\begin{document}

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\title{Assignment 1}
\author{Chistyakov Gleb\\ group 164}
\date{24th May 2019}

\maketitle

\subsection*{Task №1}
$\forall x \forall y: \ Point(x) \wedge Point(y) \wedge \neg (x = y) \Rightarrow$

\begin{enumerate}
    \item
    $\exists!l: \ Line(l) \wedge LiesOn(x, l) \wedge LiesOn(y, l)$

    \item
    $\forall l_1, \forall l_2: \ Line(l_1) \wedge Line(l_2) \ - \ l_1 \parallel l_2 \Longleftrightarrow \neg \exists x: \ Point(x) \wedge LiseOn(x, l_1) \wedge LiesOn(x, l_2)$

    \item
    $\forall x, \forall y: Line(x) \wedge Point(y) \wedge \neg LiesOn(y, x) \Rightarrow \exists!z: Line(z) \wedge LiesOn(y, z) \wedge (z \parallel x)$
\end{enumerate}

\subsection*{Task №2}
Let's show that
$$\forall x(\neg P(x) \rightarrow Q(x)) \rightarrow \forall y(\neg Q(y) \rightarrow P(y)) \equiv 1$$

So
$$\forall x(\neg P(x) \rightarrow Q(x)) \rightarrow \forall y(\neg Q(y) \rightarrow P(y)) =
\neg(\forall x(\neg P(x) \rightarrow Q(x))) \vee \forall y (Q(y) \vee P(y))$$

Consider the negation of the formula
$$\forall x (\neg P(x) \rightarrow Q(x)) \wedge \exists y(\neg Q(y) \wedge \neg P(y))$$

Take specific y' then there is one way when $\neg Q(y') \wedge \neg P(y') \equiv 1$ that is $P(y') = Q(y') = 0$
But for such values $\neg P(y') \rightarrow Q(y') = 0$ \\

In this way $\forall x (\neg P(x) \rightarrow Q(x)) \wedge \exists y(\neg Q(y) \wedge \neg P(y))$ is always false.

Therefore $\forall x(\neg P(x) \rightarrow Q(x)) \rightarrow \forall y(\neg Q(y) \rightarrow P(y)) \equiv 1$

\end{document}