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Are the steps I applied correct in simplifying logical statements and producing logical equivalencies?


(1) Simplify $s\land(\lnot t\lor s):$ \begin{align} s\land(\lnot t\lor s) &\Leftrightarrow s & \text{by absorption} \\ \end{align}


(2) Simplify $s\land[\neg(\neg s \lor t)]\lor(s \land t):$ \begin{align} s\land[\neg(\neg s \lor t)]\lor(s \land t) &\Leftrightarrow s\land[\neg(\neg s)\land \neg t]\lor (s\land t) & \text{by DeMorgan's}\\ &\Leftrightarrow s\land(s \land \neg t)\lor (s\land t) & \text{by double negation}\\ &\Leftrightarrow [(s\land s) \land \neg t]\lor (s\land t) &\text{by associativity of $\land$} \\ &\Leftrightarrow (s\land\neg t)\lor(s\land t) &\text{by idempotence}\\ &\Leftrightarrow s \land(\neg t\lor t) & \text{by distributivity of $\land$ over $\lor$}\\ &\Leftrightarrow s\land T &\text{by inverse}\\ &\Leftrightarrow s & \text{by identity} \end{align}


(3) Simplify $\neg(s \lor \neg t)\lor (\neg s \land \neg t):$ \begin{align} \neg(s \lor \neg t)\lor (\neg s \land \neg t) &\Leftrightarrow \neg s\lor\neg(\neg t)\lor(\neg s\land\neg t) & \text{by DeMorgan's}\\ &\Leftrightarrow \neg s\land t\lor(\neg s\land\neg t) & \text{by double negation}\\ &\Leftrightarrow (\neg s\land t) \lor(\neg s\land\neg t) &\text{} \\ &\Leftrightarrow \neg s\land(t\lor\neg t) &\text{by distributivity of $\land$ over $\lor$}\\ &\Leftrightarrow \neg s\land T& \text{by inverse}\\ &\Leftrightarrow \neg s &\text{by identity}\\ \end{align}


Thanks in advance.

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Yes, the simplifications and equivalences you've applied are right. The big deal here is recognizing cases where you can apply De Morgan's laws and the correct usage of associativity and distributivity.

Just a thing: I can't see the relation between the first step and the others. Is it another proposition that you didn't state in the question title? Anyway, great job. :-)

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  • $\begingroup$ Awesome! Yup didn't include the first one on the question title as it was getting convoluted.. Thanks! $\endgroup$
    – user144809
    Commented Feb 4, 2015 at 9:05

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