Write the following propositions strictly using quantifiers and give also strictly the negation of the propositions.

  • All the students of Calculus are athletes.
  • Each fish has gills.
  • Some dogs have spots.

Firstly, taking the first proposition into consideration, I have thought that it is equivalent to saying that if $x$ is a student of Calculus, then $x$ is an athlete.

But how do we use the quantifiers?

I have thought of the following:

$\forall x$:$x$ is a student of Calculus, $x$ is an athlete.

But is this enough? Or do we have to build a proposition with more quantifiers ?

Could you give a hint so that I tell you also my efforts for the other two propositions?


As an example for the first question

Define predicates such that

$C(x):x$ is a student of Calculus.

$A(x):x$ is an athlete.

All the students of Calculus are athletes.

$$\forall x(C(x)\to A(x))$$ Try fill the blanks

$F(x):x$ is a fish

$G(x):x$ has gills

Each fish has gills.

$$\forall x(?~\to~?)$$

$D(x):x$ is a dog.

$S(x):x$ has spots.

Some dogs have spots.

$$\exists x(?~\land~?)$$

  • $\begingroup$ As for the first proposition, the negation is $\exists x (C(x) \land \lnot A(x))$, right? $$$$ The equivalent expression with quantifiers to the second proposution is $\forall x (F(x) \to G(x))$, and the negation $\exists x (F(x) \land \lnot G(x))$. $$$$ As for the third proposition, the equivalent expression with quantifiers is $\exists x (D(x) \to S(x))$, and the negation is $\forall x(D(x) \land \lnot S(x))$. $$$$ Am I right? Or is something of the above wrong? $\endgroup$
    – pingu
    May 22 '20 at 22:22
  • $\begingroup$ @pingu For the third proposition, the equivalent expression with quantifiers should be $\exists x(D(x)\land S(x))$, everything else is correct. $\endgroup$
    – Manx
    May 22 '20 at 22:35
  • $\begingroup$ Also your translation $∃x(D(x)→S(x))$ is equivalent to $∀x(D(x))→∃x(S(x))$, which says: 'If everything is a dog, then something has spots.' that is not what we want to say. $\endgroup$
    – Manx
    May 22 '20 at 22:40
  • $\begingroup$ Could you explain to me why $\exists x (D(x) \to S(x))$ is equivalent to $\forall x (D(x)) \to \exists x (S(x))$ ? $$$$ Also the negation of $\exists x (D(x) \land S(x))$ is $\forall x (\lnot D(x) \lor \lnot S(x))$, right? $\endgroup$
    – pingu
    May 24 '20 at 15:06
  • $\begingroup$ @pingu This is one of quantifier distributive law, basicly it wroks because of $\exists$ is distributive over $\lor$, that $\color{red}{∃x(¬D(x)∨S(x))}\equiv\color{blue}{(∃x¬D(x)∨∃xS(x))}$ (also note $\color{red}{∃x(¬D(x)∨S(x))}\equiv∃x(D(x)→S(x))$ and $\color{blue}{(∃x¬D(x)∨∃xS(x))}\equiv\forall xD(x)\to∃xS(x)$ by conditional equivalence.) And yes, your negation is correct! $\endgroup$
    – Manx
    May 24 '20 at 17:15

Let us speak the set Theory language.

We denote the sets of students of calculus and athletes by S and A.

then, we will say

$$(\forall x\in S) \;\; x\in A$$

or if we call E, the set of all students, we could write

$$(\forall x\in E)\;\; (x\in S \implies x\in A)$$

  • $\begingroup$ Thank you! @hamam_Abdallah $\endgroup$
    – pingu
    May 30 '20 at 20:30

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