For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka

# Logic and Truth Tables - Solved Examples

I greet you this day,
First: read the notes. Second: view the videos. Third: solve the questions/solved examples. Fourth: check your solutions with my thoroughly-explained solutions. Fifth: check your solutions with the technologies as seen in the Videos. Sixth: check your solutions with the calculators as applicable.
The technologies used are Google Spreadsheet and Microsoft Excel (many thanks to the developers). The calculators used are the Wolfram Alpha widgets (many thanks to the developers).
Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome. You may contact me.
If you are my student, please do not contact me here. Contact me via the school's system. Thank you for visiting!!!

Samuel Dominic Chukwuemeka (Samdom For Peace) B.Eng., A.A.T, M.Ed., M.S

## Solved Examples - Truth Tables

Draw truth tables for the compound propositions.
Show all work.

(1.) $p \lor \neg q$

$p \lor \neg q \\[3ex] Main\:\: connective:\:\: \lor \\[3ex] First:\:\: p \\[3ex] Second:\:\: \neg q \\[3ex]$
$p$ $q$ $\neg q$ $p \lor \neg q$
$T$ $T$ $F$ $T$
$T$ $F$ $T$ $T$
$F$ $T$ $F$ $F$
$F$ $F$ $T$ $T$

(2.) $\neg p \lor \neg q$

$\neg p \lor \neg q \\[3ex] Main\:\: connective:\:\: \lor \\[3ex] First:\:\: \neg p \\[3ex] Second:\:\: \neg q \\[3ex]$
$p$ $q$ $\neg p$ $\neg q$ $\neg p \lor \neg q$
$T$ $T$ $F$ $F$ $F$
$T$ $F$ $F$ $T$ $T$
$F$ $T$ $T$ $F$ $T$
$F$ $F$ $T$ $T$ $T$

(3.) $\neg(p \land \neg q)$

$\neg(p \land \neg q) \\[3ex] Main\:\: connective:\:\: \neg \\[3ex] Other\:\: connective:\:\: \land \\[3ex] First:\:\: p \\[3ex] Second:\:\: \neg q \\[3ex]$
$p$ $q$ $\neg q$ $p \land \neg q$ $\neg(p \land \neg q)$
$T$ $T$ $F$ $F$ $T$
$T$ $F$ $T$ $T$ $F$
$F$ $T$ $F$ $F$ $T$
$F$ $F$ $T$ $F$ $T$

(4.) $p \rightarrow \neg p$

$p \rightarrow \neg p \\[3ex] Main\:\: connective:\:\: \rightarrow \\[3ex] First:\:\: p \\[3ex] Second:\:\: \neg p \\[3ex]$
$p$ $\neg p$ $p \rightarrow \neg p$
$T$ $F$ $F$
$F$ $T$ $T$

(5.) $p \leftrightarrow \neg p$

$p \leftrightarrow \neg p \\[3ex] Main\:\: connective:\:\: \leftrightarrow \\[3ex] First:\:\: p \\[3ex] Second:\:\: \neg p \\[3ex]$
$p$ $\neg p$ $p \leftrightarrow \neg p$
$T$ $F$ $F$
$F$ $T$ $F$

(6.) $p \rightarrow \neg q$

$p \rightarrow \neg q \\[3ex] Main\:\: connective:\:\: \rightarrow \\[3ex] First:\:\: p \\[3ex] Second:\:\: \neg q \\[3ex]$
$p$ $q$ $\neg q$ $p \rightarrow \neg q$
$T$ $T$ $F$ $F$
$T$ $F$ $T$ $T$
$F$ $T$ $F$ $T$
$F$ $F$ $T$ $T$

(7.) $p \leftrightarrow \neg q$

$p \leftrightarrow \neg q \\[3ex] Main\:\: connective:\:\: \leftrightarrow \\[3ex] First:\:\: p \\[3ex] Second:\:\: \neg q \\[3ex]$
$p$ $q$ $\neg q$ $p \leftrightarrow \neg q$
$T$ $T$ $F$ $F$
$T$ $F$ $T$ $T$
$F$ $T$ $F$ $T$
$F$ $F$ $T$ $F$

