Exercises
In the following exercises we list a number of patterns of inference, only some of which are valid. For each pattern, determine whether it is valid. If it is, explain why it is valid, appealing to the truth tables for the connectives involved. If it is not, give a specific example of how the step could be used to get from true premises to a false conclusion.
5.1
The simple forms of inference discussed in the last section are all instances of the principle that you can use already established cases of logical consequence in informal proofs. But the Boolean connectives also give rise to two entirely new methods of proof, methods that are explicitly applied in all types of rigorous reasoning. The first of these is the method of proof by cases. In our formal system F , this method will be called disjunction elimination, but don’t be misled by the ordinary sounding name: it is far more significant than, say, disjunction introduction or conjunction elimination.
We begin by illustrating proof by cases with a well-known piece of math-ematical reasoning. The reasoning proves that there are irrational numbers b and c such that bc is rational. First, let’s review what this means. A number is said to be rational if it can be expressed as a fraction n/m, for integers n and m. If it can’t be so expressed, then it is irrational. Thus 2 is rational (2 = 2/1), but√
2 is irrational. (We will prove this latter fact in the next sec-tion, to illustrate proof by contradiction; for now, just take it as a well-known truth.) Here now is our proof:
Proof: To show that there are irrational numbers b and c such that bc is rational, we will consider the number√
2
√2
. We note that this number is either rational or irrational.
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132 /Methods of Proof for Boolean Logic
If√ 2
√2
is rational, then we have found our b and c; namely, we take b = c =√
2.
Suppose, on the other hand, that √ 2
Thus, we see that in this case, too, bc is rational.
Consequently, whether√ 2
√2
is rational or irrational, we know that there are irrational numbers b and c such that bc is rational.
What interests us here is not the result itself but the general structure of the argument. We begin with a desired goal that we want to prove, say S, and a disjunction we already know, say P ∨ Q. We then show two things: that S proof by cases
follows if we assume that P is the case, and that S follows if we assume that Q is the case. Since we know that one of these must hold, we then conclude that S must be the case. This is the pattern of reasoning known as proof by cases.
In proof by cases, we aren’t limited to breaking into just two cases, as we did in the example. If at any stage in a proof we have a disjunction containing n disjuncts, say P1∨ . . . ∨ Pn, then we can break into n cases. In the first we assume P1, in the second P2, and so forth for each disjunct. If we are able to prove our desired result S in each of these cases, we are justified in concluding that S holds.
Let’s look at an even simpler example of proof by cases. Suppose we want to prove that Small(c) is a logical consequence of
(Cube(c) ∧ Small(c)) ∨ (Tet(c) ∧ Small(c))
This is pretty obvious, but the proof involves breaking into cases, as you will notice if you think carefully about how you recognize this. For the record, here is how we would write out the proof.
Proof: We are given
(Cube(c) ∧ Small(c)) ∨ (Tet(c) ∧ Small(c))
as a premise. We will break into two cases, corresponding to the two disjuncts. First, assume that Cube(c) ∧ Small(c) holds. But then (by
Proof by cases/ 133
conjunction elimination, which we really shouldn’t even mention) we have Small(c). But likewise, if we assume Tet(c) ∧ Small(c), then it follows that Small(c). So, in either case, we have Small(c), as desired.
Our next example shows how the odd step of disjunction introduction (from P infer P ∨ Q) can be used fruitfully with proof by cases. Suppose we know that either Max is home and Carl is happy, or Claire is home and Scruffy is happy, i.e.,
(Home(max) ∧ Happy(carl)) ∨ (Home(claire) ∧ Happy(scruffy)) We want to prove that either Carl or Scruffy is happy, that is,
Happy(carl) ∨ Happy(scruffy)
A rather pedantic, step-by-step proof would look like this:
Proof: Assume the disjunction:
(Home(max) ∧ Happy(carl)) ∨ (Home(claire) ∧ Happy(scruffy)) Then either:
Home(max) ∧ Happy(carl) or:
Home(claire) ∧ Happy(scruffy).
