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I only know how to prove that $A'+AB'+B$ is always true with a truth table, but not boolean algebra. I haven't found a suitable law to solve this.

A little guidance is much appreciated~

Edit: Thank you so much all for your answers! Really helped me understand more about how to use those laws.

I personally prefer the complement law approach, but the double-negative then De Morgan's is really useful too. Wish I could mark both of them helpful!

Vero
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  • Hint: show that $A'+AB'=A'+B'$. – Jimmy R. May 30 '20 at 14:53
  • Wait a second, you're to use a truth table - isn't that just a table with columns A, B, A', B', AB', A'+AB'+B (and four rows) in this case? You don't need no law, just the truth tables / definitions of ◌', ◌◌, and ◌+◌. – retzler May 30 '20 at 23:04
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    @retzler I read it the other way. V C knows how to do it with a truth table, and asks how to do it with Boolean algebra. – quid May 30 '20 at 23:35

4 Answers4

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You have $$AB^\prime = (AB^\prime)^{\prime \prime}=(A^\prime+B^{\prime \prime})^\prime = (A^\prime + B)^\prime$$ by double negation and de Morgan law and therefore

$$A^\prime+AB^\prime+B=(A^\prime+B) + AB^\prime = (A^\prime+B)+(A^\prime + B)^\prime = 1$$

mathcounterexamples.net
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A hint: by de Morgan,

$$A'+AB'+B = (A(A'+B)B')'$$

Now see if the expression inside the outer parentheses simplifies.

lisyarus
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\begin{align} A'+AB'+B \tag{1}\label{1} \end{align}

Since

\begin{align} A+A'=B+B'&=1 \tag{2}\label{2} , \end{align}

we can rewrite \eqref{1} as

\begin{align} &A'(B+B')+AB'+B(A+A') \\ &=A'B+A'B'+AB'+AB \tag{3}\label{3} \\ &=A'(B+B')+A(B'+B) \\ &=A'+A=1 \tag{4}\label{4} . \end{align}

g.kov
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If you're looking for an intuitive answer, make the substitution $C = A'$. Then the equation becomes

$$B + C + B'C'$$

which, in English, says that "Either B is true, C is true, or they're both false"

BlueRaja - Danny Pflughoeft
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