(8.) $(p \rightarrow q) \rightarrow (q \rightarrow p)$

$(p \rightarrow q) \rightarrow (q \rightarrow p) \\[3ex] Main\:\: connective:\:\: \rightarrow \\[3ex] First:\:\: p \rightarrow q \\[3ex] Second:\:\: q \rightarrow p \\[3ex]$
$p$ $q$ $p \rightarrow q$ $q \rightarrow p$ $(p \rightarrow q) \rightarrow (q \rightarrow p)$
$T$ $T$ $T$ $T$ $T$
$T$ $F$ $F$ $T$ $T$
$F$ $T$ $T$ $F$ $F$
$F$ $F$ $T$ $T$ $T$

(9.) $\neg q \lor (p \land r)$

$\neg q \lor (p \land r) \\[3ex] Main\:\: connective:\:\: \lor \\[3ex] First:\:\: \neg q \\[3ex] Second:\:\: p \\[3ex] Third:\:\: r \\[3ex]$
$p$ $q$ $r$ $\neg q$ $p \land r$ $\neg q \lor (p \land r)$
$T$ $T$ $T$ $F$ $T$ $T$
$T$ $T$ $F$ $F$ $F$ $F$
$T$ $F$ $T$ $T$ $T$ $T$
$T$ $F$ $F$ $T$ $F$ $T$
$F$ $T$ $T$ $F$ $F$ $F$
$F$ $T$ $F$ $F$ $F$ $F$
$F$ $F$ $T$ $T$ $F$ $T$
$F$ $F$ $F$ $T$ $F$ $T$

(10.) $(r \lor \neg p) \land \neg q$

$(r \lor \neg p) \land \neg q \\[3ex] Main\:\: connective:\:\: \land \\[3ex] First:\:\: r \\[3ex] Second:\:\: \neg p \\[3ex] Third:\:\: \neg q \\[3ex]$
$p$ $q$ $r$ $\neg p$ $r \lor \neg p$ $\neg q$ $(r \lor \neg p) \land \neg q$
$T$ $T$ $T$ $F$ $T$ $F$ $F$
$T$ $T$ $F$ $F$ $F$ $F$ $F$
$T$ $F$ $T$ $F$ $T$ $T$ $T$
$T$ $F$ $F$ $F$ $F$ $T$ $F$
$F$ $T$ $T$ $T$ $T$ $F$ $F$
$F$ $T$ $F$ $T$ $T$ $F$ $F$
$F$ $F$ $T$ $T$ $T$ $T$ $T$
$F$ $F$ $F$ $T$ $T$ $T$ $T$

(11.) $[(q \land \neg r) \land (\neg p \lor \neg q)] \lor (p \lor \neg r)$

$[(q \land \neg r) \land (\neg p \lor \neg q)] \lor (p \lor \neg r) \\[3ex] Main\:\: connective:\:\: \lor \\[3ex] First:\:\: q \land \neg r \\[3ex] Second:\:\: \neg p \lor \neg q \\[3ex] Third:\:\: p \lor \neg r \\[3ex]$
$p$ $q$ $r$ $\neg p$ $\neg q$ $\neg r$ $(q \land \neg r)$ $(\neg p \lor \neg q)$ $(p \lor \neg r)$ $[(q \land \neg r) \land (\neg p \lor \neg q)]$ $[(q \land \neg r) \land (\neg p \lor \neg q)] \lor (p \lor \neg r)$
$T$ $T$ $T$ $F$ $F$ $F$ $F$ $F$ $T$ $F$ $T$
$T$ $T$ $F$ $F$ $F$ $T$ $T$ $F$ $T$ $F$ $T$
$T$ $F$ $T$ $F$ $T$ $F$ $F$ $T$ $T$ $F$ $T$
$T$ $F$ $F$ $F$ $T$ $T$ $F$ $T$ $T$ $F$ $T$
$F$ $T$ $T$ $T$ $F$ $F$ $F$ $T$ $F$ $F$ $F$
$F$ $T$ $F$ $T$ $F$ $T$ $T$ $T$ $T$ $T$ $T$
$F$ $F$ $T$ $T$ $T$ $F$ $F$ $T$ $F$ $F$ $F$
$F$ $F$ $F$ $T$ $T$ $T$ $F$ $T$ $T$ $F$ $T$

## Solved Examples - Logical Equivalences

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

Solve each question.
Use least two methods for each question as applicable.
Show all work.

