The given circuit is equivalent to

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(D) The symbolic form of the given circuit is $(p \vee \sim q \vee \sim r) \wedge (p \vee (q \wedge r))$.Applying the distributive law,we get $p \vee

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(D) The symbolic form of the given circuit is $(p \vee \sim q \vee \sim r) \wedge (p \vee (q \wedge r))$.Applying the distributive law,we get $p \vee [(\sim q \vee \sim r) \wedge (q \wedge r)]$.Using De Morgan's law,this becomes $p \vee [\sim (q \wedge r) \wedge (q \wedge r)]$.Applying the complement law,we get $p \vee F$.Finally,by the identity law,this simplifies to $p$.Thus,the circuit is equivalent to a circuit with only switch $S_1$.

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