$\overline{(a+\overline{b})\cdot (\overline{a}+b)}$ simplification boolean algebra

Question

For context: I am learning Boolean Algebra by myself for fun and one of the questions in the book I am reading was a long boolean expression and the task was to simplify it to be the XOR boolean expression. I have managed to come quite a bit but cannot progress any further.

I am unable to simplify $\overline{(a+\overline{b})\cdot (\overline{a}+b)}$ to be $a\overline{b} + \overline{a}b$.

There seem to be no laws associated with this type of boolean multiplication. Can anybody point me in the right direction? My original equation is of an XOR logical operation, but not in its "final form" so to say.

Answer 1

If you develop your product you get: $$(a+b')(a'+b)=\overbrace{aa'}^0+ab+a'b'+\overbrace{bb'}^0=ab+a'b'$$

Yet you want $a\oplus b=(ab'+a'b)$ to appear not $(ab+a'b')$.

But notice that for any $x$ then $x+x'=1$ therefore:

$$1=a'(b+b')+a(b+b')=(ab'+a'b)+(ab+a'b')$$

So in the end $(ab+a'b')=(ab'+a'b)\ '$

Answer 2

XOR in conjuctive normal form is $(a+b)\cdot (\bar a+\bar b) = a\bar a + a\bar b + b\bar a + b\bar b$ by distributivity. But $a\bar a = 0 = b\bar b$ and $0+x = x$ for any Boolean expression $x$. Hence, the result is $a\bar b + \bar a b$ by commutativity.

Answer 3

Note that $(a+b)\cdot(\overline{a} + \overline{b}) = (a+b)\cdot\overline{a} + (a+b)\cdot\overline{b}$ by the distributive property. Applying that property again to each term gives $$(a+b)\cdot\overline{a} + (a+b)\cdot\overline{b} = a\cdot\overline{a} + b\cdot\overline{a} + a\cdot\overline{b} + b\cdot\overline{b},$$ which as noted in other answers simplifies to $b\cdot\overline{a} + a\cdot\overline{b}$.