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Could you please help me with the following question, posed in the first chapter of Jaynes' Probability Theory? Take $$C\equiv(A+\bar B)(\bar A+A\bar B)+\bar AB(A+B)$$
The verification of $$C=(B \Rightarrow \bar A)$$ is left to the reader to show that these two are the same. I started by using the basic identities: $$C\equiv A\bar A+A\bar B+\bar A\bar B+A\bar A\bar B+\bar AB$$ I am a beginner, but I know the first and the fourth terms are always false, and based on the passage, $B \Rightarrow \bar A$ means only that the propositions $B$ and $B \bar A$ have the same truth value. I analyzed all the four possible points of true and false for both the first and second expressions of C. However, they yield different results. I simply don't know how to proceed.

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    $\begingroup$ Have you looked at the truth table of $B \Rightarrow \bar A$? It is the same as the one of $A\bar B+\bar A\bar B+\bar AB$. $\endgroup$
    – posilon
    Commented Sep 19, 2020 at 22:03
  • $\begingroup$ Thank you for your reply. How about the point where A is true and B is false? $B\bar A$ will be false while $A\bar B+\bar A\bar B+\bar AB$ is true. Is the same truth table enough for verifying that the two statements are the same? Thanks. $\endgroup$
    – Nicki Bood
    Commented Sep 19, 2020 at 22:15
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    $\begingroup$ @NickiBood: Yes, they are logically equivalent precisely when they have the same truth table. $\endgroup$
    – Brian M. Scott
    Commented Sep 19, 2020 at 22:18
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    $\begingroup$ If you want to do it strictly algebraically, you can do it as follows: $$\begin{align*}A\bar B+\bar A\bar B+\bar AB&\equiv A\bar B+\bar A\bar B+\bar A\bar B+\bar AB\\&\equiv(A+\bar A)\bar B+\bar A\bar B+\bar AB\\&\equiv\bar B+\bar A(\bar B+B)\\&\equiv\bar B+\bar A\\&\equiv B\Rightarrow\bar A\end{align*}$$ $\endgroup$
    – Brian M. Scott
    Commented Sep 19, 2020 at 22:18
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    $\begingroup$ $B\bar A$ is not the same as $B \Rightarrow \bar A$. Whenever $B$ is false, $B \Rightarrow \text{anything}$ is true. $\endgroup$
    – posilon
    Commented Sep 19, 2020 at 22:20

1 Answer 1

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Since the first and fourth terms are always false, you can drop them. So, you get

$C \equiv A\bar{B} + \bar{A}\bar{B} + \bar{A}B$.

You can the factorize the two occurrences of $\bar{B}$ (i.e. you use the distributive law of conjunction), which yields

$C \equiv (A + \bar{A})\bar{B} + \bar{A}B$.

Since $A + \bar{A}$ is always true, you can omit it and obtain

$C \equiv \bar{B} + \bar{A}B$.

Now, you can use the distributive law of disjunction, thus you get

$C \equiv (\bar{B} + \bar{A})(\bar{B} + B)$.

Again, since $\bar{B} + B$ is always true, you can omit it and obtain

$C \equiv \bar{B} + \bar{A} $.

As in general $\bar{D} + E \equiv D \to E$, we have

$C \equiv B \to \bar{A}$.

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