JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Mathematical Reasoning *

For $S_1$, take $p=F,q=T,r=F$: then $p\vee q$ is true while $r$ is false, so $(p\vee q)\Rightarrow r$ is false, whereas $p\Rightarrow r=F\Rightarrow F=T$; the biconditional is false, so $S_1$ is not a tautology. For $S_2$, take $p=T,q=F,r=F$: here $(p\vee q)\Rightarrow r$ is false, but $(p\Rightarrow r)\vee (q\Rightarrow r)=F\vee T=T$, so this biconditional is also false and $S_2$ is not a tautology either. Hence neither $S_1$ nor $S_2$ is a tautology.