NIELIT 2017 July Scientist B (IT) - Section B: 13

Question

Which of the following statements is false?

• A. $(P\land Q)\lor(\sim P\land Q)\lor(P \land \sim Q)$ is equal to $\sim Q\land \sim P$
• B. $(P\land Q)\lor(\sim P\land Q)\lor(P \wedge \sim Q)$ is equal to $Q\lor P$
• C. $(P\wedge Q)\lor (\sim P\land Q)\lor(P \wedge \sim Q)$ is equal to $Q\lor (P\wedge \sim Q)$
• D. $(P\land Q)\lor(\sim P\land Q)\lor (P \land \sim Q)$ is equal to $P\lor (Q\land \sim P)$

Answer 1

• A. $(P\land Q)\lor(\sim P\land Q)\lor(P \land \sim Q)$ is equal to $\sim Q\land \sim P$
  
  We can write above left hand expression , in the digital logic form$:PQ + \overline{P}\;Q + P\;\overline{Q} = Q(P + \overline{P}) +P\; \overline{Q}  = Q +P \;\overline{Q} = (Q+\overline{Q})\cdot (Q + P) = P + Q \equiv P \vee Q \not\equiv  \sim Q\land \sim P\;\text{(False)}$
   
• B. $(P\land Q)\lor(\sim P\land Q)\lor(P \wedge \sim Q)$ is equal to $Q\lor P\;\text{(True as shown in option A)}$
   
• C. $(P\wedge Q)\lor (\sim P\land Q)\lor(P \wedge \sim Q)$ is equal to $Q\lor (P\wedge \sim Q)$
  
  We can write the right hand expression, in terms of digital logic, $Q\lor (P\wedge \sim Q) \equiv Q + P\;\overline{Q}  = (Q+\overline{Q})\cdot (Q + P) = P + Q \equiv P \vee Q$
  
  LHS and RHS are equal. (True)
   
• D. $(P\land Q)\lor(\sim P\land Q)\lor (P \land \sim Q)$ is equal to $P\lor (Q\land \sim P)$

  We can write the right hand expression, in terms of digital logic, $P\lor (Q\wedge \sim P) \equiv P + Q\;\overline{P}  = (P+\overline{P})\cdot (P + Q) = P + Q \equiv P \vee Q$

  LHD and RHS are equal. (True)

So, the correct answer is $(A).$

Comments

Nice Explanation But Quite Lengthy

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Sir, please see my approach

Answer 2

Ans :A

After  solving with truth table

(P∧Q)∨(∼P∧Q)∨(P∧∼Q)(P∧Q)∨(∼P∧Q)∨(P∧∼Q) is  not equal to  ∼Q∧∼P

Comments

see my approach

Answer 3

Only if we are provided AND or OR operation will apply this technique.

Comments

Why only for AND and OR? You’re basically doing a partial truth table construction right?

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yes Sir a partial truth table construction is easy to visualize 

it becomes more complex for some operations like implies, iff , if we have 3-4 values in LHS or RHS

suppose we have 4 elements in LHS then we will have 2^4 VALUES 

going for partial TT will be cumbersome...

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yes, not a universal approach but still can work for most questions. Moreover expected accuracy is 100% :)

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(P∧Q)∨(∼P→Q)→(P→∼Q) 

We can Simplify as P → Q is logically equivalent to ¬ P ∨ Q, This is made up of OR operations :)