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.

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 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 \(1/2^i < \epsilon\).

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) | \(true\) |

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.

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".