Mathematical-Reasoning · Unit 3 · Video 4 · Interactive Practice

The Negation of an If-Then Is an And: Negating Conditionals and Proving Equivalences by Chains

IKey Formulas

Equivalence	Name	What it gives
$\lnot(P \to Q) \equiv P \land \lnot Q$	Negation of a conditional	The negation of an if-then is an and
$P \to Q \equiv \lnot P \lor Q$	Conditional as an or	The usual first link of a chain
$P \to (Q \lor R) \equiv (P \land \lnot Q) \to R$	Conditional with a disjunction in the conclusion	One extra assumption, $\lnot Q$
$(P \lor Q) \to R \equiv (P \to R) \land (Q \to R)$	Conditional with a disjunction in the hypothesis	$R$ must follow from $P$ and from $Q$

Key Insight: An equivalence can be proven without a truth table: a chain $A \equiv B \equiv C \equiv \cdots$ in which every step is a known equivalence. The list of reasons beside the lines is the proof.

IIWhich Statement Is the Negation?

A negation of $P \to Q$ must have the opposite truth value in every one of the four rows.

Candidate for ¬(P → Q)

IIIIf $a < b$ , Then $a^2 < b^2$ ?

The negation is $a < b$ and $a^2 \ge b^2$ , because $\lnot(a^2 < b^2)$ means $a^2 \ge b^2$ , not $a^2 > b^2$ .

a −5 b −3

💡 Challenge: find $a < b$ with $a^2 = b^2$ . There the conditional is false, and only the $\ge$ form of the negation is true.

IVProof by a Chain of Equivalences

Each line of a chain has the same truth value as the line before it in every row, so the first and last lines are equivalent.

Chain Line 0

\lnot(P \to Q)

\equiv \lnot(\lnot P \lor Q) \qquad \text{[}P \to Q \equiv \lnot P \lor Q\text{, negate both sides]}

\equiv \lnot(\lnot P) \land \lnot Q \qquad \text{[De Morgan, negation of an or]}

\equiv P \land \lnot Q \qquad \text{[double negation]}

(P \land \lnot Q) \to R

\equiv \lnot(P \land \lnot Q) \lor R \qquad \text{[}P \to Q \equiv \lnot P \lor Q\text{]}

\equiv \big(\lnot P \lor \lnot(\lnot Q)\big) \lor R \qquad \text{[De Morgan, negation of an and]}

\equiv (\lnot P \lor Q) \lor R \qquad \text{[double negation]}

\equiv \lnot P \lor (Q \lor R) \qquad \text{[regroup the or]}

\equiv P \to (Q \lor R) \qquad \text{[}P \to Q \equiv \lnot P \lor Q\text{, right to left]}

VQuiz Questions

Problem 1 · Negate a Numerical Conditional

Given: the conditional “If $-2 < 1$ , then $(-2)^2 < 1^2$ .” Which statement is its negation?

✅ Correct!

\lnot(P \to Q) \equiv P \land \lnot Q

, and

\lnot\big((-2)^2 < 1^2\big)

(-2)^2 \ge 1^2

. Here

4 \ge 1

, so the negation is true and the original conditional is false.

❌ That negates only the conclusion.

P \to \lnot Q

is not the negation of

P \to Q

: when

P

is false both conditionals are true. The negation of an if-then is an and.

❌ Not quite. The hypothesis is kept, not negated:

\lnot(P \to Q) \equiv P \land \lnot Q

. A conditional fails only when its hypothesis is true and its conclusion is false.

❌ Close, but the inequality is wrong. The negation of

x < y

x \ge y

, not

x > y

. For

x = y = 9

9 < 9

is false and

9 > 9

is false too.

❌ Not quite. Use

\lnot(P \to Q) \equiv P \land \lnot Q

Show solution

Let $P$ be $-2 < 1$ and $Q$ be $(-2)^2 < 1^2$ . The negation of $P \to Q$ is $P \land \lnot Q$ .

$\lnot Q$ says it is not the case that $(-2)^2 < 1^2$ , which means $(-2)^2 \ge 1^2$ . So the negation is

-2 < 1 \ \text{ and } \ (-2)^2 \ge 1^2.

Check: $-2 < 1$ is true, and $(-2)^2 = 4 \ge 1 = 1^2$ is true, so the negation is true. That matches the original: its hypothesis is true and its conclusion $4 < 1$ is false, so the conditional is false.

Problem 2 · Negating When the Conclusion Is an Or

Given: statements $P$ , $Q$ and $R$ . Which statement is logically equivalent to $\lnot\big(P \to (Q \lor R)\big)$ ?

✅ Correct!

\lnot\big(P \to (Q \lor R)\big) \equiv P \land \lnot(Q \lor R) \equiv P \land (\lnot Q \land \lnot R)

, by the negation of a conditional and then De Morgan.

❌ That is still a conditional. Negating only the conclusion fails when

P

is false: then

P \to (\lnot Q \land \lnot R)

is true, while

\lnot\big(P \to (Q \lor R)\big)

is false.

❌ Check De Morgan. The negation of an or is the and of the negations:

\lnot(Q \lor R) \equiv \lnot Q \land \lnot R

. With

P

Q

true and

R

false your option is true, but

P \to (Q \lor R)

is true there, so its negation is false.

❌ Not quite. The hypothesis

P

is kept, not negated. With

P

Q

R

all false your option is true, but

P \to (Q \lor R)

is true there (false hypothesis), so its negation is false.

❌ Not quite. Apply

\lnot(A \to B) \equiv A \land \lnot B

with

B = Q \lor R

, then De Morgan.

