Foundations and Proof
Notes by Dr. Lynne H. Walling
and Dr. Steffi Zegowitz
September 2016
2
Truth Tables, Equivalences and the Contrapositive
2.1
Truth Tables
In a mathematical system, true and false statements are the statements of the system,
and the label ‘true’ or ‘false’ associated with a given statement is its truth value.
Notation.
• We use the symbol ¬ to mean not.
• We use the symbol ∧ to mean and.
• We use the symbol ∨ to mean or.
• We use the symbol =⇒ to mean implies.
• We use the symbol ⇐⇒ to mean if and only if. That is, for statements P, Q,
P ⇐⇒ Q means that ‘P =⇒ Q and Q =⇒ P ’. When P ⇐⇒ Q, we say P and
Q are equivalent.
Example. Let x ∈ Z, let P be the statement ‘ x ≥ 5’ and Q be the statement ‘ x ≤ 7’.
Then P ∧ Q represents the statements ‘ x ≥ 5 and x ≤ 7’, that is ‘ 5 ≤ x ≤ 7’.
Example. Let P and Q be two statements. P =⇒ Q is the statement that ‘P implies
Q’, or equivalently it is the statement that ‘If P , then Q’. Note that P =⇒ Q allows
for both P and Q to be true or for P to be false and Q to be true or for P and Q to both
be false. However, P =⇒ Q does not allow for P to be true and Q to be false.
We can represent this scenario by using a truth table, where we consider all possible
combinations of the truth values of P and Q and the consequent truth value of P =⇒ Q:
P
T
T
F
F
Q
T
F
T
F
P =⇒ Q
T
F
T
T
Remark. Here it is important to note that P =⇒ Q does not mean that P causes Q. It
merely says that P is a sufficient condition for Q.
Example. Let P and Q be two statements. We have that P ∧ Q is true exactly when P
and Q are both true. The corresponding truth table is as follows.
1
2 Truth Tables, Equivalences and the Contrapositive
P
T
T
F
F
Q
T
F
T
F
2
P∧Q
T
F
F
F
Example. Let P and Q be two statements. We have that P ∨ Q is true exactly when P
or Q is true. Hence, we have the following corresponding truth table.
P
T
T
F
F
Q
T
F
T
F
P∨Q
T
T
T
F
Example. Let P and Q be two statements. The corresponding truth table for (¬P ) ∨ Q
is as follows.
¬P
F
F
T
T
P
T
T
F
F
Q
T
F
T
F
(¬P ) ∨ Q
T
F
T
T
Example. Let P and Q be two statements. The corresponding truth table for (¬Q) =⇒
(¬P ) is as follows.
P
T
T
F
F
2.2
Q
T
F
T
F
¬P
F
F
T
T
¬Q
F
T
F
T
(¬Q) =⇒ (¬P )
T
F
T
T
Logical Equivalence
If we take two statements which are logically equivalent, say P is equivalent to Q, then
proving P to be true is equivalent to proving Q to be true. Similarly, proving Q to be true
is equivalent to proving P to be true.
We have the following theorem.
Theorem 2.1. Let P and Q be two statements.
(i) P =⇒ Q is equivalent to (¬Q) =⇒ (¬P ).
(ii) P =⇒ Q is equivalent to (¬P ) ∨ Q.
Proof. We prove part (i) and leave part (ii) as an exercise. We have the following truth
table.
P
T
T
F
F
¬P
F
F
T
T
Q
T
F
T
F
¬Q
F
T
F
T
P =⇒ Q
T
F
T
T
(¬Q) =⇒ (¬P )
T
F
T
T
Hence, for all truth values of P and Q, we have that P =⇒ Q and (¬Q) =⇒ (¬P ) have
the same truth values. Therefore, P =⇒ Q is equivalent to (¬Q) =⇒ (¬P ).
2 Truth Tables, Equivalences and the Contrapositive
2.3
3
The Contrapositive
We have the following definition.
Definition 2.2. Let P and Q be two statements. The contrapositive of the statement
P =⇒ Q is given by the statement (¬Q) =⇒ (¬P ).
As seen in Theorem 2.1, we have that the statement P =⇒ Q is equivalent to its
contrapositive.
Definition 2.3. Let P and Q be two statements. The statement Q =⇒ P is called the
converse of the statement P =⇒ Q.
Remark. The converse Q =⇒ P of the statement P =⇒ Q is not equivalent to the
original statement P =⇒ Q.
