PHIL12A
Section answers, 23 February 2011
Julian Jonker
1
How much do you know?
1. The following questions are adapted form exercises 5.1-5.6. Decide whether each pattern of inference is
valid. If it is, show that it is using truth tables. If it is not, give example sentences that show how the
conclusion can be false though the premises are true.
(a) From P and ¬Q, infer P ∧ Q.
This is invalid, as the following sentences exemplify:
1
P = Logic is fun.
True
2
¬Q = Logic is not easy.
True
3
P ∧ Q = Logic is fun and easy.
False
(b) From ¬P ∨ ¬Q and ¬P , infer ¬Q.
This is invalid, as the following sentences exemplify:
1
¬P ∨ ¬Q = Either soft drinks are unhealthy or water is unhealthy.
True
2
¬P = Soft drinks are unhealthy.
True
3
¬Q = Water is unhealthy.
False
(c) (Ex 5.6) From P ∧ Q and ¬P , infer R.
This is valid, as a truth table will show. You need to show that every row of the truth table which makes
both P ∧ Q and ¬P true also makes R true. However, because P ∧ Q and ¬P are contradictory, there
is no row of the truth table which makes both of them true (try it!). So the definition for TT validity is
(vacuously) fulfilled.
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2. Write informal proofs for the following arguments, using proof by cases. Be as explicit as possible about
each step in your proof.
(a) Suppose you know that (Cube(a)∧Tet(b))∨(Cube(c)∧Tet(b)). Show that Tet(b) follows.
There are two cases to consider. First, suppose that (Cube(a)∧Tet(b)) is true. Since both of the
conjuncts of a true conjunction are true, we can conclude that Tet(b) is true.
Second, consider that (Cube(c)∧Tet(b)) is true. Again, both the conjuncts of a true conjunction are
true, so we conclude that Tet(b) is true.
Since Tet(b) follows in each case, we conclude that Tet(b) is true.
(b) Suppose you know that (Cube(a)∧Small(b))∨(Cube(c)∧Small(c)).
Show that Small(b)∨Small(c) follows.
There are two cases to consider. First, suppose that Cube(a)∧Small(b) is true. Since both conjuncts of a conjunction are true, we know that Small(b) is true. But then we can conclude that
Small(b)∨Small(c) is true, since a disjunction is true whenever one of the disjuncts is true.
Second, suppose that Cube(c)∧Small(c) is true. Since both conjuncts of a conjunction are true, we
know that Small(c) is true. But then we can conclude that Small(b)∨Small(c) is true, since a
disjunction is true whenever one of the disjuncts is true.
It follows that Small(b)∨Small(c) is true, since it is true in every case.
(c) (Ex 5.7)
1
Home(max) ∨ Home(claire)
2
¬Home(max) ∨ Happy(carl)
3
¬Home(claire ∨ Happy(carl)
4
Happy(carl)
We want to show that if we make the premises of the argument true, the conclusion must be true. In holding
premise 1 true, there are two cases we can consider: either Max is home or Claire is home.
First, suppose that Max is home. By the second premise, Carl is happy, since at least one of the disjuncts
of the disjunction must be true. This also makes the third premise true.
Second, suppose that Claire is home. By the third premise, Carl is happy, since at least one of the disjuncts
of the disjunction must be true. This also makes the second premise true.
It follows that Carl is happy, since we can conclude this in every possible case.
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3. Write informal proofs to show that the following arguments are valid.
(a) (Ex 5.16)
1
Max or Claire is at home but either Scruffy or Carl is unhappy.
2
Either Max is not home or Carl is happy.
3
Either Claire is not home or Scruffy is unhappy.
4
Scruffy is unhappy
Premise 1 is a conjunction, and when we hold it to be true we hold both conjuncts to be true. So we accept
that Max or Claire is at home, since this is one of the conjuncts. Now we use a proof by cases to show that
Scruffy is unhappy.
There are two cases to consider. In the first case, we suppose that Max is at home. Then by the second
premise we know that Carl is happy. But not that by premise 1, it is true that either Scruffy or Carl is
unhappy. This is a disjunction, so one of the disjuncts must be true. We conclude that in this case Scruffy
is unhappy.
In the second case, we suppose that Claire is at home. The third premise is a disjunction, so one of the
disjuncts must be true. Therefore Scruffy is unhappy in this case.
It follows that Scruffy is unhappy, since this is true in every case.
