Solved Examples on Logical Equivalences: Propositional Logic

Samuel Dominic Chukwuemeka (SamDom For Peace) 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 a wrong answer.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

Solve all questions.
Show all work.
Use at least two methods including Truth Tables and the Laws of Logical Equivalences for each question as applicable.
For any law of logical equivalence that you use, please indicate the law or the logical equivalence number.

(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.


$p$: Martin trains hard;
$q$: Martin wins the race.

$p \rightarrow q$ is the Condtional statement: If Martin trains hard, then he will win the race.

We shall do this question in two ways.
You may choose any method you prefer.

First Method: Logical Equivalences
$p \rightarrow q \equiv \neg q \rightarrow \neg p$
Conditional Statement: $p \rightarrow q$: If Martin trains hard, then he will win the race.
Contrapositive Statement: $\neg q \rightarrow \neg p$: If Martin does not win the race, then he has not trained hard.
The Conditional statement is logically equivalent to the Contrapositive statement.
The valid statement is Option (iii)

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.

Conditional Statement (Main Question)
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.
$\neg q \rightarrow \neg p$

$p$ $q$ $\neg p$ $\neg q$ $\neg q \rightarrow \neg p$
$T$ $T$ $F$ $F$ $T$
$T$ $F$ $F$ $T$ $F$
$F$ $T$ $T$ $F$ $T$
$F$ $F$ $T$ $T$ $T$
As you can see, the last column of the Conditional statement is the same as the last column of Option (iii)
In other words, the Conditional statement and Contrapositive statement has the same last column in their truth tables.
Option (iii) is the valid statement.
(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.) WASSCE Consider the following statements:
$p$: Landi has cholera,
$q$: Landi is in the hospital.

If $p \rightarrow q$, state whether or not the following statements are valid:
(i) If Landi is in the hospital, then he has cholera.
(ii) If Landi is not in the hospital, then he does not have cholera.
(iii) If Landi does not have cholera, then he is not in the hospital.


$p$: Landi has cholera,
$q$: Landi is in the hospital.

$p \rightarrow q$ is the Condtional statement: If Landi has cholera, then he is in the hospital.

We shall do this question in two ways.
You may choose any method you prefer.

First Method: Logical Equivalences
$p \rightarrow q \equiv \neg q \rightarrow \neg p$
Conditional Statement: $p \rightarrow q$: If Landi has cholera, then he is in the hospital.
Contrapositive Statement: $\neg q \rightarrow \neg p$: If Landi is not in the hospital, then he does not have cholera.
The Conditional statement is logically equivalent to the Contrapositive statement.
The valid statement is Option (ii)

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.

Conditional Statement (Main Question)
If Landi is in the hospital, then he has cholera.
$p \rightarrow q$

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

(i) If Landi is in the hospital, then he has cholera.
$q \rightarrow p$

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

(ii) If Landi is not in the hospital, then he does not have cholera.
$\neg q \rightarrow \neg p$

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

(iii) If Landi does not have cholera, then he is not in the hospital.
$\neg p \rightarrow \neg q$

$p$ $q$ $\neg p$ $\neg q$ $\neg p \rightarrow \neg q$
$T$ $T$ $F$ $F$ $T$
$T$ $F$ $F$ $T$ $T$
$F$ $T$ $T$ $F$ $F$
$F$ $F$ $T$ $T$ $T$
As you can see, the last column of the Conditional statement is the same as the last column of Option (ii)
In other words, the Conditional statement and Contrapositive statement has the same last column in their truth tables.
Option (ii) is the valid statement.
(4.) ACT Given the true statement "If I live in Chicago, then I live in Illinois", which of the following statements must be true?

A. I live in Illinois.
B. I live in Chicago.
C. If I live in Illinois, then I live in Chicago.
D. If I don't live in Chicago, then I don't live in Illinois.
E. If I don't live in Illinois, then I don't live in Chicago.


If I live in Chicago, then I live in Illinois. ...conditional statement.

Let us analyze the options.
A. I live in Illinois.
B. I live in Chicago.
The main statement is a conditional statement.
Notice the "If..." I live in Illinois...
This does not mean that you live in Illinois.
Neither does it imply that you live in Chicago.
Therefore, options A. and B. are incorrect.

C. If I live in Illinois, then I live in Chicago.
This is not necessary true.
Chicago is not the only city in Illinois.
Someone lives in Springfield, Illinois.
Option C. is incorrect.

