Among the statements:(S1) quad (( p vee q ) R | vedclass

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(B) To determine if a statement is a tautology,we construct truth tables.For $(S1): ((p \vee q)$ $\Rightarrow r) \Leftrightarrow (p$ $\Rightarrow r)$T

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Neither $(S1)$ nor $(S2)$ is a tautology

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(B) To determine if a statement is a tautology,we construct truth tables.For $(S1): ((p \vee q)$ $\Rightarrow r) \Leftrightarrow (p$ $\Rightarrow r)$This is not a tautology because if $p=F, q=T, r=F$,then $(p \vee q) \Rightarrow r$ is $F$,while $p \Rightarrow r$ is $T$. Thus,the biconditional is $F$.For $(S2): ((p \vee q)$ $\Rightarrow r) \Leftrightarrow ((p$ $\Rightarrow r) \vee (q$ $\Rightarrow r))$Using logical equivalence: $(p \vee q)$ $\Rightarrow r \equiv 
eg(p \vee q) \vee r \equiv (
eg p \wedge 
eg q) \vee r \equiv (
eg p \vee r) \wedge (
eg q \vee r) \equiv (p$ $\Rightarrow r) \wedge (q$ $\Rightarrow r)$.Since $(p$ $\Rightarrow r) \wedge (q$ $\Rightarrow r)$ is not equivalent to $(p$ $\Rightarrow r) \vee (q$ $\Rightarrow r)$,$(S2)$ is not a tautology.Therefore,neither $(S1)$ nor $(S2)$ is a tautology.

Among the statements:
$(S1) \quad (( p \vee q )$ $\Rightarrow r ) \Leftrightarrow ( p$ $\Rightarrow r )$
$(S2) \quad (( p \vee q )$ $\Rightarrow r ) \Leftrightarrow (( p$ $\Rightarrow r ) \vee ( q$ $\Rightarrow r ))$

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Among the statements:$(S1) \quad (( p \vee q )$ $\Rightarrow r ) \Leftrightarrow ( p$ $\Rightarrow r )$$(S2) \quad (( p \vee q )$ $\Rightarrow r ) \Leftrightarrow (( p$ $\Rightarrow r ) \vee ( q$ $\Rightarrow r ))$