(1.) WASSCE Consider the statements:
$p$: Martin trains hard;
$q$: Martin wins the race.

If $p \rightarrow q$, state whether or not the following statements are valid:
(i) If Martin wins the race, then he has trained hard;
(ii) If Martin does not train hard then he will not win the race;
(iii) If Martin does not win the race then he has not trained hard.

We shall do this two ways.
You may use any of the methods.

First Method: Logical Equivalences
$p \rightarrow q \equiv \neg q \rightarrow \neg p$
$p \rightarrow q$: If Martin trains hard, then he will win the race.
$\neg q \rightarrow \neg p$: If Martin does not win the race, then he has not trained hard.

Second Method: Truth Tables Let us draw the truth table of all four statements.
Then, we shall look at the last column of the conditional statement (the main question) and compare it to the last column of the truth tables of the three other statements. The statement that has the same last column as the last column of the main conditional statement is the equivalent one to the main conditional statement.

Main Conditional Statement
If Martin trains hard, then he will win the race.
$p \rightarrow q$

$p$ $q$ $p \rightarrow q$
$T$ $T$ $T$
$T$ $F$ $F$
$F$ $T$ $T$
$F$ $F$ $T$

(i) If Martin wins the race, then he has trained hard.
$q \rightarrow p$

$p$ $q$ $q \rightarrow p$
$T$ $T$ $T$
$T$ $F$ $T$
$F$ $T$ $F$
$F$ $F$ $T$

(ii) If Martin does not train hard, then he will not win the race.
$\neg p \rightarrow \neg q$

$p$ $q$ $\neg p$ $\neg q$ $\neg p \rightarrow \neg q$
$T$ $T$ $F$ $F$ $T$
$T$ $F$ $F$ $T$ $T$
$F$ $T$ $T$ $F$ $F$
$F$ $F$ $T$ $T$ $T$

(iii) If Martin does not win the race, then he has not trained hard.
$\neg q \rightarrow \neg p$

$p$ $q$ $\neg p$ $\neg q$ $\neg q \rightarrow \neg p$
$T$ $T$ $F$ $F$ $T$
$T$ $F$ $F$ $T$ $F$
$F$ $T$ $T$ $F$ $T$
$F$ $F$ $T$ $T$ $T$

(2.) ACT Let $p$ and $q$ be statements.
Statement $S$: If $p$, then $q$
Converse of $S$: If $q$, then $p$
Inverse of $S$: If not $p$, then not $q$
Contrapositive of $S$: If not $q$, then not $p$

One of the following statements is the converse of the statement. "If the lights are on, then the store is open."
Which one is it?

F. If the store is open, then the lights are on.
G. If the lights are not on, then the store is not open.
H. If the store is not open, then the lights are not on.
J. The lights are not on.
K. The store is not open.

Statement: If the lights are on, then the store is open.
Converse of the Statement: If the store is open, then the lights are on.

(3.) ACT A news anchor made the true statement below.
If it is raining, then the parade is canceled.
Which one of the following statements is logically equivalent to the news anchor's statement?

F. If it is not raining, then the parade is not canceled.
G. The parade is canceled if and only if it is raining.
H. If it is not raining, then the parade is canceled.
J. If the parade is canceled, then it is raining.
K. If the parade is not canceled, then it is not raining.

Let:
$p$: It is raining
$q$: The parade is canceled.
Conditional Statement: If it is raining, then the parade is canceled.
Conditional Statement in Symbolic Logic: $p \rightarrow q$
The equivalent of the Conditional Statement is the Contrapositive Statement.
Contrapositive Statement in Symbolic Logic: $\neg q \rightarrow \neg p$
Contrapositive Statement: If the parade is not canceled, then it is not raining.

(13.) JEE The Boolean expression $\neg(p \lor q) \lor (\neg p \land q)$ is equivalent to

$(1)\:\: \neg p \\[3ex] (2)\:\: p \\[3ex] (3)\:\: q \\[3ex] (4)\:\: \neg q$

We shall do this two ways.
You may use any of the methods.
Because the JEE is a timed exam, choose whichever method you prefer that is faster for you.