Show solution

A chain of known equivalences, with $Q \lor R$ playing the role of the conclusion:

\lnot\big(P \to (Q \lor R)\big) \equiv P \land \lnot(Q \lor R) \qquad \text{[negation of a conditional]}

\equiv P \land (\lnot Q \land \lnot R) \qquad \text{[De Morgan]}

Since an and of three statements is true exactly when all three are true, the grouping does not matter: $P \land \lnot Q \land \lnot R$ . It is true in exactly one of the eight rows, $P$ true, $Q$ false, $R$ false, which is the only row where $P \to (Q \lor R)$ is false.

Problem 3 · Complete the Chain

Given: a chain proving $(P \lor Q) \to R \equiv (P \to R) \land (Q \to R)$ :

\begin{aligned} (P \lor Q) \to R &\equiv \lnot(P \lor Q) \lor R && \text{[conditional as an or]}\\ &\equiv \ \ \textbf{?} && \text{[De Morgan]}\\ &\equiv (\lnot P \lor R) \land (\lnot Q \lor R) && \text{[ ? ]}\\ &\equiv (P \to R) \land (Q \to R) && \text{[conditional as an or, twice]} \end{aligned}

What is the missing second line?

Which law justifies the third line?

✅ Correct! De Morgan turns

\lnot(P \lor Q)

into

\lnot P \land \lnot Q

, and or distributes over and:

(\lnot P \land \lnot Q) \lor R \equiv (\lnot P \lor R) \land (\lnot Q \lor R)

❌ Check De Morgan. The negation of an or is an and:

\lnot(P \lor Q) \equiv \lnot P \land \lnot Q

. Writing

\lnot P \lor \lnot Q

is De Morgan for an and.

❌ Not quite. Only

\lnot(P \lor Q)

changes in this step; De Morgan replaces it by

\lnot P \land \lnot Q

, and the

\lor R

stays.

❌ Not quite. The step turns

(\text{something}) \lor R

, with an and inside, into an and of two ors, the pattern of

P \lor (Q \land R) \equiv (P \lor Q) \land (P \lor R)

Show solution

Line 2. De Morgan's law for an or: $\lnot(P \lor Q) \equiv \lnot P \land \lnot Q$ . Replacing it inside the line gives $(\lnot P \land \lnot Q) \lor R$ .

Line 3. The distributive law $P \lor (Q \land R) \equiv (P \lor Q) \land (P \lor R)$ , applied with $R$ as the outside statement (an or can be written in either order):

(\lnot P \land \lnot Q) \lor R \equiv R \lor (\lnot P \land \lnot Q) \equiv (R \lor \lnot P) \land (R \lor \lnot Q) \equiv (\lnot P \lor R) \land (\lnot Q \lor R).

Line 4. $\lnot P \lor R \equiv P \to R$ and $\lnot Q \lor R \equiv Q \to R$ , the first equivalence read right to left.

In words: $R$ follows from $P \lor Q$ exactly when $R$ follows from $P$ and $R$ follows from $Q$ .

Problem 4 · Choosing What to Prove

Given: $a$ and $b$ are integers, and we want to prove: “If $a \cdot b$ is even, then $a$ is even or $b$ is even.” Which statement is logically equivalent to it, so that proving it proves the original?

✅ Correct! With

P

a \cdot b

is even,

Q

a

is even,

R

b

is even, the statement is

P \to (Q \lor R)

, equivalent to

(P \land \lnot Q) \to R

. For an integer, “not even” is “odd”.

❌ The extra assumption is $\lnot Q$ , not $Q$ .

(P \land Q) \to R

is not equivalent: for

a = 2

b = 1

it is false (hypothesis true,

b

odd), while the original is true there since

a

is even.

❌ That is the converse.

(Q \lor R) \to P

happens to be a true statement about integers, but it is not logically equivalent to

P \to (Q \lor R)

, so proving it proves nothing about the original.

❌ The conclusion is $R$ , not $\lnot R$ . This is

(P \land \lnot Q) \to \lnot R

. For

a = 1

b = 2

its hypothesis is true (

2

is even and

1

is odd) and its conclusion “

b

is odd” is false, yet the original is true there since

b

is even.

❌ Not quite. Name the parts

P

Q

R

and use

P \to (Q \lor R) \equiv (P \land \lnot Q) \to R

Show solution

Let $P$ be “ $a \cdot b$ is even”, $Q$ be “ $a$ is even” and $R$ be “ $b$ is even”. The statement has the form $P \to (Q \lor R)$ , and

P \to (Q \lor R) \equiv (P \land \lnot Q) \to R.

$\lnot Q$ is “ $a$ is not even”, which for an integer means “ $a$ is odd”. So it is enough to prove: if $a \cdot b$ is even and $a$ is odd, then $b$ is even. The gain is one more assumption to work with, “ $a$ is odd”, exactly as “ $3$ is not a factor of $a$ ” was in the video.

The other options have the forms $(P \land Q) \to R$ , $(Q \lor R) \to P$ and $(P \land \lnot Q) \to \lnot R$ ; each differs from $P \to (Q \lor R)$ in some row of the truth table, so none is equivalent.

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The Negation of an If-Then Is an And: Negating Conditionals and Proving Equivalences by Chains

IKey Formulas

IIWhich Statement Is the Negation?

IIIIf a<ba < ba<b, Then a2<b2a^2 < b^2a2<b2?

IVProof by a Chain of Equivalences

VQuiz Questions

IIIIf $a < b$ , Then $a^2 < b^2$ ?