# Formal proof for a set of closed logical sentences.

I am currently taking a course in logic for computer science and reading a book by M. R. A. Huth and M. D. Ryan, Logic in Computer Science. I came across an exercise where I understand but need help writing in formal language and providing formal proof.

Here is the problem:

A set of logical sentences $$x_1 \Longrightarrow y_1$$, $$x_2 \Longrightarrow y_2$$, ..., $$x_n \Longrightarrow y_n$$ is called closed if $$x_1, x_2, ..., x_n$$ are exhaustive, i.e. $$x_1 \vee x_2 \vee ... \vee x_n$$ and $$y_1, y_2,...,y_n$$ are mutually exclusive, i.e. for $$i \neq j,$$ $$\neg(y_i \wedge y_j).$$ Prove that if $$x_1 \Longrightarrow y_1, ..., x_n \Longrightarrow y_n$$ are closed, then we have $$y_1 \Longrightarrow x_1, y_2 \Longrightarrow x_2, ..., y_n \Longrightarrow x_n.$$

Now, there are two concepts that I know are mainly from probability theory, namely, exhaustive and mutually exclusive. However, I need to learn how to apply that in the given logical problem and form a formal proof. I appreciate any step-by-step guidelines.

To prove $$y_i \Longrightarrow x_i$$, it is enough to show that $$\lnot x_i \Longrightarrow \lnot y_i$$. So assume $$\lnot x_i$$. Because the $$x_k$$ are exhaustive, $$x_j$$ is true for some $$j \neq i$$. As we are given $$x_j \Longrightarrow y_j$$, we must have that $$y_j$$ is true. But because the $$y_k$$ are mutually exclusive, that means that $$y_i$$ is false, i.e., that $$\lnot y_i$$ holds. Discharging our assumption that $$\lnot x_i$$, we find that $$\lnot x_i \Longrightarrow \lnot y_i$$, giving us that $$y_i \Longrightarrow x_i$$.