In Exercises (5) and (6) from Section 2.1, we observed situations where two different statements have the same truth tables. Basically, this means these statements are equivalent, and we make the following definition:
Two expressions are logically equivalent provided that they have the same truth value for all possible combinations of truth values for all variables appearing in the two expressions. In this case, we write \(X \equiv Y\) and say that \(X\) and \(Y\) are logically equivalent.
Sometimes we actually use logical reasoning in our everyday living! Perhaps you can imagine a parent making the following two statements:
We now define two important conditional statements that are associated with a given conditional statement.
If \(P\) and \(Q\) are statements, then
In Preview Activity \(\PageIndex\), we introduced the concept of logically equivalent expressions and the notation \(X \equiv Y\) to indicate that statements \(X\) and \(Y\) are logically equivalent. The following theorem gives two important logical equivalencies. They are sometimes referred to as De Morgan’s Laws.
For statements \(P\) and \(Q\),
The first equivalency in Theorem 2.5 was established in Preview Activity \(\PageIndex\). Table 2.3 establishes the second equivalency.
| \(P\) | \(Q\) | \(P \vee Q\) | \(\urcorner (P \vee Q)\) | \(\urcorner P\) | \(\urcorner Q\) | \(\urcorner (P \vee Q) \equiv \urcorner P \wedge \urcorner Q\) |
|---|---|---|---|---|---|---|
| T | T | T | F | F | F | F |
| T | F | T | F | F | T | F |
| F | T | T | F | T | F | F |
| F | F | F | T | T | T | T |
It is possible to develop and state several different logical equivalencies at this time. However, we will restrict ourselves to what are considered to be some of the most important ones. Since many mathematical statements are written in the form of conditional statements, logical equivalencies related to conditional statements are quite important.
The first two logical equivalencies in the following theorem were established in Preview Activity \(\PageIndex\), and the third logical equivalency was established in Preview Activity \(\PageIndex\).
For statements \(P\) and \(Q\),
The logical equivalency \(\urcorner (P \to Q) \equiv P \wedge \urcorner Q\) is interesting because it shows us that the negation of a conditional statement is not another conditional statement. The negation of a conditional statement can be written in the form of a conjunction. So what does it mean to say that the conditional statement
If you do not clean your room, then you cannot watch TV,
is false? To answer this, we can use the logical equivalency \(\urcorner (P \to Q) \equiv P \wedge \urcorner Q\). The idea is that if \(P \to Q\) is false, then its negation must be true. So the negation of this can be written as
You do not clean your room and you can watch TV.
For another example, consider the following conditional statement:
This conditional statement is false since its hypothesis is true and its conclusion is false. Consequently, its negation must be true. Its negation is not a conditional statement. The negation can be written in the form of a conjunction by using the logical equivalency \(\urcorner (P \to Q) \equiv P \wedge \urcorner Q\). So, the negation can be written as follows:
However, the second part of this conjunction can be written in a simpler manner by noting that “not less than” means the same thing as “greater than or equal to.” So we use this to write the negation of the original conditional statement as follows:
This conjunction is true since each of the individual statements in the conjunction is true.
We have seen that it often possible to use a truth table to establish a logical equivalency. However, it is also possible to prove a logical equivalency using a sequence of previously established logical equivalencies. For example,
The last step used the fact that \(\urcorner (\urcorner P)\) is logically equivalent to \(P\).
When proving theorems in mathematics, it is often important to be able to decide if two expressions are logically equivalent. Sometimes when we are attempting to prove a theorem, we may be unsuccessful in developing a proof for the original statement of the theorem. However, in some cases, it is possible to prove an equivalent statement. Knowing that the statements are equivalent tells us that if we prove one, then we have also proven the other. In fact, once we know the truth value of a statement, then we know the truth value of any other logically equivalent statement. This is illustrated in Progress Check 2.7.
In Section 2.1, we constructed a truth table for \((P \wedge \urcorner Q) \to R\).
\(P \wedge \urcorner Q) \to R \equiv \urcorner (P \wedge \urcorner Q) \vee R\),
If 3 is a factor of \(a \cdot b\), then 3 is a factor of \(a\) or 3 is a factor of \(b\).