We have the following proposition.
Proposition 2.4. Suppose P is a statement. Then P ⇐⇒ (¬(¬P )).
Proof. We have the following truth table.
P
T
F
¬P
F
T
¬(¬P )
T
F
We can see from the table that the truth values of P and ¬(¬P ) always agree. Hence, the
two statements P and ¬(¬P ) are equivalent.
The next proposition shows that ∧ and ∨ are associative, that is, P ∧ Q ∧ R and
P ∨ Q ∨ R are statements that do not require parentheses.
Proposition 2.5. Suppose P, Q, R are statements.
(i) ((P ∧ Q) ∧ R) ⇐⇒ (P ∧ (Q ∧ R)).
(ii) ((P ∨ Q) ∨ R) ⇐⇒ (P ∨ (Q ∨ R)).
Proof. We prove part (i) and leave the proof of part (ii) as an exercise. We have the
following truth table.
P
T
T
T
T
F
F
F
F
Further, we have
Q
T
T
F
F
T
T
F
F
R
T
F
T
F
T
F
T
F
P∧Q
T
T
F
F
F
F
F
F
(P ∧ Q) ∧ R
T
F
F
F
F
F
F
F
2 Truth Tables, Equivalences and the Contrapositive
P
T
T
T
T
F
F
F
F
Q
T
T
F
F
T
T
F
F
R
T
F
T
F
T
F
T
F
Q∧R
T
F
F
F
T
F
F
F
4
P ∧ (Q ∧ R)
T
F
F
F
F
F
F
F
Hence, for any truth values of P, Q, R, the truth tables show that ((P ∧ Q) ∧ R) ⇐⇒
(P ∧ (Q ∧ R)).
The above proposition shows that we can write P ∧ Q ∧ R and P ∨ Q ∨ R without
parentheses and without there being confusion. As an exercise, one may also prove the
following sometimes useful equivalences.
Proposition 2.6. Let P, Q, R be statements. Then
(i) (P ∧ (Q ∧ R)) ⇐⇒ ((P ∧ Q) ∧ (P ∧ R)).
(ii) (P ∨ (Q ∨ R)) ⇐⇒ ((P ∨ Q) ∨ (P ∨ R)).
Proposition 2.7. Let P, Q, R be statements. Then
(i) (P ∧ (Q ∨ R)) ⇐⇒ ((P ∧ Q) ∨ (P ∧ R)).
(ii) (P ∨ (Q ∧ R)) ⇐⇒ ((P ∨ Q) ∧ (P ∨ R)).
Proof. We will prove part (i) and leave the proof of part (ii) as an exercise. We have the
following truth table.
P
T
T
T
T
F
F
F
F
Q
T
T
F
F
T
T
F
F
R
T
F
T
F
T
F
T
F
Q∨R
T
T
T
F
T
T
T
F
P ∧ (Q ∨ R)
T
T
T
F
F
F
F
F
Further, we have
P
T
T
T
T
F
F
F
F
Q
T
T
F
F
T
T
F
F
R
T
F
T
F
T
F
T
F
P ∧Q
T
T
F
F
F
F
F
F
P ∧R
T
F
T
F
F
F
F
F
(P ∧ Q) ∨ (P ∧ R)
T
T
T
F
F
F
F
F
2 Truth Tables, Equivalences and the Contrapositive
5
Hence, for any truth values of P, Q, R, the above truth tables show that (P ∧(Q∨R)) ⇐⇒
((P ∧ Q) ∨ (P ∧ R)).
Theorem 2.8. Suppose P, Q are two statments. Then
(i) ¬(P ∧ Q) ⇐⇒ ((¬P ) ∨ (¬Q)).
(ii) ¬(P ∨ Q) ⇐⇒ ((¬P ) ∧ (¬Q)).
(iii) ¬(P =⇒ Q) ⇐⇒ (P ∧ (¬Q)).
Proof. We will prove part (i) and leave parts (ii) and (iii) as exercises. We have the
following truth table.
P
T
T
F
F
Q
T
F
T
F
¬P
F
F
T
T
¬Q
F
T
F
T
P ∧Q
T
F
F
F
¬(P ∧ Q)
F
T
T
T
(¬P ) ∨ (¬Q)
F
T
T
T
Thus, for any truth values of P, Q, R, the truth values of ¬(P ∧ Q) and (¬P ) ∨ (¬Q) are
the same.
As an exercise, one may prove the following.
