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The Boolean Expression $(p\land \lnot q) \lor q \lor (\lnot p\land q)$ is equivalent to: $(1) p \land q \space \space (2) p \lor q \space \space (3) p \lor \lnot q\space \space (4) \lnot p \land q $

My attempt: $pq'+q+p'q = pq' + q$. Not able to proceed next.

The answer is given as $(2)$.

Though by making venn diagram or truth table or just opening the expression with the given symbols, I am able to get the answer. But I wonder how to solve it by converting AND as multiplication and OR as addition. Not able to write $pq'+q$ as $p+q$.

aarbee
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1 Answers1

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If you call false $0$ and true $1$ and note both satisfy $x^2=x$, $\land$ multiplies and $\lor$ maps $x,\,y$ to $1-(1-x)(1-y)=x+y-xy$ (rather than $x+y$, unfortunately), so $(p\land\neg q)\lor q=p(1-q)+q-pq(1-q)=p+q-pq$, i.e. $p\lor q$ (which is obvious with other methods). Similarly, $\neg p\land q=q(1-p)$, so your function is$$p+q-pq+q(1-p)-(p+q-pq)q(1-p)=p+q-pq=p\lor q.$$We can't, however, choose values for false and true so that $\lor$ is ordinary addition. For starters, $\top\lor\top=\top$ and $\bot\lor\bot=\bot$ would require both values to be $0$.

J.G.
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  • "(p∧¬q)∨q=p(1−q)+q−pq(1−q)=p+q−pq". Here, I got p(1-q)+q. But after that, not able to understand how that expression came. – aarbee Apr 11 '20 at 20:06
  • oh, got it. It is from ur explanation proceeding that expression. – aarbee Apr 11 '20 at 20:09