# Electronic – Formulas for Carry out in a Full Adder

Given a full adder with inputs $$A, B \text{ and } C_{in}$$
The formulas for the outputs are $$S = A \oplus B \oplus C_{in} \text{ , Where } \oplus \text{means XOR}$$ and $$C_{out} = AB + AC_{in} + BC_{in}$$

But when a Full Adder is created by combining two half-adders, the obtained equation for carry out is $$C_{out} = AB + (A \oplus B)C_{in}$$

The two expressions have equivalent Truth Tables, but the reason for their equality is not obvious.

Can someone please help me understand how it can easily be seen they're equal?

Start off with the second expression $$C_{out} = A B + (A \oplus B)C_{in}$$ Using the expansion, $$A \oplus B = \bar AB + A \bar B \quad \text{ for XOR}$$ The right hand side becomes $$AB + \bar A B C_{in} + A \bar B C_{in} \tag{1}$$ Taking B common between the forst two terms, we get (Property of Boolean Algebra) $$B(A + \bar AC_{in})$$ Using the fact that addition distributes over multiplication (In boolean Algebra) $$A + BC = (A + B)(A + C) \tag{2}$$ Hence the term becomes $$B((A + \bar A)(A + C_{in})) = AB + BC_{in}$$ Plugging this back into (1) and reusing property (2) with $$B+ \bar B C_{in}$$ We get the alternate form $$AB + BC_{in} + C_{in} \tag {3}A$$ Intuitive Approach
In boolean form this would be, $$AB + (A+B)C_{in}$$ Which is the same as (3)