(b) (Ex 5.17)
1
Cube(a) ∨ Tet(a) ∨ Large(a)
2
¬Cube(a) ∨ a=b ∨ Large(a)
3
¬Large(a) ∨ a=c
4
¬(c=c ∧ Tet(a))
5
a=b ∨ a=c
In order to hold premise 1 true, at least one of the disjuncts must be true. Let’s consider each disjunct:
there are three cases. In each case we wish to show that the conclusion, a=b∨a=c, holds.
Case 1: a is a cube. By premise 2 it follows that a and b are identical, or a is large, or both. Now we show
by cases that a=b∨a=c holds. Suppose first that a and b are identical. Then the disjunction a=b∨a=c is
true since the first disjunct is true. Suppose next that a is large. Then by premise 3 we have that a and c
are identical. But then it follows that the disjunction a=b∨a=c is true, since the second disjunct is true. It
follows that a=b∨a=c is true in case 1.
Case 2: a is a tetrahedron. By premise 4 we have that either c is not identical to c or a is not a tetrahedron.
In either case, we have a contradiction! But anything follows from a contradiction: including a=b∨a=c.
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Case 3: a is large. By premise 3 we have that a and c are identical. But then we know that a=b∨a=c is
true, since the second disjunct is true.
It follows that a=b∨a=c is true, since it follows in each case.
Note that the our proof contained proofs by cases embedded within a proof by cases. The structure of this
would have been much easier to follow if we had uses a formal proof!
4. Construct formal proofs for the following arguments.
(a) (Ex 6.4)
1
(A ∧ B) ∨ C
2
C ∨B
Proof:
1
(A ∧ B) ∨ C
2
(A ∧ B)
3
B
∧Elim: 2
4
C ∨B
∨Intro: 3
5
C
6
C ∨B
7
C ∨B
∨Intro: 5
∨Elim: 1, 2-4, 5-6
4
(b) (Ex 6.2)
1
P ∨ (Q ∧ R)
2
(P ∨ Q) ∧ (P ∨ R)
Proof:
1
P ∨ (Q ∧ R)
2
P
3
P ∨Q
∨Intro: 2
4
P ∨R
∨Intro: 2
5
(P ∨ Q) ∧ (P ∨ R)
∧Intro: 3,4
6
Q∧R
7
Q
∧Elim: 6
8
P ∨Q
∨Intro: 7
9
R
∧Elim: 6
10
P ∨R
∨Intro: 9
11
(P ∨ Q) ∧ (P ∨ R)
∧Intro: 8, 10
12
(P ∨ Q) ∧ (P ∨ R)
5
∨Elim: 1, 2-5, 6-11
(c) For the following proof, you should recall that you can nest subproofs within subproofs.
1
(P ∨ Q) ∧ (P ∨ R)
2
P ∨ (Q ∧ R)
(I will use Reit steps in some of these proofs to clarify things, even though not strictly necessary.)
Proof:
1
(P ∨ Q) ∧ (P ∨ R)
2
P ∨Q
3
P
4
P ∨ (Q ∧ R)
5
Q
6
P ∨R
7
P
8
P ∨ (Q ∧ R)
9
R
10
Q
Reit: 5
11
Q∧R
∧Intro: 9, 10
12
P ∨ (Q ∧ R)
∨Intro: 11
13
14
∧Elim: 1
∨Intro: 3
∧Elim: 1
P ∨ (Q ∧ R)
P ∨ (Q ∧ R)
6
∨Intro: 7
∨Elim: 6, 7-8, 9-12
∨Elim: 2, 3-4, 5-13
(d) (Ex 6.8)
1
P ∧ (Q ∨ ¬¬R)
2
¬¬P ∧ (Q ∨ R)
1
P ∧ (Q ∨ ¬¬R)
2
P
∧Elim: 1
3
¬P
4
P
Reit: 2
5
⊥
⊥Intro: 3,4
6
¬¬P
¬Intro: 3-5
7
Q ∨ ¬¬R
∧Elim: 1
8
Q
9
Q∨R
10
¬¬R
11
R
¬Elim: 10
12
Q∨R
∨Intro: 11
∨Intro: 8
13
Q∨R
∨Elim: 7, 8-9, 10-12
14
¬¬P ∧ (Q ∨ R)
∧Intro: 6,13
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(e)
1
¬(P ∨ Q)
2
¬P ∧ ¬Q
Proof:
1
¬(P ∨ Q)
2
P
3
P ∨Q
∨Intro: 2
4
¬(P ∨ Q)
Reit: 1
5
⊥
⊥Intro: 3,4
6
¬P
¬Intro: 2-5
7
Q
8
P ∨Q
∨Intro: 7
9
¬(P ∨ Q)
Reit: 1
⊥
⊥Intro: 8,9
10
11
¬Q
¬Intro: 10
12
¬P ∧ ¬Q
∧Intro: 6,11
8
(f) (Ex 6.19)
1
A∨B
2
¬B ∨ C
3
A∨C
Here’s one proof:
1
A∨B
2
¬B ∨ C
3
A
4
A∨C
5
B
∨Intro: 3
6
¬B
7
⊥
⊥Intro: 5,6
8
A∨C
⊥Elim: 7
9
C
10
11
12
A∨C
A∨C
A∨C
∨Intro: 9
∨Elim: 2, 6-8, 9-10
∨Elim: 1, 3-4, 5-11
(Note that you might have found other proofs for some of these exercises. It’s not surprising to find more
than one proof of the validity of an argument. You will also hear mathematicians and sometimes logicians
talking about one proof as being better than another, for some or other reason: a proof might be deemed
shorter, more elegant, or more explanatory. What’s debatable is whether this kind of talk has anything to
do with logic.)
