# Lecture 7: Logical statements and proofs

For each of a handful of logical connectives, we summarized how to prove them, how to use them in a proof, and how to negate them.

We have seen many of these things in passing, either in lecture or in the homework, here we bring them together in one place.

## Terminology

A proposition is a statement that is either true or false. For example, "$$5^2=25$$", "the pope is catholic", and "John can fool Mike" are all propositions.

A predicate is a statement whose truth depends on a variable. For example, "$$x^2 = 25$$", "$$x$$ is catholic", and "$$x$$ can fool Mike" are all predicates.

You can think of a predicate as a function that returns a proposition.

Note that if $$P$$ is a predicate, then $$\forall x, P(x)$$ and $$\exists x, P(x)$$ are propositions.

## Proving is recursive

Proving things is a recursive process: to prove a complicated statement involving $$P$$ and $$Q$$, you may have to prove $$P$$ or prove $$Q$$.

For example, to prove $$\forall \epsilon > 0, \exists n \in \mathbb{N}, \forall i > n, 1/2^i < \epsilon$$, You would first choose an arbitrary $$\epsilon$$. Then, you prove $$\exists n \in \mathbb{N}, \forall i > n, 1/2^n < \epsilon$$.

To do that, you give a specific $$n$$. In this case, I know that there exists some $$n > log_2(1/\epsilon)$$, so I will choose that $$n$$ (here I am using an existential statement). Now I must prove that $$\forall i > n, 1/2^n < \epsilon$$.

To do that, I choose an arbitrary $$i$$ and assume only that $$i > n$$. I must show $$1/2^n < \epsilon$$. But I know that $$n > log_2(1/\epsilon)$$ and that $$i > n$$, so $$i > log_2(1/\epsilon)$$. Then $$2^i > 1/\epsilon$$, so $$1/2^i < \epsilon$$.

## Summary of proof techniques

In the following, $$P$$ and $$Q$$ may denote either propositions or predicates.

logical statement English interpretation How to prove it How to use it Logical negation
$$P\land Q$$ $$P$$ and $$Q$$ prove $$P$$ and then prove $$Q$$. if you know $$P \land Q$$, you can conclude $$P$$. You can also conclude $$Q$$. $$(\lnot P) \lor (\lnot Q)$$
$$P\lor Q$$ $$P$$ or $$Q$$ (or both) either give a proof of $$P$$ or give a proof of $$Q$$ (your choice). You can do case analysis. If you know $$P \lor Q$$ and are trying to prove $$R$$, you can give one proof of $$R$$ assuming $$P$$ is true, and another proof of $$R$$ assuming $$Q$$ is true. This is enough to conclude that $$R$$ is always true. $$(\lnot P) \land (\lnot Q)$$
$$P ⇒ Q$$ "if $$P$$ then $$Q$$" or "$$P$$ implies $$Q$$". Be careful because $$P \implies Q$$ does not say that there is any causal relationship between $$P$$ and $$Q$$, just that if $$P$$ happens to hold, then $$Q$$ holds as well. Could be coincidence. Assume $$P$$, and then prove $$Q$$. If you know $$P \implies Q$$ and you know $$P$$ you can conclude $$Q$$. $$P \land (\lnot Q)$$
$$\forall x \in A, P(x)$$ for all $$x$$ in $$A$$, $$P(x)$$ is true. Choose an arbitrary $$x$$ (see below). Using only the fact that $$x \in A$$, prove $$P(x)$$. If you know $$\forall x \in A, P(x)$$ and you know some element $$y \in A$$, then you can conclude $$P(y)$$. $$\exists x, \lnot P(x)$$
$$\exists x \in A, P(x)$$ there is some $$x$$ in $$A$$ satisfying $$P$$. Write down a specific $$x$$. Check that $$x \in A$$, and prove $$P(x)$$. If you know that $$\exists x \in A, P(x)$$, you can use $$x$$ in the remainder of your proof, as well as the facts $$x \in A$$ and $$P(x)$$. $$\forall x, \lnot P(x)$$
$$true$$ "true" No effort required. Not much you can do with it. $$false$$
$$false$$ we have arrived at a contradiction Good luck proving that false is true! But if you come to a falsehood in your proof, you must have made a bad assumption, so you can conclude anything you want (proof by contradiction) $$true$$

## What does "Arbitrary" mean?

When you "choose an arbitrary $$x$$", you cannot assume anything about $$x$$. For example, it is wrong to say "choose an arbitrary $$x$$, say 7", because you are assuming many things about $$x$$: it is odd, it is prime, it is greater than 3, .... This means that your proof will not work if I want to apply it to a different $$x$$, so you haven't proven $$\forall x, P(x)$$, you've just proven $$P(7)$$ (or $$\exists x, P(x)$$)

Similarly, when using an existential statement, you can't assume anything about the element that exists, other than that it has the given property.

## Things to avoid

Don't write backwards proofs. A backwards proof is one that starts with what you are trying to show, and ends with something you know. Backwards proofs make it easy to prove nonsense:

Claim: $$5 = 3$$.

Proof: $5 = 3$ $3 = 5$ $5 + 3 = 3 + 5$ $8 = 8 ✓$

It is often helpful to write down what you're trying to show, but if you do, you must clearly indicate this, for example by writing "we want to show that".