Understanding direct proofs and proofs by cases?

Question

I'm reviewing my book Mathematical Proofs a Transition to Advanced Mathematics and looking to understand things at a deeper level. I will try to explain what I've considered so far in regards to this chapter on direct proofs. I'm hoping you guys can point out where I'm correct, incorrect, and any bigger picture ideas I might be missing. Also I'd appreciate any advice you all may have.

Concerning proofs of if/then statements of two open sentences P(x) and Q(x), P(x) --> Q(x), it seems what is really going on is that we're showing the solution set of P(x) is a subset of the solution set of Q(x). It makes sense that this can be done, since a subset with less elements is more restrictive than the larger set it's a part of, and if it meets those restrictions it clearly should also be a member of the larger set. I guess this is an example of the transitive property.

For the if/then statements I've been looking at, many of the premises (P(x)) involve claiming a variable (x, c^2, 5a-7, etc) is an element of some set A, and the conclusions (Q(x)) claim that after running that variable through some function it will become an element of some set B, where A and B don't have to be different. It seems to me that in direct proofs we replace our variable in the conclusion with the requirement to be an element of set A (or I suppose you could say the definition of set A). Then we use algebraic manipulation to turn the function in the conclusion into the requirement to be an element of set B (for example all elements of the even numbers can be expressed in the form 2k where k is an integer). And this would complete the direct proof, showing all solutions for P(x) must also be solutions for Q(x).

As for when to use proof by cases instead of a direct proof, it seems to me to be when set B is a subset of set A. Although the solution sets of P(x) and Q(x) may be equal, it's harder or impossible to directly substitute the requirement for a broader set into a function to prove something about its subset. It's easier to partition that larger set into subsets with higher requirements, and then show that the conclusion is true when those requirements are substituted into the function in place of the variable.

An example for when this would be the case: If t is an integer, then t^2 +3t -1 is odd.

The strategy would be to partition the integers into the set of even and odd numbers, which have higher requirements (E:={2k: k is an integer}. and O:={2k+1: k is an integer}), and prove the function t^2 +3t -1 is odd using a direct proof technique for each subset (case).

I'm curious if every function that transforms a set into its subset has to be proved via proof by cases or if there's an easier (or more complicated i guess) method.

Answer 1

Your grasp of these 'proof techniques' is roughly correct, though there are indeed clearer ways of understanding the difference between them. 'Indirect proofs' included not only 'proof by cases' but also 'proof by contradiction' and 'proof by contrapositive'. But I'll focus on the distinction between 'proof by cases' and 'proof without cases'.

A 'proof by cases' uses the following inference rule called disjunction elimination: $\def\rule#1#2{\left|\!\!\begin{array}{l}#1\\\hline#2\end{array}\right.}$

Given sentences $P,Q,R$:

  $\rule{ P \lor Q \\ \rule{P}{R} \\ \rule{Q}{R} }{R}$

We read "$\rule{X}{Y}$" to mean that if you have proven "$X$" then you can prove "$Y$". Proof by cases means that if you have proven "$P \lor Q$" and also you can prove "$R$" from either "$P$" or "$Q$", then you can prove $R$. The structure of the rule explains its name, since it allows us to 'break down' a previously proven disjunction.

There is a corresponding inference rule called disjunction introduction:

Given any sentences $P,Q$:

  $\rule{P}{ P \lor Q \\ Q \lor P}$

These two rules completely characterize "$\lor$". We can similarly give inference rules to govern all the other boolean operations. If we do so without referring to "$\lor$" anymore, we can then classify proofs according to whether or not they use disjunction elimination, and we say that a proof is a proof by cases if it uses disjunction elimination.

It is also then easy to show that some theorems cannot be proven 'without cases'. One way is to observe that all the other inference rules remain sound when we can interpret "$P \lor Q$" differently as "$\top$" (any true sentence), but then the sentence "$( ( P \lor Q ) \land \neg P ) \to Q$" would be false if "$P$" and "$Q$" are both false sentences, and so it cannot be proven without using disjunction elimination. It is easy to check that it can be proven using disjunction elimination.

In addition, we frequently use disjunction elimination together with another inference rule called the law of excluded middle (LEM):

Given sentence $P$:

  $\rule{}{P \lor \neg P}$

LEM can be proven from the other rules (which include double negation elimination), and combining it with disjunction elimination yields the following effective rule:

Given sentences $P,Q$:

  $\rule{ \rule{P}{Q} \\ \rule{\neg P}{Q} }{ Q }$

Thus by our classification any proof that uses this LEM-based rule is also a proof by cases.

Now let's take a look at your example.

Take any integer $t$. Then $t^2 +3t -1$ is odd.

Here is a 'proof' that appears to be 'without cases':

$t^2 + 3t - 1 = (t+1)(t+2) - 3$ which is odd because $(t+1)(t+2)$ is a product of consecutive integers, one of which has to be even.

This proof is not exactly wrong, but it uses a lot of non-trivial facts:

• An even number minus an odd number gives an odd number.

• Two consecutive integers has at least one even number among them.

• A product of two integers one of which is even gives an even integer.

If I use your definition of even and odd, the first is easy, but proving the second and third requires a proof by cases.

Lastly, your proof relies crucially on the fact that every integer is either even or odd, which I hope you realize is not a trivial tautology! In fact, it cannot be proven without using induction. It also requires disjunction elimination during the induction step.