Proposition 2.9. Let P, Q be two statements. Then
(P ∨ Q) ⇐⇒ ((¬P ) =⇒ Q)
Remark. Let P, Q, R be statements. Then
P =⇒ Q ⇐⇒ R
and
P ⇐⇒ Q =⇒ R
have no clear meanings. As an exercise, one may show that the statements
P =⇒ (Q ⇐⇒ R)
and
(P =⇒ Q) ⇐⇒ R
are not equivalent. Further, one may show that
P ⇐⇒ (Q =⇒ R)
and
(P ⇐⇒ Q) =⇒ R
are not equivalent. Hence, assertions such as
P =⇒ Q ⇐⇒ R =⇒ S
have no clear meaning.
2.4
The Pigeonhole Principle
We have the following theorem
Theorem 2.10 (Pigeonhole Principle). Let A be a set with n elements and B a set with
m elements where m, n ∈ N with m < n. Then there is no injection from A into B.
We will give two proofs of Theorem 2.10. The first one is a proof by contradiction,
and the second one is a proof by contrapositive.
2 Truth Tables, Equivalences and the Contrapositive
6
Proof 1. Let A = {a1 , a2 , . . . , an } and B = {b1 , b2 , . . . , bm } with m < n. Let g : A → B.
To the contrary, suppose that g : A → B is injective. Now, define a set
C = {g(ai ) : 1 ≤ i ≤ n, n ∈ N}.
Then C is a subset of B, and since g is injective, C is a set with n elements. But then C is
a subset of B which contains more elements than B, which is a contradiction. Therefore,
there is no injection from A to B.
Proof 2. We will prove that if there is an injection from A into B, then m ≥ n.
Let A = {a1 , a2 , . . . , an } and B = {b1 , b2 , . . . , bm } and suppose that g : A → B is
injective. Now, define
C = {g(ai ) : 1 ≤ i ≤ n, n ∈ N}.
Then C is a subset of B, and since g is injective, C is a set with n elements. It follows that
B must have at least n elements, that is m ≥ n. The result follows by contrapositive.
2 Truth Tables, Equivalences and the Contrapositive
7
Exercises
2.1. Suppose P, Q, R are statements. Prove that
(a) (P =⇒ Q) ⇐⇒ ((¬P ) ∨ Q).
(b) ((P ∨ Q) ∨ R) ⇐⇒ (P ∨ (Q ∨ R)).
(c) (P ∨ (Q ∧ R)) ⇐⇒ ((P ∨ Q) ∧ (P ∨ R)).
2.2. Suppose P, Q are statements. Prove that
(a) ¬(P ∨ Q) ⇐⇒ ((¬P ) ∧ (¬Q)).
(b) ¬(P =⇒ Q) ⇐⇒ (P ∧ (¬Q)).
2.3. Suppose P, Q, R are statements. Prove that
(a) (P ∧ (Q ∧ R)) ⇐⇒ ((P ∧ Q) ∧ (P ∧ R)).
(b) (P ∨ (Q ∨ R)) ⇐⇒ ((P ∨ Q) ∨ (P ∨ R)).
2.4. Suppose P, Q are statements. Prove that
(a) (P ∨ Q) ⇐⇒ ((¬P ) =⇒ Q).
(b) (¬(P =⇒ Q) =⇒ (¬P )) ⇐⇒ (P =⇒ Q).
2.5. Suppose P, Q, R are statements.
(a) Show that (P =⇒ Q) ⇐⇒ R is not equivalent to P =⇒ (Q ⇐⇒ R).
(b) Show that (P ⇐⇒ Q) =⇒ R is not equivalent to P ⇐⇒ (Q =⇒ R).
2.6. We use R2 to denote R × R, and R3 to denote R × R × R. Define f : R2 → R3 by
f ((x, y)) = (x + y, x − y, x2 + y 2 ).
(a) Prove that f is injective. (Suggestion: Suppose that (x, y), (u, v) ∈ R2 so that
f ((x, y)) = f ((u, v)). Show that (x, y) = (u, v). (So you must show that x = u
and y = v.)
(b) Prove that f is not surjective. (So you need to choose explicit values for u, v, w
and then deduce that for any choices for x, y ∈ R, it is impossible to have
f ((x, y)) = (u, v, w). Suggestion: To the contrary, suppose that f is surjective.
Carefully choose explicit values u, v, w ∈ R, and suppose that (x, y) ∈ R2 with
f ((x, y)) = (u, v, w) and derive a contradiction.)