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Question: The logical statement \([ \sim ( \sim p \vee q) \vee (p \wedge r)] \wedge ( \sim q \wedge r)\)is equ...

The logical statement [∼(∼p∨q)∨(p∧r)]∧(∼q∧r)[ \sim ( \sim p \vee q) \vee (p \wedge r)] \wedge ( \sim q \wedge r)is equivalent to
(A) (P∧r)∧∼q(P \wedge r) \wedge \sim q
(B) (p∧∼q)∨r(p \wedge \sim q) \vee r
(C) (∼p∧∼q)∧r( \sim p \wedge \sim q) \wedge r
(D) ∼p∨r \sim p \vee r

Explanation

Solution

The given logical statement needs to be simplified with the help of associative, distributive laws and Demorgan’s theorem.
Associative law: (a∧b)∧ca \wedge b) \wedge c= a∧(b∧c)a \wedge (b \wedge c)
Distributive law: a∧(b∨c)=(a∧b)∨(a∧c)a \wedge (b \vee c) = (a \wedge b) \vee (a \wedge c)
Otherwise, verify the options with eight sets of T and F in the p,qp,qand rrentries respectively.

Complete Step by Step Solution:
Given logical statement is:
[∼(∼p∨q)∨(p∧r)]∧(∼q∧r)[ \sim ( \sim p \vee q) \vee (p \wedge r)] \wedge ( \sim q \wedge r)
Here, ∼\simdenotes negation
∨\veedenotes OR gate or union
∧\wedgedenoted AND gate or intersection.
p,qp,q and rr are variables which can take only two values T and F as per truth table. and if we consider this expression in Boolean algebra, then it can take 11 and 00 in place of T and F in the entries of p,qp,q and rr.
The given logical statement can be manipulated in the given ways following different laws:
[∼(∼p∨q)∨(p∧r)]∧(∼q∧r)[ \sim ( \sim p \vee q) \vee (p \wedge r)] \wedge ( \sim q \wedge r)
=[(p∧∼q)∨(p∨r)]∧(∼q∧r)= [(p \wedge \sim q) \vee (p \vee r)] \wedge ( \sim q \wedge r) →\to [∵∼(a∨b)=(∼a∧∼b)\because \sim (a \vee b) = ( \sim a \wedge \sim b)]
=[p∧(∼q∨r)]∧(∼q∧r)]→= [p \wedge ( \sim q \vee r)] \wedge ( \sim q \wedge r)] \to [ by distributive law: a∧(b∨c)=(a∧b)∨(a∧c)a \wedge (b \vee c) = (a \wedge b) \vee (a \wedge c)]
=p∧[∼q∧(∼q∧r)]∨[r∧(∼q∧r)]= p \wedge [ \sim q \wedge ( \sim q \wedge r)] \vee [r \wedge ( \sim q \wedge r)] →\to[∵[(a∨b)∧(a∧b)]=[a∧(a∧b)∨(b∧(a∧b))]\because [(a \vee b) \wedge (a \wedge b)] = [a \wedge (a \wedge b) \vee (b \wedge (a \wedge b))]
=p∧[(∼q∧r)∨(∼q∧r)]→[∵b∧(a∧a)=(b∧a)]= p \wedge [( \sim q \wedge r) \vee ( \sim q \wedge r)] \to [\because b \wedge (a \wedge a) = (b \wedge a)]
=p∧[∼q∧r]= p \wedge [ \sim q \wedge r]
=(p∧r)∧(∼q)= (p \wedge r) \wedge ( \sim q) (by associative law)
Hence, the logical statement [∼(∼p∨q)∨(p∧r)]∧(∼q∧r)[ \sim ( \sim p \vee q) \vee (p \wedge r)] \wedge ( \sim q \wedge r)is equivalent to(p∧r)∧(∼q)(p \wedge r) \wedge ( \sim q).

Therefore, the correct option is (A).

Note:
Here, basic Boolean rules have been used to solve this problem. Alternatively, we can verify each option by taking a set of entries and then check which option satisfies all of them. One should remember all the laws to save time while solving the questions related to Boolean expression. Otherwise, it will take a lot of time if we keep verifying all the options.