The Verification of a Logical Statement in Jaynes' Book

Question

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.

Answer 1

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

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