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:
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.
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:
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 ...  (3) ... when ...  (4) ... $\,A\,$ is false ...  (5) ... or ...  (6) ... $\,B\,$ is false 
[beautiful math coming... please be patient] $\overbrace{\text{not}}$  $\overbrace{(A\text{ and }B)}$  $\overbrace{\text{is equivalent to}}$  $\overbrace{(\text{not }A)}$  $\overbrace{\text{ or }}$  $\overbrace{(\text{not }B)}$ 
$A$  $B$  $A \text{ and } B$  [beautiful math coming... please be patient] $\text{not}(A \text{ and } B)$  $\text{not } A$  $\text{not } B$  $(\text{not } A) \text{ or } (\text{not } B)$  $(\text{not}(A \text{ and } B))\iff ((\text{not } A) \text{ or } (\text{not } B))$ 
T  T  T  F  F  F  F  T 
T  F  F  T  F  T  T  T 
F  T  F  T  T  F  T  T 
F  F  F  T  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 column—the 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.
Think about this:
the reason that we can write things like
[beautiful math coming... please be patient]
$\,1 + 2 + 3\,$ 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:
Note that the truth table below involves three subsentences ($\,A\,$, $\,B\,$, and $\,C\ $),
so there are
[beautiful math coming... please be patient]
$\,2^3 = 8\,$ rows needed to cover all possible truth values.
Always list the eight rows in exactly the order that is shown here.
$A$  $B$  $C$  $A \text{ and } B$  [beautiful math coming... please be patient] $(A \text{ and } B) \text{ and } C$  $B \text{ and } C$  $A \text{ and } (B \text{ and } C)$ 
T  T  T  T  T  T  T 
T  T  F  T  F  F  F 
T  F  T  F  F  F  F 
T  F  F  F  F  F  F 
F  T  T  F  F  T  F 
F  T  F  F  F  F  F 
F  F  T  F  F  F  F 
F  F  F  F  F  F  F 
In the exercises, you will construct truth tables to prove all the following logical equivalences.
NAME/DESCRIPTION  [beautiful math coming... please be patient] $1^{\text{st}}\,$ sentence  ... is equivalent to ...  $2^{\text{nd}}\,$ sentence  INTUITION 
de Morgan's Law: negating an ‘and’ sentence 
[beautiful math coming... please be patient] $\text{not}(A \text{ and } B)$  ... is equivalent to ...  $(\text{not }A) \text{ or } (\text{not }B)$  an ‘and’ sentence is false when $\,A\,$ is false or $\,B\,$ is false 
de Morgan's Law: negating an ‘or’ sentence 
[beautiful math coming... please be patient] $\text{not}(A \text{ or }B)$  ... is equivalent to ...  $(\text{not }A) \text{ and }(\text{not }B)$  an ‘or’ sentence is false when $\,A\,$ is false and $\,B\,$ is false 
associativity of ‘and’  [beautiful math coming... please be patient] $(A \text{ and }B) \text{ and }C$  ... is equivalent to ...  $A \text{ and }(B \text{ and }C)$  the grouping doesn't matter in an ‘and’ sentence 
associativity of ‘or’  [beautiful math coming... please be patient] $(A \text{ or }B) \text{ or }C$  ... is equivalent to ...  $A \text{ or }(B \text{ or }C)$  the grouping doesn't matter in an ‘or’ sentence 
commutativity of ‘and’  [beautiful math coming... please be patient] $A \text{ and } B$  ... is equivalent to ...  $B \text{ and } A$  the order doesn't matter in an ‘and’ sentence 
commutativity of ‘or’  $A \text{ or } B$  ... is equivalent to ...  $B \text{ or } A$  the order doesn't matter in an ‘or’ sentence 
law of double negation  $\text{not}(\text{not }A)$  ... is equivalent to ...  $A$  negating twice in succession gets you back to where you started 
distributive law  [beautiful math coming... please be patient] $A \text{ or }(B \text{ and }C\,)$  ... is equivalent to ...  $(A \text{ or }B) \text{ and }(A \text{ or }C\,)$  similar to: $\,a(b + c) = ab + ac$ 
distributive law  [beautiful math coming... please be patient] $A \text{ and }(B \text{ or }C\,)$  ... is equivalent to ...  $(A \text{ and }B) \text{ or }(A \text{ and }C\,)$  similar to: $\,a(b + c) = ab + ac$ 
alternate form of an implication 
$A \Rightarrow B$  ... is equivalent to ...  $(\text{not }A) \text{ or }B$  an implication is true when the hypothesis is false or the conclusion is true 
contrapositive of an implication 
[beautiful math coming... please be patient] $A \Rightarrow B$  ... is equivalent to ...  $(\text{not }B) \Rightarrow (\text{not }A)$  an implication is equivalent to its contrapositive 
negating an implication  [beautiful math coming... please be patient] $\text{not}(A \Rightarrow B)$  ... is equivalent to ...  $A \text{ and }(\text{not }B)$  an implication is false when the hypothesis is true and the conclusion is false 
biconditional statement  [beautiful math coming... please be patient] $A \Longleftrightarrow B$  ... is equivalent to ...  $(A \Rightarrow B) \text{ and } (B \Rightarrow A)$  this is justification for the ‘double arrow’ that is used for equivalence 
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.
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:
Here's the intuition:
Whenever [beautiful math coming... please be patient]
$\,P\,$ is true, so is $\,Q\,$; and,
whenever $\,Q\,$ is true, so is $\,R\,$; so,
whenever $\,P\,$ is true, so is $\,R\,$.
Here's the intuition:
If $\,P\,$ is true; and,
whenever $\,P\,$ is true, so is $\,Q\,$; then,
$\,Q\,$ must be true.
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
