1 votes 1 votes State whether the following statements are true or false? $(P\to Q) \rightarrow (Q\to P)$ always holds, for all propositions $P, Q$. $\left ( \left ( P\vee Q \right )\rightarrow Q \right )$ $\rightarrow$ $\left ( Q\to \left ( P\vee Q \right ) \right )$ always holds, for all propositions $P, Q$. $a$ is true, $b$ is false Both $a$ and $b$ are true $a$ is false, $b$ is true. Both $a$ and $b$ are false. GATE tbb-mockgate-1 discrete-mathematics mathematical-logic propositional-logic + – Bikram asked Jan 16, 2017 edited Dec 19, 2022 by gatecse Bikram 763 views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments mehul vaidya commented May 20, 2019 reply Follow Share >= should be made --> it just create confusion 0 votes 0 votes JashanArora commented Jan 11, 2020 reply Follow Share Is $\geq$ a typo? 0 votes 0 votes toxicdesire commented Jan 28, 2020 reply Follow Share it doesn't look like a typo, it's intentional. 0 votes 0 votes Please log in or register to add a comment.
Best answer 3 votes 3 votes The statement a becomes false when P is 0 & Q is 1. The statement b is true because = (( P v Q) => Q) => (Q => (P v Q)) = 1 nandini gupta answered Jan 18, 2017 selected Jan 18, 2017 by Bikram nandini gupta comment Share Follow See all 2 Comments See all 2 2 Comments reply Shubhgupta commented Jan 30, 2019 reply Follow Share 1st statement is false for P=1 and Q=0 not P=0 and Q=1. 0 votes 0 votes Srividyaketharaju commented Feb 4, 2020 reply Follow Share 1) (P≥Q)→(Q≥P) (P ->Q) -> (Q->P) = ~(~PVQ)V(~QVP)=(P/\~Q)V(~QVP) if P =0, Q=1 (1.(~1))+((~1)+0) = 0+0 = 0(False) 2) ((P∨Q)→Q) → (Q≥(P∨Q)) => True ( P= 0, Q=1) 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Here >= is Implication 1) (P≥Q)→(Q≥P) (P ->Q) -> (Q->P) = ~(~PVQ)V(~QVP)=(P/\~Q)V(~QVP) if P =0, Q=1 (1.(~1))+((~1)+0) = 0+0 = 0(False) 2) ((P∨Q)→Q) → (Q≥(P∨Q)) => alwaysTrue ( P= 0, Q=1 Srividyaketharaju answered Feb 4, 2020 Srividyaketharaju comment Share Follow See all 0 reply Please log in or register to add a comment.