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The maximum number of compound propositions, out of \mathrm{p} \vee \mathrm{r} \vee \mathrm{s}, \mathrm{p} \vee \mathrm{r} \vee \sim \mathrm{s}$, p $\vee \sim \mathrm{q} \vee \mathrm{s}, \sim \mathrm{p} \vee \sim \mathrm{r} \vee \mathrm{s}, \sim \mathrm{p} \vee \sim \mathrm{r} \vee \sim \mathrm{s}, \sim \mathrm{p} \vee \mathrm{q} \vee \sim \mathrm{s}, \mathrm{q} \vee \mathrm{r} \vee \sim \mathrm{s}, \mathrm{q} \vee \sim \mathrm{r} \vee \sim \mathrm{s}, \sim \mathrm{p} \vee \sim \mathrm{q} \vee \sim \mathrm{s} that can be made simultaneously true by an assignment of the truth values to \mathrm{p}, \mathrm{q}, \mathrm{r}$ and $\mathrm{s}, is equal to____________.

Option: 1

9


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

1. If all p,q,r,s are true, then \sim p \vee \sim r V \sim s \text { and } \sim p \vee \sim q \vee \sim s are false. So 7 are true

2.If only one is false.

(i) If p is false, rest are true then all 9 statements are true, which is the maximum. So no need to check further.

\therefore Answer is all 9 are true when p is false and q, r, s are true

Hence answer is 9

Posted by

Sanket Gandhi

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