How to simplify ab + a'b'? Is there a gate for this?
If you are trying to use the various laws of Boolean algebra to go from the equation you have to a pair of XOR gates, it is not too tricky, you just have to spot similarities.

Lets look at your equation, what can we do?

$$x \cdot \overline y \cdot \overline c + \overline x \cdot \overline y \cdot c + x \cdot y \cdot c + \overline x \cdot y \cdot \overline c$$


Try taking out factors:

Notice there are some \$c\$ terms in common:

$$\begin{align}
  x \cdot \overline y \cdot \overline c + \overline x \cdot y \cdot \overline c &amp;= (x \cdot \overline y + \overline x \cdot y) \cdot \overline c\\
  \overline x \cdot \overline y \cdot c + x \cdot y \cdot c &amp;= (\overline x \cdot \overline y + x \cdot y) \cdot c\\
  \end{align}$$

That would leave us with:

$$(\overline x \cdot \overline y + x \cdot y) \cdot c + (x \cdot \overline y + \overline x \cdot y) \cdot \overline c$$
Using the standard table of Boolean algebra functions, we can easily spot the XNOR and XOR gates:


$$(\overline x \cdot \overline y + x \cdot y) \cdot c + (x \cdot \overline y + \overline x \cdot y) \cdot \overline c = \overline{(x \oplus y)}\cdot c + (x \oplus y)\cdot\overline c$$


Lets add a dummy variable to see things more clearly. Let \$d = (x \oplus y)\$. Then substitute it in to the above:


$$\overline{(x \oplus y)}\cdot c + (x \oplus y)\cdot\overline c = \overline d \cdot c + d \cdot \overline c$$


Spot the XOR gate there?


$$\overline d \cdot c + d \cdot \overline c = d \oplus c$$


Finally, lets plop back in \$d\$.


$$d \oplus c = (a \oplus b) \oplus c $$