24 votes 24 votes The proposition $p \wedge (\sim p \vee q)$ is: a tautology logically equivalent to $p \wedge q$ logically equivalent to $p \vee q$ a contradiction none of the above Mathematical Logic gate1993 mathematical-logic easy propositional-logic + – Kathleen asked Sep 29, 2014 • recategorized Apr 22, 2021 by Lakshman Bhaiya Kathleen 8.3k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply debasree88 commented Aug 29, 2019 reply Follow Share why not option a? it is modus ponens clearly and hence it is a tautology... pls pls clear my doubt... 0 votes 0 votes ankitgupta.1729 commented Aug 29, 2019 reply Follow Share modus ponens is an inference rule to conclude something.. It is not used to show tautology. Here, put 'P' as False. It is not a tautology. 0 votes 0 votes Please log in or register to add a comment.
Best answer 31 votes 31 votes $p \wedge (\sim p \vee q)$ $\equiv (p \wedge \sim p) \vee (p \wedge q)$ $\equiv F \vee (p \wedge q)$ $\equiv (p \wedge q)$ Hence, Option(B) logically equivalent to $ (p \wedge q)$. LeenSharma answered May 16, 2017 • edited Jun 14, 2018 by kenzou LeenSharma comment Share Follow See all 3 Comments See all 3 3 Comments reply debasree88 commented Aug 29, 2019 reply Follow Share why not option a? it is modus ponens clearly and hence it is a tautology... pls pls clear my doubt... –1 votes –1 votes Amcodes commented Oct 4, 2020 reply Follow Share Just substitute values debasree , it’s not ALWAYS TRUE, so not tautology. 0 votes 0 votes chokostar commented Apr 14, 2023 reply Follow Share By modus ponens, p p → q q Now, q can either be True or False hence a contingency and not a tautology! 0 votes 0 votes Please log in or register to add a comment.
12 votes 12 votes p ^ (~p v q) = (p ^ ~p) v (p ^ q) = False V (p^q) = (p^q) Mridul Sachan answered Mar 11, 2016 Mridul Sachan comment Share Follow See all 0 reply Please log in or register to add a comment.
3 votes 3 votes $P \wedge (\neg P \vee Q ) \equiv ( P \wedge \neg P ) \vee ( P \wedge Q )$ $\equiv F \vee ( P \wedge Q )\:\:\: \bigg[\because (P \wedge \neg P) \: \text{ is always False and} \: (P \vee \neg P)\:\text{ is always True} \bigg]$ $\equiv ( P \wedge Q )$ Lakshman Bhaiya answered Feb 21, 2018 • edited Dec 7, 2019 by Lakshman Bhaiya Lakshman Bhaiya comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes So option-b is correct akshay_123 answered Sep 22, 2023 akshay_123 comment Share Follow See all 0 reply Please log in or register to add a comment.