Explain why we will have proven this statement if we prove the following:
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As we will see, it is often difficult to construct a direct proof for a conditional statement of the form \(P \to (Q \vee R)\). The logical equivalency in Progress Check 2.7 gives us another way to attempt to prove a statement of the form \(P \to (Q \vee R)\). The advantage of the equivalent form, \(P \wedge \urcorner Q) \to R\), is that we have an additional assumption, \(\urcorner Q\), in the hypothesis. This gives us more information with which to work.
Theorem 2.8 states some of the most frequently used logical equivalencies used when writing mathematical proofs.
For statement \(P\), \(Q\), and \(R\),
De Morgan's Laws \(\urcorner (P \wedge Q) \equiv \urcorner P \vee \urcorner Q\).
\(\urcorner (P \vee Q) \equiv \urcorner P \wedge \urcorner Q\).
Conditional Statement. \(P \to Q \equiv \urcorner Q \to \urcorner P\) (contrapositive)
\(P \to Q \equiv \urcorner P \vee Q\)
\(\urcorner (P \to Q) \equiv P \wedge \urcorner Q\)
Biconditional Statement \((P leftrightarrow Q) \equiv (P \to Q) \wedge (Q \to P)\)
Double Negation \(\urcorner (\urcorner P) \equiv P\)
Distributive Laws \(P \vee (Q \wedge R) \equiv (P \vee Q) \wedge (P \vee R)\)
\(P \wedge (Q \vee R) \equiv (P \wedge Q) \vee (P \wedge R)\)
Conditionals withDisjunctions \(P \to (Q \vee R) \equiv (P \wedge \urcorner Q) \to R\)
\((P \vee Q) \to R \equiv (P \to R) \wedge (Q \to R)\)
We have already established many of these equivalencies. Others will be established in the exercises.
If \(f\) is differentiable at \(x = a\), then \(f\) is continuous at \(x = a\). Which of the following statements have the same meaning as this conditional statement and which ones are negations of this conditional statement? Note: This is not asking which statements are true and which are false. It is asking which statements are logically equivalent to the given statement. It might be helpful to let P represent the hypothesis of the given statement, \(Q\) represent the conclusion, and then determine a symbolic representation for each statement. Instead of using truth tables, try to use already established logical equivalencies to justify your conclusions.
If \(a\) divides \(bc\), then \(a\) divides \(b\) or \(a\) divides \(c\). Which of the following statements have the same meaning as this conditional statement and which ones are negations of this conditional statement? The note for Exercise (10) also applies to this exercise.
If \(x^3 - x = 2x^2 +6\), then \(x = -2\) or \(x = 3\).
Which of the following statements have the same meaning as this conditional statement and which ones are negations of this conditional statement? Explain each conclusion. (See the note in the instructions for Exercise (10).)
(a) If \(x \ne -2\) and \(x \ne 3\), then \(x^3 - x \ne 2x^2 +6\).
(b) If \(x = -2\) or \(x = 3\), then \(x^3 - x = 2x^2 +6\).
(c) If \(x \ne -2\) or \(x \ne 3\), then \(x^3 - x \ne 2x^2 +6\).
(d) If \(x^3 - x = 2x^2 +6\) and \(x \ne -2\), then \(x = 3\).
(e) If \(x^3 - x = 2x^2 +6\) or \(x \ne -2\), then \(x = 3\).
(f) \(x^3 - x = 2x^2 +6\), \(x \ne -2\), and \(x \ne 3\).
(g) \(x^3 - x \ne 2x^2 +6\) or \(x = -2\) or \(x = 3\).
If \(x \cdot y\) is even, then \(x\) is even or \(y\) is even. We notice that we can write this statement in the following symbolic form: \(P \to (Q \vee R)\),
where \(P\) is“\(x \cdot y\) is even,” \(Q\) is“\(x\) is even,”and \(R\) is “\(y\) is even.”
(a) Write the symbolic form of the contrapositive of \(P \to (Q \vee R)\). Then use one of De Morgan’s Laws (Theorem 2.5) to rewrite the hypothesis of this conditional statement.
(b) Use the result from Part (13a) to explain why the given statement is logically equivalent to the following statement:
If \(x\) is odd and \(y\) is odd, then \(x \cdot y\) is odd.
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This page titled 2.2: Logically Equivalent Statements is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Ted Sundstrom (ScholarWorks @Grand Valley State University) via source content that was edited to the style and standards of the LibreTexts platform.
