4 4 votes The statement $(p \wedge(\sim q)) \vee((\sim p) \wedge q) \vee((\sim p) \wedge(\sim q))$ is equivalent to $\_\_\_\_$ .$(\sim p) \vee(\sim q)$$p \vee(\sim q)$$p \vee q$$(\sim p) \vee q$ Mathematical Logic discrete-mathematics goclasses goclasses-cs-dpp goclasses-cs-dpp-day-226 goclasses-dm-practice-questions propositional-logic + – GO Classes 344 views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
2 2 votes Correct Ans is Option A Last Step is 1.(p'+ q') akash_kumar 9 answered Mar 23 akash_kumar 9 comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes (p ∧ ¬q) ∨ (¬p ∧ q) ∨ (¬p ∧ ¬q)Case:1 p = TRUE => (T ∧ ¬q) ∨ (F ∧ q) ∨ (F ∧ ¬q) = ¬q ∨ F ∨ F = ¬qa. ¬p ∨ ¬q = F ∨ ¬q = ¬qb. p ∨ ¬q = T ∨ ¬q = Tc. p ∨ q = T ∨ q = Td. ¬p ∨ q = F ∨ q = qCase:2 p = FALSE => (F ∧ ¬q) ∨ (T ∧ q) ∨ (T ∧ ¬q) = F ∨ q ∨ ¬q = TRUEa. ¬p ∨ ¬q = T ∨ ¬q = TRUEb. p ∨ ¬q = F ∨ ¬q = ¬qc. p ∨ q = F ∨ q = qd. ¬p ∨ q = T ∨ q = TRUE Answer: A chidambareswar23 answered Mar 23 chidambareswar23 comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes Answer: A Meticulous_March answered Mar 23 Meticulous_March comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes A IS THE ANSWER Aparna Bharti answered Apr 2 Aparna Bharti comment Share Follow 0 reply Please log in or register to add a comment.