Proving a boolean algebra relation

Question

I just started reading the book Probability Theory the Logic of Science by Jaynes and on pg. 13 he includes this exercise, which I'm having trouble proving:

$C\equiv(A+\bar B)(\bar A+A \bar B)+\bar AB(A +B)$

"... it is left for the reader to verify that $C$ is logically the same statement as the implication $C=(B\implies \bar A)$"

I'm getting $C=(\bar B + \bar AB)$ when I think $(B\implies \bar A)$ means I should be getting $C=(\bar B + \bar A)$.

Would someone please show me how to prove this?

I think I'm stuck on the specific part, $\bar B + \bar AB == \bar B + \bar A$?

Answer 1

$$\begin{align} C &\equiv (A+\bar B)(\bar A+A \bar B)+\bar AB(A +B)\tag 1\\ \\ &\equiv \underbrace{A(\bar A)}_{\large \text F}+ A(A\bar B) + \bar B(\bar A) +\bar B A\bar B + \underbrace{\bar AB A}_{\large\text{F}}+ \bar ABB \tag {(2) distribution} \\ \\ &\equiv A\bar B + \bar A \bar B+ \bar AB\tag {(3) complement.}\\ \\ &\equiv (\underbrace{A+\bar A}_{\large\text{T}})\bar B + \bar AB \tag {(4) distribution}\\ \\ &\equiv \bar B + \bar AB \tag {(5) idempotence}\\ \\ &\equiv (\bar B + \bar A)(\underbrace{\bar B +B}_{\large T})\tag{(6) distribution} \\ \\ &\equiv \bar B + \bar A \tag{(7) idempotence} \end{align}$$

Answer 2

By complementation and idempotence: $\bar B+\bar AB ~{= (A+\bar A)\bar B+\bar AB \\= (A+\bar A+\bar A)\bar B+\bar AB \\= (A+\bar A)\bar B+ \bar A(\bar B+\bar B)\\=\bar B+\bar A}$

By distribution: $\bar B+(\bar A)(B) ~{= (\bar B+\bar A)(\bar B+B) \\= \bar A+ \bar B}$

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Note also: $(\bar A +B)\bar B ~{= \bar A\bar B+B\bar B \\= \bar A \bar B}$