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2
Something slightly harder, if there’s time.
1. (Ex 5.24) Assume that n + m is odd. Prove that n × m is even.
We assume the facts that were proved in lecture. In particular, recall that a natural number is either even or odd,
but not both. If a number n is even, then there is a number k such that n = 2k. If a number m is odd, there is a
number k 0 such that n = 2k 0 + 1. Here are some more facts:
Fact 1. If the sum of m and n is odd, then either m is odd or n is odd, but not both.
Proof. We prove the fact by contradiction. We will assume the negation of the conclusion, and derive a contradiction. The negation of the conclusion is: either m and n are both odd, or m and n are both even. There are
two cases to consider.
Case 1. Suppose both m and n are odd. Then there are numbers k and k 0 such that m = 2k + 1 and n = 2k 0 + 1.
Then m + n = 2k + 1 + 2k 0 + 1 = 2k + 2k 0 + 2 = 2(k + k 0 + 1). In other words m + n is even.
Case 2. Suppose both m and n are even. Then there are numbers l and l0 such that m = 2l and m = 2l0 . But
then m + n = 2l + 2l0 = 2(l + l0 ). In other words, m + n is even.
In either case, we find that m + n is even. But this is a contradiction, so our assumption must have been false.
That is, exactly one of m and n is odd.
Fact 2. If m is odd and n is even, then m × n is even.
Proof. Since m is odd, there is a number k such that m = 2k + 1. Since n is even, there is a number l such that
n = 2l. Then m × n = 2l × (2k + 1) = 4lk + 2l = 2(2lk + l). But then m × n is even.
Now we are ready to prove our main claim.
Proof. If m + n is odd, then by Fact 1 exactly one of n, m is odd. Suppose, without loss of generality, that m
is odd.1 Then n is even. But then the product of m and n is even, by Fact 2.
1 We
say ‘without loss of generality’ because if we picked one number as m and it turned out to be even, we could just pick the other number as
m. It would be odd, so our proof would still go through.
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2. (Based on Ex 5.14) Give an informal proof that if S is a tautological consequence of P then S is a tautological consequence of P ∧ Q.
I think this is harder than it looks. For any informal proof, you should always start from your most basic
definitions. Explain why the result holds in terms of those definitions. That is essentially what we did in the
last problem, starting with the meaning of the predicates ‘is even’ and ‘is odd’. Now we should start from the
meaning of the predicate ‘is a tautological consequence of’.
It’s also hard because hidden in this problem (and similar ones) is a question about what makes truth tables work.
The answer is, of course, that they evaluate the logical connectives, and these connectives are truth-functional.
I’m going to give you a proof that shows how to generalize this way of thinking about truth tables. Then I’ll
give you the quick and easy proof.
Proof. Consider all the atomic sentences in a language. There could be infinitely many of them. Each of these
sentences can be either true or false. Let’s describe a truth assignment as an assignment of truth values to all of
these sentences. There are lots of ways to make assignments: just think of all the different combinations you
could have for an infinite set of atomic sentences.
