Mathematical-Reasoning · Unit 3 · Video 3 · Interactive Practice

Logical Equivalence: De Morgan's Laws and the Contrapositive

IKey Formulas

Equivalence	Name	What it says
$X \equiv Y$	Logical equivalence	Same truth value in every row of the truth table
$\neg(P \wedge Q) \equiv \neg P \vee \neg Q$ $\neg(P \vee Q) \equiv \neg P \wedge \neg Q$	De Morgan's Laws	The negation moves onto each part and the connective switches
$P \to Q \equiv \neg P \vee Q$ $\neg(P \to Q) \equiv P \wedge \neg Q$	Conditional and its negation	The negation of a conditional is a conjunction
$P \to Q \equiv \neg Q \to \neg P$	Contrapositive	The converse is different: $P \to Q \not\equiv Q \to P$

Key Insight: Equivalence is a row-by-row test: the columns must agree in all four rows, and a single disagreeing row rules it out.

IIThe Row-by-Row Test

Two expressions are logically equivalent exactly when their columns agree in every row.

X Y

IIIConverse and Contrapositive at Each $x$

With $P$ : $x = 3$ and $Q$ : $x^2 = 9$ , each real $x$ lands in one row of the truth table.

x 0

💡 A conditional and its contrapositive stand or fall together, but proving $P \to Q$ says nothing about the converse $Q \to P$ : the converse needs its own proof.

IVQuiz Questions

Problem 1 · De Morgan's Second Law

Given: statements $P$ and $Q$ . Which expression is logically equivalent to $\neg(P \vee Q)$ ?

✅ Correct! The negation moves onto each part and

\vee

switches to

\wedge

: both columns read F, F, F, T.

❌ Close, but the connective must switch.

\neg P \vee \neg Q

reads F, T, T, T, while

\neg(P \vee Q)

reads F, F, F, T; they disagree in rows 2 and 3. In fact

\neg P \vee \neg Q \equiv \neg(P \wedge Q)

, the first law.

❌ Not quite.

\neg P \vee Q

negates only

P

and keeps

\vee

: it reads T, F, T, T (it is equivalent to

P \to Q

), while

\neg(P \vee Q)

reads F, F, F, T; they disagree in rows 1 and 3.

❌ Not quite.

\neg P \wedge Q

switches the connective but negates only

P

: it reads F, F, T, F, while

\neg(P \vee Q)

reads F, F, F, T; they disagree in rows 3 and 4. The negation must move onto both parts.

Show solution

$P \vee Q$ is false only when both parts are false (row 4), so $\neg(P \vee Q)$ reads F, F, F, T.

$\neg P \wedge \neg Q$ is true only when $\neg P$ and $\neg Q$ are both true, which is again row 4: it reads F, F, F, T.

The columns agree in every row, so $\neg(P \vee Q) \equiv \neg P \wedge \neg Q$ . The other options each disagree in at least one row: $\neg P \vee \neg Q$ in rows 2 and 3, $\neg P \vee Q$ (T, F, T, T) in rows 1 and 3, and $\neg P \wedge Q$ (F, F, T, F) in rows 3 and 4.

Problem 2 · Negating a Conditional

Given: the statement "If it rains, then the game is cancelled." Which sentence is its negation?

✅ Correct! With

P

: it rains and

Q

: the game is cancelled,

\neg(P \to Q) \equiv P \wedge \neg Q

, and the negation of a conditional is a conjunction.

❌ Not quite. The negation of a conditional is not a conditional.

P \to \neg Q

reads F, T, T, T, but

\neg(P \to Q)

reads F, T, F, F; they disagree in rows 3 and 4.

❌ Not quite. This is

\neg P \vee Q

, which is logically equivalent to

P \to Q

itself, so it says the same thing as the original statement rather than denying it.

❌ Not quite.

\neg P \to \neg Q

reads T, T, F, T, but

\neg(P \to Q)

reads F, T, F, F; they disagree in rows 1 and 4.

Show solution

Let $P$ be "it rains" and $Q$ be "the game is cancelled". The statement is $P \to Q$ , which is false only when $P$ is true and $Q$ is false (row 2). So $\neg(P \to Q)$ reads F, T, F, F.

$P \wedge \neg Q$ is true only when $P$ is true and $Q$ is false, again row 2, so it also reads F, T, F, F. Hence

\neg(P \to Q) \equiv P \wedge \neg Q,

and the negation is "It rains and the game is not cancelled."