First Method: Truth Tables

$\neg(p \lor q) \lor (\neg p \land q) \\[3ex] Main\:\: Connective:\:\: \lor \\[3ex] First:\:\: \neg(p \lor q) \\[3ex] Second:\:\: \neg p \land q \\[3ex]$
$p$ $q$ $p \lor q$ $\neg(p \lor q)$ $\neg p$ $\neg p \land q$ $\neg(p \lor q) \lor (\neg p \land q)$
$T$ $T$ $T$ $F$ $F$ $F$ $F$
$T$ $F$ $T$ $F$ $F$ $F$ $F$
$F$ $T$ $T$ $F$ $T$ $T$ $T$
$F$ $F$ $F$ $T$ $T$ $F$ $T$
First Same Second Same

$\neg(p \lor q) \lor (\neg p \land q) \equiv \neg p \\[3ex]$ Second Method: Laws of Logical Equivalence

$\neg(p \lor q) \lor (\neg p \land q) \\[3ex] \neg(p \lor q) \equiv \neg p \land \neg q ...De\:\: Morgan's\:\: Law \\[3ex] \implies (\neg p \land \neg q) \lor (\neg p \land q) \\[3ex] (\neg p \land \neg q \lor \neg p) \land (\neg p \land \neg q \lor q)...Distributive\:\: Law \\[3ex] \neg q \lor q \equiv T ...Negation\:\: Law \\[3ex] \implies (\neg p \land \neg q \lor \neg p) \land (\neg p \land T) \\[3ex] \neg p \land T \equiv \neg p...Identity\:\: Law \\[3ex] \implies (\neg p \land \neg q \lor \neg p) \land \neg p \\[3ex] (\neg p \land \neg q \lor \neg p) \equiv [\neg p \land (\neg q \lor \neg p)] \\[3ex] \neg q \lor \neg p \equiv \neg p \lor \neg q...Commutative\:\: Law \\[3ex] \implies [\neg p \land (\neg p \lor \neg q)] \land \neg p \\[3ex] [\neg p \land (\neg p \lor \neg q)] \equiv \neg p ...Absorption\:\: Law \\[3ex] \implies \neg p \land \neg p \equiv \neg p ...Idempotent\:\: Law$

(14.) JEE The following statement $(p \rightarrow q) \rightarrow [(\neg p \rightarrow q) \rightarrow q]$ is

$(1)\:\: Equivalent \:\: to\:\: \neg p \rightarrow q \\[3ex] (2)\:\: Equivalent\:\: to\:\: p \rightarrow \neg q \\[3ex] (3)\:\: A\:\: fallacy \\[3ex] (4)\:\: A\:\: tautology$

We shall do this two ways.
You may use any of the methods.
Because the JEE is a timed exam, choose whichever method you prefer that is faster for you.

First Method: Truth Tables

$(p \rightarrow q) \rightarrow [(\neg p \rightarrow q) \rightarrow q] \\[3ex] Main\:\: Connective:\:\: \rightarrow \\[3ex] First:\:\: p \rightarrow q \\[3ex] Second:\:\: (\neg p \rightarrow q) \rightarrow q \\[3ex]$
$p$ $q$ $p \rightarrow q$ $\neg p$ $\neg p \rightarrow q$ $(\neg p \rightarrow q) \rightarrow q$ $(p \rightarrow q) \rightarrow [(\neg p \rightarrow q) \rightarrow q]$
$T$ $T$ $T$ $F$ $T$ $T$ $T$
$T$ $F$ $F$ $F$ $T$ $F$ $T$
$F$ $T$ $T$ $T$ $T$ $T$ $T$
$F$ $F$ $T$ $T$ $F$ $T$ $T$
First Second Tautology

$(p \rightarrow q) \rightarrow [(\neg p \rightarrow q) \rightarrow q] \equiv T \\[3ex]$ Second Method: Laws of Logical Equivalence

