1
$\begingroup$

Analyze if the following statement is true or false: Suppose $(D,+,\cdot,',0,1)$ is a Boolean Algebra. Then $a'b+(abc)'+c(b'+a)=b'+c'$.


My guess is that the statement is false.

Let $D=D_{15}$ be the divisors of $15$. The Hasse diagram is

Hasse diagram of D_{15}

We know that $(D,+,\cdot,',0,1)$ is a Boolean Algebra since, by a property, $15$ can be expressed as a product of unique primes. So let for example $a=1$, $b=3$ and $c=15$. Then $a'=15$, $b'=5$ and $c'=1$, thus \begin{align*}a'b+(abc)'+c(b'+a)&=15\cdot3+(1\cdot3\cdot15)'+15\cdot(5+1)\\&=3+15+5\\&=15,\end{align*} but \begin{align*}b'+c'&=5+1\\&=5,\end{align*} hence the proposition is false.


Another way that occurred to me to prove that the proposition is false is through the truth tables of each expression, knowing that $\cdot$ represents the $\mathrm{AND}$ logical and $+$ is the $\mathrm{OR}$ logical. So: $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline a&b&c&a'&b'&c'&a'b&abc&(abc)'&b'+a&c(b'+a)&a'b+(abc)'+c(b'+a)\\\hline \mathrm T&\mathrm T&\mathrm T&\mathrm F&\mathrm F&\mathrm F&\mathrm F&\mathrm T&\mathrm F&\mathrm T&\mathrm T&\mathrm T\\\hline \mathrm T&\mathrm T&\mathrm F&\mathrm F&\mathrm F&\mathrm T&\mathrm F&\mathrm F&\mathrm T&\mathrm T&\mathrm F&\mathrm T\\\hline \mathrm T&\mathrm F&\mathrm T&\mathrm F&\mathrm T&\mathrm F&\mathrm F&\mathrm F&\mathrm T&\mathrm T&\mathrm T&\mathrm T\\\hline \mathrm T&\mathrm F&\mathrm F&\mathrm F&\mathrm T&\mathrm T&\mathrm F&\mathrm F&\mathrm T&\mathrm T&\mathrm F&\mathrm T\\\hline \mathrm F&\mathrm T&\mathrm T&\mathrm T&\mathrm F&\mathrm F&\mathrm T&\mathrm F&\mathrm T&\mathrm F&\mathrm F&\mathrm T\\\hline \mathrm F&\mathrm T&\mathrm F&\mathrm T&\mathrm F&\mathrm T&\mathrm T&\mathrm F&\mathrm T&\mathrm F&\mathrm F&\mathrm T \\\hline \mathrm F&\mathrm F&\mathrm T&\mathrm T&\mathrm T&\mathrm F&\mathrm F&\mathrm F&\mathrm T&\mathrm T&\mathrm T&\mathrm T \\\hline \mathrm F&\mathrm F&\mathrm F&\mathrm T&\mathrm T&\mathrm T&\mathrm F&\mathrm F&\mathrm T&\mathrm T&\mathrm F&\mathrm T \\\hline \end{array}$$ but $$\begin{array}{|c|c|c|c|c|} \hline b&c&b'&c'&b'+c' \\\hline \mathrm T&\mathrm T&\mathrm F&\mathrm F&\mathrm F \\\hline \mathrm T&\mathrm F&\mathrm F&\mathrm T&\mathrm T \\\hline \mathrm F&\mathrm T&\mathrm T&\mathrm F&\mathrm T \\\hline \mathrm F&\mathrm F&\mathrm T&\mathrm T&\mathrm T \\\hline \end{array}$$ As we can see, both truth tables do not produce the same truth values, hence the proposition is false.

Are both options correct to prove that each one separately proves the falsehood of the proposition?

Thanks!!

$\endgroup$
2
  • 1
    $\begingroup$ Your decision is right, the table is correct. $\endgroup$
    – user376343
    Feb 22 '19 at 21:37
  • 1
    $\begingroup$ @user376343 thank you!! Would you like to post an answer? $\endgroup$
    – manooooh
    Feb 22 '19 at 22:41
1
$\begingroup$

The statement is false. Besides the methods you used, there are various other.

Karnaugh map is full of 1s.

BA rules applied to the LHS give $$\begin{aligned} a'b+(abc)'+c(b'+a)&=a'b+ a'+b'+c'+cb'+ca\\&=(a'b+ a')+(b'+cb')+c'+ac\\ &=a'+b'+c'+ac\\ &=(a'+a)(a'+c)+b'+c'\\ &=a'+c+b'+c'=1\end{aligned}$$ which is clearly not equal to RHS.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.