Question

Which of the following statements is not a tautology?
(a)$\left( p\wedge q \right)\to p$
(b)$\left( p\wedge q \right)\to \left( \sim p \right)\vee q$
(c)$p\to \left( p\vee q \right)$
(d)$\left( p\vee q \right)\to \left( p\vee \left( \sim q \right) \right)$

Answer 1

Hint: Tautology of the statement means that in every case the statement is true so the correct option is one which has at least one time becomes false. To find the statement whether it is always true or not we have to construct the truth table for the given logic and then see the result of the truth table if the result of the truth table is always true it means the statement is a tautology otherwise not a tautology.

Complete step by step answer:
We have given the various logics in the options which are given as statements and we have to mark the
option which is not a tautology.
Now, tautology of a statement means no matter what condition we put on the statement the answer is
always true. To find the tautology of the given options, we are going to write their corresponding truth

tables and then see the result of the statement if the result of the statement in every condition is true
then that statement is a tautology.
As you can see that in each option implies that the $\left( \to \right)$ operator is there so we are
showing the truth table of this operator. Let us consider p and q as our sample propositions so the truth
table of $p\to q$ as follows:
p q $\left( p\wedge q \right)\to p $
T T T
T F F
F T T
F F T
In the above table, we have represented “T” as True and “F’ as False.
Checking option (a) $\left( p\wedge q \right)\to p$
In the below, we are drawing a truth table for this logic:
$\left( p\wedge q \right)$ p $\left( p\wedge q \right)\to p$
T T T
F T T
F F T

Now, you might think, how we have written the true/false in the above table. You can understand the
T/F written in the third column of the above truth table from the “implies that” table that we have
shown above and the truth table of first two tables you can understand as When p is true then $\left(
p\wedge q \right)$ can be true or false and if p is false then $\left( p\wedge q \right)$ is always false
because if p is false then whether q is true or false the answer is always false. To get more clarification,
you can see the truth table for $\left( p\wedge q \right)$ that we have shown below:
p q $\left( p\wedge q \right)$
T T T
T F F
F T F
F F F

As you can see the truth table for option (a) is always giving the true answer so option (a) is a tautology.
Hence, option (a) is incorrect.
Checking option (b) $\left( p\wedge q \right)\to \left( \sim p \right)\vee q$ by constructing the truth
table as follows:
p q $\left(
p\wedge\right)

\left( \sim
p \right)\vee$

$\left( p\wedge q
\right)\to \left( \sim p

q \right)$ $\left(q \right)\vee q$

T T T T T
T F F F T
F T F T T
F F F T T

In the above truth table, $\sim p$ is not p means if p is true then $\sim p$ is false. To understand how
we get the truth value for the fourth column we have shown below the truth table of $p\vee q$.
p q $\left( p\vee q \right)$
T T T
T F T
F T T
F F F

As you can see that the truth table for option (b) is always showing the true value it means option (b) is
a tautology. Hence, option (b) is incorrect.
Checking option (c) $p\to \left( p\vee q \right)$ by constructing its truth table we get,
q p $\left( p\vee q \right)$ $p\to \left( p\vee q

\right)$
T T T T
T F T T
F T T T
F F F T

As you can see that the truth table of the above logic is always true so this option is a tautology. Hence,
option (c) is not correct.

Checking option (d) $\left( p\vee q \right)\to \left( p\vee \left( \sim q \right) \right)$ by constructing the
truth table we get,
p q $p\vee q$ $\left( p\vee \left(
\sim q \right)
\right)$

$\left( p\vee q
\right)\to \left(
p\vee \left( \sim q
\right) \right)$

T T T T T
T F T T T
F T T F F
F F F T T
In the above table, $\sim q$ represents not q.
As you can see that in the above truth table we are getting one false, this means that this option is not a
tautology. Hence, option (d) is correct.
From the above solution, the correct option is (d).
Note:
 The above question demands the knowledge of truth tables for $\left( p\wedge q \right),\left(
p\vee q \right),\left( p\to q \right)$. If you know the truth tables for these logics then this question is a
cup of tea for you. And another thing that you should know is what tautology is.
The mistake that you could make in this problem is that the question is asking we should mark the
option which is “not” a tautology but in the haste of solving the questions we miss the “not” so don’t
ignore the “not” in the question.