If the first alternative holds, then Happy(carl), and so we have Happy(carl) ∨ Happy(scruffy)
by disjunction introduction. Similarly, if the second alternative holds, we have Happy(scruffy), and so
Happy(carl) ∨ Happy(scruffy)
So, in either case, we have our desired conclusion. Thus our conclu-sion follows by proof by cases.
Arguing by cases is extremely useful in everyday reasoning. For example, one of the authors (call him J) and his wife recently realized that their parking meter had expired several hours earlier. J argued in the following way that there was no point in rushing back to the car (logicians argue this way; don’t marry one):
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134 /Methods of Proof for Boolean Logic
Proof: At this point, either we’ve already gotten a ticket or we haven’t. If we’ve gotten a ticket, we won’t get another one in the time it takes us to get to the car, so rushing would serve no purpose.
If we haven’t gotten a ticket in the past several hours, it is extremely unlikely that we will get one in the next few minutes, so again, rushing would be pointless. In either event, there’s no need to rush.
J’s wife responded with the following counterargument (showing that many years of marriage to a logician has an impact):
Proof: Either we are going to get a ticket in the next few minutes or we aren’t. If we are, then rushing might prevent it, which would be a good thing. If we aren’t, then it will still be good exercise and will also show our respect for the law, both of which are good things. So in either event, rushing back to the car is a good thing to do.
J’s wife won the argument.
The validity of proof by cases cannot be demonstrated by the simple truth table method introduced in Chapter 4. The reason is that we infer the con-clusion S from the fact that S is provable from each of the disjuncts P and Q. It relies on the principle that if S is a logical consequence of P, and also a logical consequence of Q, then it is a logical consequence of P ∨ Q. This holds because any circumstance that makes P ∨ Q true must make at least one of P or Q true, and hence S as well, by the fact that S is a consequence of both.
Remember
Proof by cases: To prove S from P1∨ . . . ∨ Pn using this method, prove S from each of P1, . . . , Pn.
Exercises
The next two exercises present valid arguments. Turn in informal proofs of the arguments’ validity. Your proofs should be phrased in complete, well-formed English sentences, making use of first-order sentences as convenient, much in the style we have used above. Whenever you use proof by cases, say so. You don’t have to be explicit about the use of simple proof steps like conjunction elimination. By the way, there is typically more than one way to prove a given result.
Proof by cases/ 135
Assume the same four premises as in Exercise 5.8. Is LeftOf(b, c) a logical consequence of these premises? If so, turn in an informal proof of the argument’s validity. If not, submit a counterexample world.
5.10
✎
Suppose Max’s favorite basketball team is the Chicago Bulls and favorite football team is the Denver Broncos. Max’s father John is returning from Indianapolis to San Francisco on United Airlines, and promises that he will buy Max a souvenir from one of his favorite teams on the way. Explain John’s reasoning, appealing to the annoying fact that all United flights between Indianapolis and San Francisco stop in either Denver or Chicago. Make explicit the role proof by cases plays in this reasoning.
5.11
✎
Suppose the police are investigating a burglary and discover the following facts. All the doors to the house were bolted from the inside and show no sign of forced entry. In fact, the only possible ways in and out of the house were a small bathroom window on the first floor that was left open and an unlocked bedroom window on the second floor. On the basis of this, the detectives rule out a well-known burglar, Julius, who weighs two hundred and fifty pounds and is arthritic. Explain their reasoning.
5.12
✎
In our proof that there are irrational numbers b and c where bc is rational, one of our steps was to assert that√
2
√2
is either rational or irrational. What justifies the introduction of this claim into our proof ?
5.13
✎
Describe an everyday example of reasoning by cases that you have performed in the last few days.
5.14
✎?
Give an informal proof that if S is a tautological consequence of P and a tautological conse-quence of Q, then S is a tautological conseconse-quence of P ∨ Q. Remember that the joint truth table for P ∨ Q and S may have more rows than either the joint truth table for P and S, or the joint truth table for Q and S. [Hint: Assume you are looking at a single row of the joint truth table for P ∨ Q and S in which P ∨ Q is true. Break into cases based on whether P is true or Q is true and prove that S must be true in either case.]
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Section 5.3