Equation 1.15 in Jaynes Probability Theaory - The Logic of Science

Question

The equation 1.15 states:

$C ≡ (A + \overline{B})(\overline{A} + A\overline{B}) + \overline{A}B(A + B)$

Further the text it is suggested to the reader to verify that the equation is equal to C = (B ⇒ $\overline{A}$)

I can't grasp the idea behind it:(

For me it: $(A + \overline{B})(\overline{A} + A\overline{B}) + \overline{A}B(A + B) = A\overline{A} +AA\overline{B} +\overline{A}\overline{B} + A\overline{B}\overline{B} + \overline{A}AB +\overline{A}BB = A\overline{B} + \overline{A}\overline{B} + \overline{A}B =\overline{B} +\overline{A}B$

Please help me to undestend how an implication can be derived from this equation.

upd. Ok, I think I see why it's true (from Venn diagram): the only way B can be True is in conjunction with $\overline{A}$, so if $B$ is True then $\overline{A}$ is also True.

Answer 1

For me it was: $(A + \overline{B})(\overline{A} + A\overline{B}) + \overline{A}B(A + B) \\= A\overline{A} +AA\overline{B} +\overline{A}\,\overline{B} + A\overline{B}\,\overline{B} + \overline{A}AB +\overline{A}BB \\= A\overline{B} + \overline{A}\,\overline{B} + \overline{A}B \\=\overline{B} +\overline{A}B$

Yes, so far so good. From there, on to $C = \overline B+\overline A$ by the Absorption Law.

Or from the second last line, use idempotence. $C~{= A\overline{B} + \overline{A}\,\overline{B} + \overline{A}B \\ = A\overline{B} + \overline{A}\,\overline{B}~+~\overline{A}\,\overline{B}+\overline{A}B\\ =\overline B+\overline A}$

Therefore $C= (B\to\overline A)$ by Implication Equivalence.