2 votes 2 votes Which of the following statements is false? $(P\land Q)\lor(\sim P\land Q)\lor(P \land \sim Q)$ is equal to $\sim Q\land \sim P$ $(P\land Q)\lor(\sim P\land Q)\lor(P \wedge \sim Q)$ is equal to $Q\lor P$ $(P\wedge Q)\lor (\sim P\land Q)\lor(P \wedge \sim Q)$ is equal to $Q\lor (P\wedge \sim Q)$ $(P\land Q)\lor(\sim P\land Q)\lor (P \land \sim Q)$ is equal to $P\lor (Q\land \sim P)$ Mathematical Logic nielit2017july-scientistb-it mathematical-logic propositional-logic + – admin asked Mar 30, 2020 • retagged Aug 22, 2020 by Lakshman Bhaiya admin 881 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 1 votes 1 votes $(P\land Q)\lor(\sim P\land Q)\lor(P \land \sim Q)$ is equal to $\sim Q\land \sim P$ We can write above left hand expression , in the digital logic form$:PQ + \overline{P}\;Q + P\;\overline{Q} = Q(P + \overline{P}) +P\; \overline{Q} = Q +P \;\overline{Q} = (Q+\overline{Q})\cdot (Q + P) = P + Q \equiv P \vee Q \not\equiv \sim Q\land \sim P\;\text{(False)}$ $(P\land Q)\lor(\sim P\land Q)\lor(P \wedge \sim Q)$ is equal to $Q\lor P\;\text{(True as shown in option A)}$ $(P\wedge Q)\lor (\sim P\land Q)\lor(P \wedge \sim Q)$ 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) $(P\land Q)\lor(\sim P\land Q)\lor (P \land \sim Q)$ 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 $(A).$ Lakshman Bhaiya answered Aug 22, 2020 • selected Dec 29, 2021 by Arjun Lakshman Bhaiya comment Share Follow See all 2 Comments See all 2 2 Comments reply ShivangiChauhan commented Mar 1, 2021 reply Follow Share Nice Explanation But Quite Lengthy 0 votes 0 votes Mohitdas commented Oct 11, 2021 reply Follow Share Sir, please see my approach 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes 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 anjli answered Feb 20, 2021 anjli comment Share Follow See 1 comment See all 1 1 comment reply Mohitdas commented Oct 11, 2021 reply Follow Share see my approach 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Only if we are provided AND or OR operation will apply this technique. Mohitdas answered Oct 11, 2021 • edited Dec 30, 2021 by Mohitdas Mohitdas comment Share Follow See all 4 Comments See all 4 4 Comments reply Arjun commented Dec 29, 2021 reply Follow Share Why only for AND and OR? You’re basically doing a partial truth table construction right? 0 votes 0 votes Mohitdas commented Dec 30, 2021 reply Follow Share 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 RHSsuppose we have 4 elements in LHS then we will have 2^4 VALUES going for partial TT will be cumbersome... 0 votes 0 votes Arjun commented Dec 30, 2021 reply Follow Share yes, not a universal approach but still can work for most questions. Moreover expected accuracy is 100% :) 2 votes 2 votes Mohitdas commented Dec 30, 2021 i edited by Mohitdas Dec 30, 2021 reply Follow Share (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 :) 0 votes 0 votes Please log in or register to add a comment.