D. If I don't live in Chicago, then I don't live in Illinois.
This is also incorrect because someone lives in Springfield, Illinois.

Let:
$p$: I live in Chicago
$q$: I live in Illinois.
Conditional Statement: If I live in Chicago, then I live in Illinois.
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 I do not live in Illinois, then I do not live in Chicago.
This means: E. If I don't live in Illinois, then I don't live in Chicago.
Option E. is the correct option.
(5.) ACT Cassandra has a collection of animal figurines of various solid colors.
She makes the following true statement about the collection: "If a figurine is blue, then the figurine is a cat."
Which of the following statements about the collection is logically equivalent to Cassandra's statement?

F. "A figurine is a cat if and only if it is blue."
G. "If a figurine is a cat, then the figurine is blue."
H. "If a figurine is a cat, then the figurine is NOT blue."
J. "If a figurine is NOT blue, then the figurine is NOT a cat."
K. "If a figurine is NOT a cat, then the figurine is NOT blue."


Let:
$p$: A figurine is blue.
$q$: The figurine is a cat.
Conditional Statement: If a figurine is blue, then the figurine is a cat.
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 a figurine is not a cat, then the figurine is not blue.
(6.) 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.
(7.) Simplify $(p \land q) \lor \neg(\neg p \lor q)$


We can solve this question in at least two ways.
Use any method you prefer.

First Method: Laws of Logical Equivalence

$ (p \land q) \lor \neg(\neg p \lor q) \\[3ex] Main\:\:connective:\:\: \lor \\[3ex] *** \\[3ex] \underline{RHS} \\[3ex] \neg(\neg p \lor q) \\[3ex] \equiv \neg\neg p \land \neg q ...De\;Morgan's\;\;Law \\[3ex] ... \\[3ex] \neg\neg p \equiv p ...Double\;\;Negation\;\;Law \\[3ex] ... \\[3ex] \equiv p \land \neg q \\[3ex] *** \\[3ex] \underline{Statement} \\[3ex] \equiv (p \land q) \lor (p \land \neg q) \\[3ex] \equiv [(p \land q) \lor p] \land [(p \land q) \lor \neg q]...Distributive\:\:Law \\[3ex] @@@ \\[3ex] \underline{LHS} \\[3ex] [(p \land q) \lor p] \\[3ex] \equiv [p \lor (p \land q)]...Commutative\:\:Law \\[3ex] \equiv p ... Absorption\:\:Law \\[3ex] \underline{RHS} \\[3ex] [(p \land q) \lor \neg q] \\[3ex] \equiv [\neg q \lor (p \land q)]...Commutative\:\:Law \\[3ex] \equiv [(\neg q \lor p) \land (\neg q \lor q)] \\[3ex]...Distributive\:\:Law \\[3ex] \neg q \lor q \equiv T...Negation\:\:Law \\[3ex] \equiv [(\neg q \lor p) \land T] \\[3ex] \equiv \neg q \lor p ...Identity\:\:Law \\[3ex] \equiv p \lor \neg q ...Commutative\:\:Law \\[3ex] @@@ \\[3ex] \equiv p \land (p \lor \neg q) \\[3ex] \equiv p...Absorption\:\:Law \\[3ex] $ Second Method: Truth Tables

$p$ $q$ $p \land q$ $\neg p$ $\neg p \lor q$ $\neg(\neg p \lor q)$ $(p \land q) \lor \neg(\neg p \lor q)$
$T$ $T$ $T$ $F$ $T$ $F$ $T$
$T$ $F$ $F$ $F$ $F$ $T$ $T$
$F$ $T$ $F$ $T$ $T$ $F$ $F$
$F$ $F$ $F$ $T$ $T$ $F$ $F$
As you can see: the last column is equivalent to the first column.
Therefore: $(p \land q) \lor \neg(\neg p \lor q) \equiv p$
(8.)


(9.)


(10.)


(11.)


(12.)


(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 \\[3ex] $

We shall do this question in two ways.
You may choose any method you prefer.
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.)

(15.) 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 \\[3ex] $

We shall do this question in two ways.
You may choose any method you prefer.
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 $
(16.)


(17.) 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 \\[3ex] $

We shall do this question in two ways.
You may choose any method you prefer.
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 $
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