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A direct proof works by applying a sequence of axiomatic steps to transform the hypothesis into the result. Direct proof is pervasive, being used internally within almost every other proof technique.

An **axiom** is something we take as true without proof. \lnot \lnot A \equiv A is an example of an axiom. During proof writing, we commonly treat as axiomatic a much larger set of truths than we wold call axiomatic when designing a logic: effectively, we treat any step we expect our readers can follow and not doubt as if it were an axiom. Sometimes the larger truths used as axioms during a proof are called **proof rules**.

Direct-proof axioms are of two types:

- Equivalences
If \mathcal A \equiv \mathcal B then \mathcal A is true whenever \mathcal B is true and false whenever it is false.

When proving new equivalences, our direct proof steps must all themselves by equivalences. Equivalences are commutative, so we can use them in either direction in our proof: from \mathcal A to \mathcal B or from \mathcal B to \mathcal A, as whichever is helpful.

Note that \mathcal A \equiv \mathcal B and \mathcal A \leftrightarrow \mathcal B \equiv \top are two ways of saying the same thing.

Any expression that is equivalent to \top is called a

**tautology**. Any expression that is equivalent to \bot is called a**contradiction**.- Entailments
If \mathcal A \vDash \mathcal B then \mathcal A being true is sufficient to establish the truth of \mathcal B, but not vice-versa: \mathcal B could be true even when \mathcal A is false.

When proving entailments, we may use any mix of entailments and equivalences in our proof. Entailments are not commutative: we must use them in the direction indicated in our proof.

Note that \mathcal A \vDash \mathcal B and \mathcal A \rightarrow \mathcal B \equiv \top are two ways of saying the same thing.

If \mathcal A \equiv \mathcal B then both \mathcal A \vDash \mathcal B and \mathcal B \vDash \mathcal A, and vice-versa.

We have a list of axioms you may find helpful, and will add to it from time to time as we discuss new kinds of axioms during class.

Direct proofs are a bit like a puzzle: You look at where you are, find all the pieces that could fit, and then pick one that seems most likely to help make progress.

At any given point in a proof, you have some things you know (are are assuming) are true. Initially, that is just the given of the proof. Every time you apply a proof step, you add one more thing to the set of things you know.

Prove that P \land (P \rightarrow Q) is equivalent to P \land Q.

Given P \land (P \rightarrow Q), we apply the definition of implication to get P \land (\lnot P \lor Q); …

At this point in the proof, we know two things: P \land (P \rightarrow Q) and P \land (\lnot P \lor Q). The next proof step can proceed from either part.

In general, you want to proceed from the last step, but in some cases (particularly some more complicated entailments) you may want to have several chains you combine later on.

There are only a few axioms that apply in most situations, generally dictated by the connectives in a what you know.

You can always use double negation, and in many places, since it does not depend on anything; you can also always add an \lor \bot or an \land \top, though those are rarely helpful. And there are a few axioms that can be used for each logical connective.

Prove that P \land (P \rightarrow Q) is equivalent to P \land Q.

Let’s assume we’re continuing from the last step: P \land (\lnot P \lor Q). We can

- double-negate the first P, the second P, the Q, the parenthesized \lor, or the entire expression
- distribute the \land over the \lor
- apply the definition of implication to the \lor

Not all axioms are equally useful; you want to use the one that will get you closer to your goal. While not always sufficient, the following can help:

- don’t undo what you’ve done
- turn everything to ands and ors, work with those, then change back
- De Morgan changes ands to ors and vice versa, and can be enabled by double negation
- implications come from disjunctions, possibly with a double negation to get the negated term for the antecedent

When none of these obviously apply, it’s generally wise to pick a step that changes the formula a lot (such as distribution) or simplifies it.

Prove that P \land (P \rightarrow Q) is equivalent to P \land Q.

Given P \land (P \rightarrow Q),

We convert to ands and ors first, as we have more rules for them:

we apply the definition of implication to get P \land (\lnot P \lor Q);

We listed options above. Of those, distribution will get us a P \land Q as one term; we want it on its own, but having it as a term seems closer than not having it as a term

distribution gives us (P \land \lnot P) \lor (P \land Q),

Now we can simplify, almost always a good idea

which can be simplified as \bot \lor (P \land Q), which simplifies to P \land Q.

and that finished the proof! Of course, many proofs are quite a bit longer, but the same guidelines still hold.

When you what to prove that \mathcal A \equiv \mathcal B, you can start with \mathcal A and show it is equivalent to \mathcal B, but you can also work backwards: start with \mathcal B and show it is equivalent to \mathcal A. Sometimes one is a bit easier than the other, though generally not by much.

Putting this another way, a direct proof of \mathcal A \equiv \mathcal B looks like \mathcal A \equiv X_1 \equiv X_2 \equiv \dots \equiv X_n \equiv \mathcal B; it doesn’t matter what order you show these: \mathcal A \equiv X_1 first or X_n \equiv \mathcal B first. You could even start in the middle and work out, though that would be pretty unusual.

Working backwards, or from the middle, is also possible (but uncommon) with proofs of entailment. A direct proof of \mathcal A \vDash \mathcal B looks like \mathcal A \vDash X_1 \vDash X_2 \vDash \dots \vDash X_n \vDash \mathcal B; it’s usually easiest to work forwards from \mathcal A to \mathcal B, but you can do the other direction, or both and meet in the middle, instead. Note, though, that if you start from \mathcal B your next step needs to be that some new expression entails \mathcal B, not that $B entails something else.

While working backwards can make some problems easier, it can also make them harder. We recommend only trying a backwards approach if you already tried going forwards and it didn’t work.