audio read-through Logical Equivalences and Practice with Truth Tables

Before studying this section, you may want to review:

A logical equivalence is a statement that two mathematical sentence forms are completely interchangeable: if one is true, so is the other; if one is false, so is the other.

For example, we could express that an implication is equivalent to its contrapositive in either of the following ways:

$A \Rightarrow B\,$
is (logically) equivalent to
$\,(\text{not }B) \Rightarrow (\text{not }A)$

or

‘$(A \Rightarrow B) \ \ \Longleftrightarrow \ \ ((\text{not }B) \Rightarrow (\text{not } A))$’
is a logical equivalence

In this section, you will construct truth tables involving implications, equivalences, negations, and the mathematical words ‘and’ and ‘or’. In the process, you'll be introduced to many useful logical equivalences, and will start to develop intuition for investigating logical equivalences.

Negating ‘and’ and ‘or’ Sentences:  De Morgan's Laws

How can a sentence ‘$A \text{ and } B\,$’ be false? The only time an ‘and’ sentence is true is when both subsentences are true. So, an ‘and’ sentence is false when at least one of the subsentences is false.

Precisely, the truth table below shows that:

$\text{not}(A \text{ and } B)$
is equivalent to
$ (\text{not } A) \text{ or } (\text{not } B)$

Here's the intuition you should have when looking at this sentence. Start with (1) and work your way to (6):

(2) ... is false ... (1) an ‘and’ sentence ...
$\overbrace{\text{not}}$ $\overbrace{(A\text{ and }B)}$
(3) ... when ...
$\overbrace{\text{is equivalent to}}$
(4) ... $\,A\,$ is false ... (5) ... or ... (6) ... $\,B\,$ is false
$\overbrace{(\text{not }A)}$ $\overbrace{\text{ or }}$ $\overbrace{(\text{not }B)}$

In particular, note that you say ‘is false’ when you see the word ‘not’.

$A$ $B$ $A \text{ and } B$ $\text{not}(A \text{ and } B)$
T T T F
T F F T
F T F T
F F F T
$\text{not } A$ $\text{not } B$ $(\text{not } A) \text{ or } (\text{not } B)$
F F F
F T T
T F T
T T T
$(\text{not}(A \text{ and } B))\iff ((\text{not } A) \text{ or } (\text{not } B))$
T
T
T
T

Here's the critical observation: the columns for ‘$\,\text{not}(A \text{ and } B)\,$’ and ‘$\,(\text{not } A) \text{ or } (\text{not } B)\,$’ are identical! Some people like to think of this as a sort of distributive law, except that ‘and’ changes to ‘or’ in the distribution process.

For simplicity, in the subsequent truth tables we will not include the last columnthe one that shows that the ‘is equivalent to’ statement is always true. This last column takes up a lot of space, and makes things look more complicated than they really are. It is equally convincing to compare the two relevant columns, to show that they are identical.

Importance of Associative Laws

Think about this: the reason that we can write things like $\,1 + 2 + 3\,$ without ambiguity, is because both $\,(1 + 2) + 3\,$ and $\,1 + (2 + 3)\,$ give the same result. If they produced different results, then we'd have to write the parentheses every time we worked with these expressions.

You'll prove in the exercises that:

$(A \text{ or } B) \text{ or } C\,$
is equivalent to
$\,A \text{ or } (B \text{ or } C)$

Below, it is shown that:

$(A \text{ and } B) \text{ and } C\,$
is equivalent to
$\,A \text{ and } (B \text{ and } C)$

Thus (hooray!) we can write

‘$\,A \text{ or } B \text{ or } C\,$’
and
‘$\,A \text{ and } B \text{ and } C\,$’

without ambiguity.

Note that the truth table below involves three subsentences ($\,A\,,$ $\,B\,,$ and $\,C\,$), so there are $\,2^3 = 8\,$ rows needed to cover all possible truth values. Always list the eight rows in exactly the order that is shown here.

$A$ $B$ $C$ $A \text{ and } B$ $(A \text{ and } B) \text{ and } C$
T T T T T
T T F T F
T F T F F
T F F F F
F T T F F
F T F F F
F F T F F
F F F F F
$B \text{ and } C$ $A \text{ and } (B \text{ and } C)$
T T
F F
F F
F F
T F
F F
F F
F F

In the exercises, you will construct truth tables to prove all the following logical equivalences.