$(p \rightarrow q) \rightarrow [(\neg p \rightarrow q) \rightarrow q] \\[3ex] \neg p \rightarrow q \equiv \neg(\neg p) \lor q \\[3ex] \implies \neg p \rightarrow q \equiv p \lor q \\[3ex] \implies [(\neg p \rightarrow q) \rightarrow q] \equiv [(p \lor q) \rightarrow q] \\[3ex] (p \lor q) \rightarrow q \equiv \neg(p \lor q) \lor q \\[3ex] \neg(p \lor q) \equiv \neg p \land \neg q ...De\:\: Morgan's \:\: Law \\[3ex] \implies \neg(p \lor q) \lor q \equiv (\neg p \land \neg q) \lor q \\[3ex] (\neg p \land \neg q) \lor q \equiv q \lor (\neg p \land \neg q) ...Commutative\:\: Law \\[3ex] q \lor (\neg p \land \neg q) \equiv (q \lor \neg p) \land (q \lor \neg q) ...Distributive\:\: Law \\[3ex] q \lor \neg q \equiv T ...Negation\:\: Law \\[3ex] \implies (q \lor \neg p) \land (q \lor \neg q) \equiv (q \lor \neg p) \land T \\[3ex] (q \lor \neg p) \land T \equiv q \lor \neg p ...Identity\:\: Law \\[3ex] p \rightarrow q \equiv \neg p \lor q \\[3ex] (p \rightarrow q) \rightarrow [(\neg p \rightarrow q) \rightarrow q] \\[3ex] \implies (\neg p \lor q) \rightarrow (q \lor \neg p) \\[3ex] \implies \neg(\neg p \lor q) \lor (q \lor \neg p) \\[3ex] \neg(\neg p \lor q) \equiv p \land \neg q ...De\:\: Morgan's\:\: Law \\[3ex] \implies (p \land \neg q) \lor (q \lor \neg p) \\[3ex] p \land \neg q \lor q \lor \neg p \\[3ex] \neg q \lor q \equiv T ...Negation\:\: Law \\[3ex] \implies p \land T \lor \neg p \\[3ex] p \land T \equiv p ...Identity\:\: Law \\[3ex] \implies p \lor \neg p \equiv T ... Negation\:\: Law$

(20.) JEE The compound statement $(\neg C \land A \land B) \lor (\neg C \land \neg A \land B) \lor (C \land B)$ is equivalent to

$(1)\:\: B \\[3ex] (2)\:\: A \\[3ex] (3)\:\: \neg A \\[3ex] (4)\:\: C$

We shall do this two ways.
You may use any of the methods.
Because the JEE is a timed exam, choose whichever method you prefer that is faster for you.

First Method: Truth Tables

$Let: \\[3ex] A = p \\[3ex] B = q \\[3ex] C = r \\[3ex] (\neg C \land A \land B) \lor (\neg C \land \neg A \land B) \lor (C \land B) \\[3ex] (\neg r \land p \land q) \lor (\neg r \land \neg p \land q) \lor (r \land q) \\[3ex] Main\:\: Connectives:\:\: \lor, \lor \\[3ex] First:\:\: \neg r \land p \land q \\[3ex] Second:\:\: \neg r \land \neg p \land q \\[3ex] Third:\:\: r \land q \\[3ex]$
$p$ $q$ $r$ $\neg r$ $\neg r \land p$ $\neg r \land p \land q$ $\neg p$ $\neg r \land \neg p$ $\neg r \land \neg p \land q$ $r \land q$ $(\neg r \land p \land q) \lor (\neg r \land \neg p \land q) \lor (r \land q)$
$T$ $T$ $T$ $F$ $F$ $F$ $F$ $F$ $F$ $T$ $T$
$T$ $T$ $F$ $T$ $T$ $T$ $F$ $F$ $F$ $F$ $T$
$T$ $F$ $T$ $F$ $F$ $F$ $F$ $F$ $F$ $F$ $F$
$T$ $F$ $F$ $T$ $T$ $F$ $F$ $F$ $F$ $F$ $F$
$F$ $T$ $T$ $F$ $F$ $F$ $T$ $F$ $F$ $T$ $T$
$F$ $T$ $F$ $T$ $F$ $F$ $T$ $T$ $T$ $F$ $T$
$F$ $F$ $T$ $F$ $F$ $F$ $T$ $F$ $F$ $F$ $F$
$F$ $F$ $F$ $T$ $F$ $F$ $T$ $T$ $F$ $F$ $F$
Same First Second Third Same

$(\neg r \land p \land q) \lor (\neg r \land \neg p \land q) \lor (r \land q) \equiv q \\[3ex] (\neg C \land A \land B) \lor (\neg C \land \neg A \land B) \lor (C \land B) \equiv B$

## Solved Examples - Arguments

Determine the validity or invalidity of the arguments.
Use at least two methods for each argument.