Problem 3 · Contrapositive Versus Converse

Given: an integer $n$ is a multiple of 4 provided $n = 4k$ for some integer $k$ . Let $S$ be the true statement "If $n$ is a multiple of 4, then $n$ is even." Find the contrapositive of $S$ and the truth value of the converse of $S$ .

Which statement is the contrapositive of S?

What is the truth value of the converse of S?

✅ Correct! The contrapositive

\neg Q \to \neg P

is equivalent to

S

, but the converse

Q \to P

is not, and

n = 2

is even without being a multiple of 4.

❌ Check the contrapositive. For

P \to Q

the contrapositive is

\neg Q \to \neg P

: the parts trade places and each one is negated. Option (a) only swaps, option (b) only negates, and option (c) is the negation of

S

❌ Check the converse. The converse "If

n

is even, then

n

is a multiple of 4" needs its own check; test a small even integer.

❌ Not quite.

P \to Q \not\equiv Q \to P

: a true conditional (and its equally true contrapositive) says nothing about whether the converse is true. Test the converse directly on a small even integer.

❌ Not quite. A counterexample to "If

n

is even, then

n

is a multiple of 4" must make the hypothesis true and the conclusion false. Since

8 = 4 \cdot 2

, the conclusion holds for

n = 8

Show solution

Let $P$ be " $n$ is a multiple of 4" and $Q$ be " $n$ is even", so $S$ is $P \to Q$ . ( $S$ is true: if $n = 4k$ , then $n = 2(2k)$ with $2k$ an integer.)

Contrapositive $\neg Q \to \neg P$ : "If $n$ is not even, then $n$ is not a multiple of 4." It is logically equivalent to $S$ , so it is also true.

Converse $Q \to P$ : "If $n$ is even, then $n$ is a multiple of 4." Take $n = 2$ : the hypothesis is true since $2 = 2 \cdot 1$ , but $2 = 4k$ would need $k = \tfrac{1}{2}$ , which is not an integer. So the conclusion is false and the converse is false. This is row 3 of the table ( $P$ false, $Q$ true), where $P \to Q$ is true and $Q \to P$ is false.

Problem 4 · An Equivalent of the Converse

Given: statements $P$ and $Q$ . Which expression is logically equivalent to the converse $Q \to P$ ?

✅ Correct!

\neg P \to \neg Q

is the contrapositive of

Q \to P

, so the two always share a truth value: both read T, T, F, T.

❌ Not quite.

\neg Q \to \neg P

is the contrapositive of

P \to Q

, so it is equivalent to the original, which disagrees with the converse in rows 2 and 3.

❌ Not quite. Rewriting a conditional as a disjunction negates its hypothesis:

Q \to P \equiv \neg Q \vee P

, not

Q \vee P

Q \vee P

reads T, T, T, F against the converse's T, T, F, T; they disagree in rows 3 and 4.

❌ Not quite.

Q \wedge \neg P

is the negation of the converse, since

\neg(Q \to P) \equiv Q \wedge \neg P

. It reads F, F, T, F, the opposite of the converse's T, T, F, T in every row.

Show solution

Apply the contrapositive equivalence to the conditional $Q \to P$ itself: its hypothesis is $Q$ and its conclusion is $P$ , so trading places and negating each part gives

Q \to P \equiv \neg P \to \neg Q.

Check with the table. $Q \to P$ is false only when $Q$ is true and $P$ is false (row 3): T, T, F, T. $\neg P \to \neg Q$ is false only when $\neg P$ is true and $\neg Q$ is false, that is $P$ false and $Q$ true, again row 3: T, T, F, T. The columns agree in every row.

Option (a) is the contrapositive of $P \to Q$ , so it reads T, F, T, T and disagrees with the converse in rows 2 and 3. Option (c), $Q \vee P$ , drops the negation that the disjunction form needs ( $Q \to P \equiv \neg Q \vee P$ ) and disagrees in rows 3 and 4. Option (d), $Q \wedge \neg P$ , is the negation of the converse and disagrees in every row.

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Logical Equivalence: De Morgan's Laws and the Contrapositive

IKey Formulas

IIThe Row-by-Row Test

IIIConverse and Contrapositive at Each xxx

IVQuiz Questions

IIIConverse and Contrapositive at Each $x$