The statement p - - q r - r q is

Name: JEE- Grade 11- Mathematics- Mathematical Reasoning- Session 03- W04
Uploaded: 2022-07-04T10:36:42+05:30
Description: [ q amp; r amp; ∼ q nbsp; amp;∼ r amp; ∼ q nbsp; r =x amp; y=∼ r q amp; x y nbsp;; T amp; T ...

[ q amp; r amp; ∼ q nbsp; amp;∼ r amp; ∼ q nbsp; r =x amp; y=∼ r q amp; x y nbsp;; T amp; T ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Method of Contradiction

The statement...

Question

The statement $p \land (\sim q \lor r) \lor (\sim r \lor q)$ is

logically equivalent to

x \lor (\sim x)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a contradiction

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a tautology

No worries! We‘ve got your back. Try BYJU‘S free classes today!

neither a tautology nor a contradiction

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D neither a tautology nor a contradiction
$\begin{matrix} q & r & \sim q & \sim r & \sim q \lor r = x & y =\sim r \lor q & x \lor y T & T & F & F & T & T & T T & F & F & T & F & T & T F & T & T & F & T & F & T F & F & T & T & T & T & T \end{matrix}$

Now, $p \land (\sim q \lor r) \lor (\sim r \lor q) = p \land T = p$ , which is neither a tautology nor contradiction.
$[∵ (\sim q \lor r) \lor (\sim r \lor q) = x \lor y is a tautology]$

Suggest Corrections

Similar questions

[p \land (\sim q)] \land [(\sim p) \lor q]

Q. If

(p \land \sim q) \land (p \land r) \to \sim p \lor q

is false, then the truth values of

p, q

and

r

are, repectively :

Q. The statement

(p \lor q) \land (\sim p \land \sim q)

Q. Given the following two statements:

(S_{1}) : (q \lor p) \to (p \leftrightarrow\sim q)

is a tautology.

(S_{2}) :\sim q \land (\sim p \leftrightarrow q)

is a fallacy. Then

Q. The Boolean expression

\sim (p \lor q) \lor (\sim p \land q)

is equivalent to

Join BYJU'S Learning Program

Explore more

Method of Contradiction

Standard XII Mathematics

Join BYJU'S Learning Program

The statement p∧(∼q ∨r)∨(∼r∨q) is

The correct option is D neither a tautology nor a contradictionqr∼q∼r∼q∨r=xy=∼r∨qx∨yTTFFTTTTFFTFTTFTTFTFTFFTTTTT Now, p∧(∼q∨r)∨(∼r∨q)= p∧T=p, which is neither a tautology nor contradiction. [ ∵(∼q∨r)∨(∼r∨q)=x∨y is a tautology]

The statement $p \land (\sim q \lor r) \lor (\sim r \lor q)$ is