I did it in two different ways (I looked at this post but I can not get anything useful because I do not understand if it represents a $T, C$ or contingency).

I have to say if the following complex propositional function (in $\mathbb R$) represents a tautology, contradiction or contingency:

$$\exists xP(x)\wedge\neg\exists yQ(y)\wedge\forall z(P(z)\Rightarrow Q(z)).$$

I tried to do it in two different ways, and both give me that it represents a contradiction:

  • I singled out everything in one variable: $ a $. In this way you can remove the quantifiers: $$P(a)\wedge\neg Q(a)\wedge(P(a)\Rightarrow Q(a))\quad\underbrace{\Leftrightarrow}_{\textrm{Modus Tollens}}\\P(a)\wedge\neg P(a)\Leftrightarrow \textrm{Contradiction}.$$

  • Transform them into premises and see their truth values: $$\begin{array}{lc} \exists xP(x)&(1)\\\neg\exists {\color{red}y}Q(y)&(2)\\\forall z(P(z)\Rightarrow Q(z))&(3) \end{array}$$ First of $(2)$ we can conclude $\forall y\neg Q (y)$. Then, because they are premises, $\exists xP(x)$ must be true, and since $\forall y\neg Q (y)$ must also be true, we conclude that $Q(y)$ must be false. Therefore, since the third premise speaks of "for all $z$" we can consider $P(x)\Rightarrow Q(y)$ as particular cases of $z$, that is, $T\Rightarrow F$ (which is false) and therefore, there is a contradiction in the third premise, so the whole expression is a contradiction.

Are my reasoning very bad? If so, please, how could you solve the exercise (without much difficulty)?


  • 1
    $\begingroup$ You're right that it's a contradiction. But $\lnot\exists yQ(y)$ does not mean $\lnot Q(a)$ for some $a$. It means $\forall y\lnot Q(y)$. You can see the sentence is a contradiction simply by translating it into English: it says there is a P, and there's no Q, and that every P is a Q. But if there's a P and every P is a Q, then there's a Q; contradiction. $\endgroup$ – symplectomorphic Apr 7 '18 at 18:50
  • $\begingroup$ Wow thank you! I think you do not mention the quantifiers. Can I know why it is like that? Why did not you publish it as an answer? Just curious $\endgroup$ – manooooh Apr 7 '18 at 18:56
  • $\begingroup$ But why $\mathbb R$ ? $\endgroup$ – Mauro ALLEGRANZA Apr 7 '18 at 19:00
  • $\begingroup$ @MauroALLEGRANZA The statement says it's like that. $\endgroup$ – manooooh Apr 7 '18 at 19:03
  • $\begingroup$ What do you mean I "do not mention the quantifiers"? Of course I did: "there is" is a quantifier; so is "every." $\endgroup$ – symplectomorphic Apr 7 '18 at 19:31

It is a contradiction.

And your argument is basically correct.

We have $∃xP(x)$ and thus, for some $a$, $P(a)$ holds.

We have also $∀z(P(z) \to Q(z))$. The quantified formula holds for every object, and thus also for $a$ : $P(a) \to Q(a)$.

Thus, by modus ponens : $Q(a)$.

But $¬∃yQ(y)$ is equivalent to $∀y¬Q(y)$. Again, the quantified formula holds for every object, and thus also for $a$ : $¬Q(a)$.

Having derived both : $Q(a)$ and $¬Q(a)$, we have to conclude that the formula is unsatisfiable, i.e. a contradiction.

  • $\begingroup$ Since the other user did not publish it as an answer, I give it to you. I like the elegance with which you say it. Thank you! $\endgroup$ – manooooh Apr 7 '18 at 19:02

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