(8.) It is wrong to smoke in public if secondary cigarette smoke is a health risk. If secondary cigarette smoke were not a health risk, the American Lung Association would not say that it is. The American Lung Association says that secondary cigarette smoke is a health risk. Therefore, it is wrong to smoke in public.
Use $p$ to represent "Secondary cigarette smoke is a health risk," use $q$ to represent "It is wrong to smoke in public," and use $r$ to represent "The American Lung Association says that secondary cigarette smoke is a health risk."

$p$: Secondary cigarette smoke is a health risk
$q$: It is wrong to smoke in public
$r$: The American Lung Association says that secondary cigarette smoke is a health risk

$Premise\: 1: p \rightarrow q \\ Premise\: 2: \neg p \rightarrow \neg r \\ Premise\: 3: r \\ \rule{2.5in}{0.5pt} \\ Conclusion \therefore q \\[3ex]$ First Method: Valid Forms and Invalid Forms of Arguments
Does Not Apply

Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $r$ $p \rightarrow q$ $\neg p$ $\neg r$ $\neg p \rightarrow \neg r$
$T$ $T$✓ $T$✓ $T$✓ $F$ $F$ $T$✓
$T$ $T$ $F$ $T$ $F$ $T$ $T$
$T$ $F$ $T$ $F$ $F$ $F$ $T$
$T$ $F$ $F$ $F$ $F$ $T$ $T$
$F$ $T$ $T$ $T$ $T$ $F$ $F$
$F$ $T$ $F$ $T$ $T$ $T$ $T$
$F$ $F$ $T$ $T$ $T$ $F$ $F$
$F$ $F$ $F$ $T$ $T$ $T$ $T$
Conclusion Premise $3$ Premise $1$ Premise $2$

In the case where all premises are true (only one case), the conclusion is also true.
Argument is valid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2) \land (premise\: 3)] \rightarrow conclusion$
$[(p \rightarrow q) \land (\neg p \rightarrow \neg r) \land r] \rightarrow q$
$p$ $q$ $r$ $p \rightarrow q$ $\neg p$ $\neg r$ $\neg p \rightarrow \neg r$ $(p \rightarrow q) \land (\neg p \rightarrow \neg r)$ $(p \rightarrow q) \land (\neg p \rightarrow \neg r) \land r$ $[(p \rightarrow q) \land (\neg p \rightarrow \neg r) \land r] \rightarrow q$
$T$ $T$ $T$ $T$ $F$ $F$ $T$ $T$ $T$ $T$
$T$ $T$ $F$ $T$ $F$ $T$ $T$ $T$ $F$ $T$
$T$ $F$ $T$ $F$ $F$ $F$ $T$ $F$ $F$ $T$
$T$ $F$ $F$ $F$ $F$ $T$ $T$ $F$ $F$ $T$
$F$ $T$ $T$ $T$ $T$ $F$ $F$ $F$ $F$ $T$
$F$ $T$ $F$ $T$ $T$ $T$ $T$ $T$ $F$ $T$
$F$ $F$ $T$ $T$ $T$ $F$ $F$ $F$ $F$ $T$
$F$ $F$ $F$ $T$ $T$ $T$ $T$ $T$ $F$ $T$
Tautology

The formula is a tautology.
Argument is valid.

## References

Chukwuemeka, S.D (2016, April 30). Samuel Chukwuemeka Tutorials - Math, Science, and Technology. Retrieved from https://www.samuelchukwuemeka.com
Blitzer, R. (2015). Thinking Mathematically ($6^{th}$ ed.). Boston: Pearson
Rosen, K. H. (2013). Discrete mathematics and its applications ($8^{th}$ ed.). New York: McGraw-Hill.
Tan, S. (2015). Finite Mathematics for the Managerial, Life, and Social Sciences (Revised/Expanded ed.). Boston: Cengage Learning.