $\forall x\in\Omega$ and $\exists x\in\Omega$

$L(x)$ : x is a lion

$C(x)$ : x drinks coffee


$\forall x(L(x)\rightarrow C(x))\quad$ all lions drink coffee

$\forall x(L(x)\rightarrow \lnot C(x))\quad$ no lion drink coffee $\iff$ all lions do not drink coffee

$\exists x(L(x)\wedge C(x))\quad$ some lions drink coffee

$\exists x(L(x)\wedge \lnot C(x))\quad$ some lions do not drink coffee $\iff$ not all lions drink coffee

$\forall x(L(x)\wedge C(x))\quad$ all animals are lions and drink coffee

$\exists x(L(x)\rightarrow C(x))\quad$ some animals are not lions or drink coffee $\iff$ not all animals are non-coffee drinking lions

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    I think you meant to say $\forall x (X \wedge Y)$ means $X \cap Y$. Oh!, it's fixed. – 2017-02-20
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$\forall x~\big(P (x) \to Q(x)\big)$ reads, "If anything satisfies P, $\overset\to{\textit{then}}$ it satisfies Q."   Such as "If any thing is a lion, then it will be fierce."   Ie: "Any lion is fierce," "All lions are fierce," et c.

$\exists x~\big(P (x) \wedge \neg R(x)\big)$ reads, "There is something, which satisfies P, $\overset\wedge{\textit{and}}$ also does not satisfy R."   Such as "There is some thing, which is a lion, and does not drink coffee."   Ie: "There is some lion which does not drink coffee," "Some lions don't drink coffee," et c.

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The symbol $\to$ represents a material conditional, whereas $\land$ merely represents conjunction without a necessary (causal) connection. For example, the sentence $\forall x(P(x) \to Q(x))$ says that all lions are fierce, and furthermore, it implies that one need only know that a given animal is a lion before concluding that it is also fierce.

On the other hand, sentences 2 and 3 are only claiming that 'some' given individuals do not drink coffee. Thus, the sentence $\exists x(P(x) \land \lnot R(x))$ literally says "there exists an animal $x$ such that $x$ is both a lion and does not drink coffee", which is certainly very different to saying that all lions do not drink coffee. In this case, the sentence would be given as $\forall x(P(x) \to \lnot R(x)) $.

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Let's start from the existential quantifier:

$\exists x, (P (x) \wedge \lnot R(x))$ means we can find at least one lion and it is also true that he does not drink coffee. Similarly for $\exists x, (Q(x) \wedge \lnot R(x))$.

Now take a look a look at the universal quantifier. Suppose you were to say $ \forall x(P (x) \wedge Q(x)) $ this would mean that all lions are fierce(Yay!), but...what about if an $x$ is not a lion? Since you've used conjunction($\wedge $), this means all non-lions are not fierce(Oops!). Hence, the conditional is the correct choice.

$A \implies B$ means, if $A$ is true then so is $B$. The conditional(in its entirety) evaluates to true in that case. But if $A$ is true and $B$ is false, the conditional does not hold- it is false. The last case is that of vacuous truth, which simply means that if $A$ is false $B$ could be either true or false; the conditional evaluates to true in either case. It is easy to see why in this particular case, if $x$ is not a lion it may or may not be fierce(We never said all fierce $x$ are lions). It is this property of vacuous truth, due to which the conditional is indispensable here.

If you're still confused about the above it helps to imagine that "the set of all cases when $B$ is true" to be contained within "the set of all cases when A is true".