Question

Which of the following is not a tautology?
A. $p \to (p \vee q)$
B. $(p \wedge q) \to p$
C. \[(p \vee q) \vee (p \wedge \sim q)\]
D. $(p\, \vee \sim p)$

Answer 1

Hint:
In these types of the question first, we will try to simplify the statement to get the required answer and if we can’t find the answer through simplifying, we have to make the truth tables of every statement to find out the required answer.

Complete step by step solution:
According to the question, we have to find the choice which is not a tautology, and we know that a tautology is a statement that is true for every possible interpretation i.e. it is always true
So, checking for option A
 $p \to (p \vee q)$
We know that \[(p \to q)\] is logically equivalent to \[ \sim p \vee q\]
 \[ \Rightarrow \sim p \vee (p \vee q)\]
On using distributive law, we get,
 $ \Rightarrow ( \sim p \vee p) \vee ( \sim p \vee q)$
We also know $( \sim p \vee p)$ is equal to a universal set which is always true.
 \[(True) \vee ( \sim p \vee q)\]
And the union of a universal set with any other set is also a universal set which means it is always
True
So, this option A is a tautology
Checking for option B
 $(p \wedge q) \to p$
We know that \[(p \to q)\] is logically equivalent to \[ \sim p \vee q\]
 $ \Rightarrow \sim (p \wedge q) \vee p$
Since $ \sim (a \wedge b) = ( \sim a \vee \sim b)$ , so we get,
 $ \Rightarrow ( \sim p \vee \sim q) \vee p$
On using distributive law, we get,
 $ \Rightarrow ( \sim p \vee p) \vee ( \sim q \vee p)$
 We also know $( \sim p \vee p\,\,)$ is equal to a universal set which is always true.
 $True\,\, \vee ( \sim q \vee p)$
And the union of a universal set with any other set is also a universal set which means it is always
True
So, this option B is a tautology
Checking for option C
 \[(p \vee q) \to (p \wedge ( \sim q))\]
Since, we know that \[a \to b = \sim a \vee b\] , we get
 \[ \Rightarrow \sim (p \vee q) \vee (p \wedge ( \sim q))\]
Using De Morgan’s law, we get
 \[ \Rightarrow ( \sim p \wedge \sim q) \vee (p \wedge \sim q)\]
Taking \[ \sim q\] common, we get
 \[ \Rightarrow ( \sim p \vee p) \wedge \sim q\]
Let \[( \sim p \vee p) = t\] , so we have
 \[ \Rightarrow t \wedge \sim q\]
As,
 \[ \Rightarrow \sim q \ne t\]
Since we cannot say that \[ \sim q\] is always true as if q is true then \[ \sim q\] is false, hence
 \[ \sim q\, \ne \,True\]
So, this option C is not a tautology
Checking for option D
 $(p\,\, \vee \sim p)$
We know $(p\,\, \vee \sim p)$ is equal to a universal set which is always $True$.
True
So, this option D is a tautology.

Note:
Alternate Solution
We can make truth tables and if all the columns of the final statement are true then it is a tautology
For Option A

p	q	$(p\,\, \vee q)$	$p \to (p\,\, \vee \,q)$
T	T	T	T
T	F	T	T
F	T	T	T
F	F	F	T

Since all the columns of the final statement are true, it is a tautology

For Option B

p	q	$(p\,\, \vee q)$	$p \to (p\,\, \vee q)$
T	T	T	T
T	F	T	T
F	T	T	T
F	F	F	T

Since all the columns of the final statement are true, it is a tautology

For Option C

p	q	$(p\,\, \vee q)$	$ \sim q$	$(p \wedge \sim q)$	\[(p \vee q) \vee (p\,\, \wedge \sim q)\]
T	T	T	F	F	T
T	F	T	T	T	T
F	T	T	F	F	T
F	F	F	T	F	F

Since all the columns of the final statement are NOT true, it is a NOT tautology

For Option D

p	q	$(p\,\, \vee q)$
T	F	T
F	T	T

Since all the columns of the final statement are true, it is a tautology

Hence, we get the answer as Option D.