Logical Equivalences

Name/Description
$1^{\text{st}}\,$ sentence $\iff$ $2^{\text{nd}}\,$ sentence
Intuition
de Morgan's Law: Negating an ‘and’ sentence
$\text{not}(A \text{ and } B)$ $\iff$ $(\text{not }A) \text{ or } (\text{not }B)$
An ‘and’ sentence is false when $\,A\,$ is false or $\,B\,$ is false
de Morgan's Law: Negating an ‘or’ sentence
$\text{not}(A \text{ or }B)$ $\iff$ $(\text{not }A) \text{ and }(\text{not }B)$
An ‘or’ sentence is false when $\,A\,$ is false and $\,B\,$ is false
Associativity of ‘and’
$(A \text{ and }B) \text{ and }C$ $\iff$ $A \text{ and }(B \text{ and }C)$
The grouping doesn't matter in an ‘and’ sentence
Associativity of ‘or’
$(A \text{ or }B) \text{ or }C$ $\iff$ $A \text{ or }(B \text{ or }C)$
The grouping doesn't matter in an ‘or’ sentence
Commutativity of ‘and’
$A \text{ and } B$ $\iff$ $B \text{ and } A$
The order doesn't matter in an ‘and’ sentence
Commutativity of ‘or’
$A \text{ or } B$ $\iff$ $B \text{ or } A$
The order doesn't matter in an ‘or’ sentence
Law of Double Negation
$\text{not}(\text{not }A)$ $\iff$ $A$
Negating twice in succession gets you back to where you started
Distributive Law
$A \text{ or }(B \text{ and }C\,)$ $\iff$ $(A \text{ or }B) \text{ and }(A \text{ or }C\,)$
Similar to: $\,a(b + c) = ab + ac$
Distributive Law
$A \text{ and }(B \text{ or }C\,)$ $\iff$ $(A \text{ and }B) \text{ or }(A \text{ and }C\,)$
Similar to: $\,a(b + c) = ab + ac$
Alternative Form of an Implication
$A \Rightarrow B$ $\iff$ $(\text{not }A) \text{ or }B$
An implication is true when the hypothesis is false or the conclusion is true
Contrapositive of an Implication
$A \Rightarrow B$ $\iff$ $(\text{not }B) \Rightarrow (\text{not }A)$
An implication is equivalent to its contrapositive
Negating an Implication
$\text{not}(A \Rightarrow B)$ $\iff$ $A \text{ and }(\text{not }B)$
An implication is false when the hypothesis is true and the conclusion is false
Biconditional Statement
$A \Longleftrightarrow B$ $\iff$ $(A \Rightarrow B) \text{ and } (B \Rightarrow A)$
This is justification for the ‘double arrow’ that is used for equivalence

Tautologies

A tautology is a mathematical sentence form that is always true. For example, ‘$\,(A\text{ and }B)\Rightarrow A\,$’ is a tautology, as the truth table below confirms:

$A$ $B$ $A\text{ and }B$ $(A\text{ and }B)\Rightarrow A$
T T T T
T F F T
F T F T
F F F T

Here's the intuition for the tautology ‘$(A\text{ and } B)\Rightarrow A\,$’: the only way an ‘and’ sentence is true is when both subsentences are true. Thus, if ‘$A\text{ and }B\,$’ is true, then (in particular) $\,A\,$ must be true. Check that ‘$(A\text{ and }B)\Rightarrow B\,$’ is also a tautology.

Note that a logical equivalence is a particular type of tautology—one that tells us that two different sentence forms always have the same truth values, regardless of the truth values of the components.

Difference in Usage Between the Words ‘Tautology’ and ‘Identity’

Although the words ‘identity’ and ‘tautology’ are both used in mathematics to refer to sentences that are always true, there is a slight difference in usage.

The word ‘tautology’ tends to be used in logic, where the ‘pieces’ of the (always true) sentence have the choice of being either true or false. For example, ‘$(A\text{ and }B)\Rightarrow A\,$’ is a tautology: the sentence is always true; the ‘pieces’ $\,A\,$ and $\,B\,$ can be true or false. The sentence ‘$A\text{ or } (\text{not }A)\,$’ is another tautology: the sentence is always true; the ‘piece’ $\,A\,$ can be true or false.

The word ‘identity’, on the other hand, tends to be used for mathematical sentences outside of the realm of logic that are always true. Here are some examples from algebra and trigonometry (and don't worry if something is unfamiliar here):

We finish with two important tautologies that allow us to ‘chain’ results together:

LAW OF SYLLOGISM
The following compound mathematical sentence is true, for all possible truth values of $\,P\,,$  $\,Q\,,$ and $\,R\,$:

If $\,((P\Rightarrow Q)\text{ and }(Q\Rightarrow R))\,$ then $\,(P\Rightarrow R)\,$

Here's the intuition: Whenever $\,P\,$ is true, so is $\,Q\,$; and, whenever $\,Q\,$ is true, so is $\,R\,$; so, whenever $\,P\,$ is true, so is $\,R\,.$

LAW OF DETACHMENT
The following compound mathematical sentence is true, for all possible truth values of $\,P\,$ and $\,Q\,$:

$(P\text{ and }(P\Rightarrow Q))\ \Rightarrow\ Q$

Here's the intuition: If $\,P\,$ is true; and, whenever $\,P\,$ is true, so is $\,Q\,$; then, $\,Q\,$ must be true.

Concept Practice