Now consider the joint truth table for P and S. Since a truth table evaluates the truth value of a complex sentence
in terms of its composition from atomic sentences and truth-functional connectives, the truth value of P depends
entirely and only upon the truth values assigned to the atomic sentences upon which it depends (we could call
them P1 , ..., Pn ). Similarly, the truth value of S depends entirely and only upon the truth values assigned to some
sentences S1 , ..., Sn , upon which it depends. A complete truth table lists all the different possible assignments
to these atomic sentences. Think of these assignments as abbreviations for truth assignments to all the atomic
sentences in a language. The truth table lists only the relevant atomic sentences. Why? Because the truth values
of P and S do not depend upon the truth values of other atomic sentences. In other words, if there is some
atomic sentence R which features in neither P nor S, there is no need to know whether a truth assignment
values it true or false.
So this presents us with a new way of describing tautological consequence. A sentence S is a tautological
consequence of a sentence P if every truth assignment that makes P true also makes S true.
Now consider the truth assignments that make P true. Since ∧ is truth functional, the truth value of P ∧ Q
depends only upon the truth values of P and Q, and is defined in every case in which the truth values of P and
Q are defined. Furthermore, P ∧ Q is true under exactly those truth assignments which make both P and Q
true. So whenever an assignment makes P ∧ Q true, it also makes P true.
But we know that every truth assignment that makes P true also makes S true. So it follows that every truth
assignment that makes P ∧ Q true also makes S true, and so S is a tautological consequence of P ∧ Q. (The
relevant bit of each such assignment is represented by a row in the joint truth table of P ∧ Q and S in which
both P ∧ Q and S are true.)
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Quick and dirty:
Proof. Construct a truth table for P , P ∧ Q and S. Every row of this truth table assigns truth values to atomic
sentences in such a way that the truth values of P and S can be computed. (Indeed, the table does not omit any
combination of truth values for the atomic sentences upon which the truth values of P and S depend.)
Now consider the rows of the truth table in which P ∧ Q is true. Since ∧ is truth-functional, we know from the
definition of ∧ that P is true in each such row. But then this is a row containing an assignment of truth values to
the atomic sentences that makes S true. Thus S is a tautological consequence of P ∧ Q.
3. Give an informal proof of the following argument. What is strange about the argument?
1
Home(max) ∨ Home(claire)
2
¬Home(max)
3
¬Home(claire)
4
Home(max) ∧ Happy(carl)
Here’s one proof:
Proof. Let’s consider each case in which the first premise is true. Case 1: Max is home. Then by the second
premise, Max is not home. Contradiction.
Case 2: Claire is home. By the third premise, Claire is not home. Contradiction.
We find that there is a contradiction in each case. Anything follows from a contradiction, so we hold that Max
is home and Carl is happy.
Yes, what’s weird about the argument is that the premises are inconsistent, so the argument is valid. But the
above proof is not satisfying unless you already believe principles such as: anything follows from a contradiction. Here’s another proof that you might prefer:
Proof. The first premise tells us that either Max is home or Claire is home. The third premise tells that Claire is
not home. So in order to make the first premise true, it must be the case that Max is home.
Now again, we know from the first premise that either Max is home or Claire is home. But the second premise
tells us that Max is not home. So Claire must be home. But then it’s true that Claire is home or Carl is happy
(since the first disjunct is true). But by the third premise we know Claire is not home, so it is the case that Carl
is happy.
Since we have shown that Max is home and that Carl is happy, we have proved the conclusion.
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Illogical, right? If you were beginning to get comfortable with the idea that anything follows from a contradiction, I hope this makes you uncomfortable again.
3
Challenge question
(Based on Ex 5.26) Prove that
√
3 is irrational, using the proof that
√
2 is irrational as a model. Now prove that
the square root of any prime number is irrational.
I’ll do the general case. The basic fact you need is:
Fact 3. If n2 is divisible by a prime number k, then n is divisible by k.
I’m not going to give you a convincing proof, but consider that if k is prime and it divides any product m × n then
it must divide either m or n. (Since it is prime, it cannot be factored into further facts, some of which divide m and
some of which divide n). In the case of n2 , we have that k therefore divides n.
Now let’s prove our claim:
√
Proof. Consider an arbitrary prime number k. We show that k is irrational.
√
√
Suppose that k is rational. Then there are two numbers n and m, not both divisible by k, such that k = n/m.
(If both were divisible by k we could simplify the fraction until they were not.)
By simple algebra we have that k = n2 /m2 , and so m2 = kn2 . So m2 is divisible by k. But by Fact 3, this means
that m is divisible by k. But then n2 is divisible by k, and so n is divisible by k. Contradiction. Our assumption that
√
k is rational must have been wrong.
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