Distributive Laws for Logical Connectives

De Morgan's Laws show how negation changes logical connectives. The next major family of equivalences explains how conjunction and disjunction distribute through each other. Distributive laws for logical connectives are central in simplification, switching networks, normal forms, and proofs of equivalence. In this lesson, distributive laws for logical connectives will be used to rewrite compound statements without changing their truth values.

Let

p

q

, and

r

be primitive statements. Then

\begin{equation*} p\land(q\lor r)\Leftrightarrow(p\land q)\lor(p\land r) \end{equation*}

and

\begin{equation*} p\lor(q\land r)\Leftrightarrow(p\lor q)\land(p\lor r). \end{equation*}

These are called the \textbf{Distributive Laws} for logical connectives.

Let

p

q

, and

r

be primitive statements. Then

\begin{equation*} p\land(q\lor r)\Leftrightarrow(p\land q)\lor(p\land r) \end{equation*}

and

\begin{equation*} p\lor(q\land r)\Leftrightarrow(p\lor q)\land(p\lor r). \end{equation*}

These are called the \textbf{Distributive Laws} for logical connectives.

Given that

p

q

, and

r

are primitive statements. To prove the distributive laws. The statement

p\land(q\lor r)

is true exactly when

p

is true and at least one of

q

and

r

is true. The statement

(p\land q)\lor(p\land r)

is true exactly when

p

and

q

are both true, or

p

and

r

are both true. These conditions are the same. Therefore,

\begin{equation*} p\land(q\lor r)\Leftrightarrow(p\land q)\lor(p\land r). \end{equation*}

The statement

p\lor(q\land r)

is true when

p

is true, or when both

q

and

r

are true. The statement

(p\lor q)\land(p\lor r)

is true in the same cases. Therefore,

\begin{equation*} p\lor(q\land r)\Leftrightarrow(p\lor q)\land(p\lor r). \end{equation*}

Hence, both distributive laws hold.

\square

For primitive statements

p

q

, and

r

\begin{equation*} p\land(q\lor r)\Leftrightarrow(p\land q)\lor(p\land r). \end{equation*}

This law distributes

\land

over

\lor

For primitive statements

p

q

, and

r

\begin{equation*} p\lor(q\land r)\Leftrightarrow(p\lor q)\land(p\lor r). \end{equation*}

This law distributes

\lor

over

\land

Use the comparator to test both logical distributive laws under the same truth-value assignment. Choose which law to study, then set values for

p

q

, and

r

. The two sides should always produce the same output when the law is valid. This is especially helpful for the second distributive law, because it has no direct ordinary arithmetic analogue.

For real numbers

a

b

, and

c

\begin{equation*} a(b+c)=ab+ac. \end{equation*}

This resembles

p\land(q\lor r)\Leftrightarrow(p\land q)\lor(p\land r)

. Logic also has

p\lor(q\land r)\Leftrightarrow(p\lor q)\land(p\lor r)

, which has no direct counterpart in ordinary arithmetic addition and multiplication.

Rewrite

p\land(q\lor r)

and

p\lor(q\land r)

using distributive laws.

Given that

p

q

, and

r

are primitive statements. Using the first distributive law,

\begin{equation*} p\land(q\lor r)\Leftrightarrow(p\land q)\lor(p\land r). \end{equation*}

Using the second distributive law,

\begin{equation*} p\lor(q\land r)\Leftrightarrow(p\lor q)\land(p\lor r). \end{equation*}

Hence, the required rewritten forms are

(p\land q)\lor(p\land r)

and

(p\lor q)\land(p\lor r)

Use laws of logic to verify

(p\lor q)\land(p\lor\neg q)\Leftrightarrow p

Given that

p

and

q

are primitive statements. To prove the equivalence, use the distributive law in reverse:

\begin{align*} (p\lor q)\land(p\lor\neg q)&\Leftrightarrow p\lor(q\land\neg q)\\ &\Leftrightarrow p\lor F_0\\ &\Leftrightarrow p. \end{align*}

Hence,

(p\lor q)\land(p\lor\neg q)\Leftrightarrow p

Simplify

\neg q\lor(\neg p\land q)

Given that

p

and

q

are primitive statements. Using distribution,

\begin{align*} \neg q\lor(\neg p\land q)&\Leftrightarrow(\neg q\lor\neg p)\land(\neg q\lor q)\\ &\Leftrightarrow(\neg q\lor\neg p)\land T_0\\ &\Leftrightarrow\neg q\lor\neg p. \end{align*}

Hence, the simplified statement is

\neg p\lor\neg q

Use distributive laws to rewrite: (1)

p\land(q\lor r)

. (2)

p\lor(q\land r)

. (3)

(p\lor q)\land(p\lor\neg q)

(1)

(p\land q)\lor(p\land r)

. (2)

(p\lor q)\land(p\lor r)

. (3)

p\lor(q\land\neg q)

, which simplifies to

p

Yes. In logic, both

\land

distributes over

\lor

and

\lor

distributes over

\land

Yes. Logical equivalence works in both directions.

They transform compound statements into forms that can often be simplified.

BMLabs Mathematics

BMLabs Mathematics

Distributive Laws for Logical Connectives

Distributive Laws for Logical Connectives

DISTRIBUTIVE LAWS

CONJUNCTION DISTRIBUTED OVER DISJUNCTION

DISJUNCTION DISTRIBUTED OVER CONJUNCTION

TESTING BOTH DISTRIBUTIVE LAWS

DISTRIBUTIVE LAW COMPARATOR

ARITHMETIC COMPARISON

FREQUENTLY ASKED QUESTIONS

USE DISTRIBUTION CAREFULLY

Introduction To Discrete Mathematics

Paths and Circuits