The correct simplified circuit diagram for th | vedclass

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(B) Let the given logical statement be $L = [\{q \wedge (\sim q \vee r)\} \wedge \{\sim p \vee (p \wedge \sim r)\}] \vee (p \wedge r)$.Using the distr

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(B) Let the given logical statement be $L = [\{q \wedge (\sim q \vee r)\} \wedge \{\sim p \vee (p \wedge \sim r)\}] \vee (p \wedge r)$.Using the distributive law,$\{q \wedge (\sim q \vee r)\} = (q \wedge \sim q) \vee (q \wedge r) = F \vee (q \wedge r) = (q \wedge r)$.Using the distributive law,$\{\sim p \vee (p \wedge \sim r)\} = (\sim p \vee p) \wedge (\sim p \vee \sim r) = T \wedge (\sim p \vee \sim r) = (\sim p \vee \sim r)$.Substituting these back into $L$,we get $L = [(q \wedge r) \wedge (\sim p \vee \sim r)] \vee (p \wedge r)$.Using the distributive law,$(q \wedge r) \wedge (\sim p \vee \sim r) = (q \wedge r \wedge \sim p) \vee (q \wedge r \wedge \sim r) = (q \wedge r \wedge \sim p) \vee F = (q \wedge r \wedge \sim p)$.Now,$L = (q \wedge r \wedge \sim p) \vee (p \wedge r) = r \wedge [(q \wedge \sim p) \vee p]$.Using the distributive law,$(q \wedge \sim p) \vee p = (q \vee p) \wedge (\sim p \vee p) = (q \vee p) \wedge T = (q \vee p)$.Thus,$L = r \wedge (p \vee q)$.In terms of switches,$r$ is $S_3$,$p$ is $S_1$,and $q$ is $S_2$.So,$L = S_3 \wedge (S_1 \vee S_2)$.This represents switch $S_3$ in series with the parallel combination of switches $S_1$ and $S_2$.

The correct simplified circuit diagram for the logical statement $[\{q \wedge (\sim q \vee r)\} \wedge \{\sim p \vee (p \wedge \sim r)\}] \vee (p \wedge r)$ where $p, q, r$ represent switches $S_1, S_2, S_3$ respectively.

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