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Prove Co = AB + C¡ (A ⊕ B) = BC¡+AC¡ + AB​

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To prove the equation
\displaystyle\sf Co = AB + C\overline{A \oplus B} = BC\overline{A} + AC\overline{B} + AB, we will manipulate the given expression using Boolean algebra laws and logic operations.

Starting with the left-hand side of the equation:


\displaystyle\sf AB + C\overline{A \oplus B}

We can expand the term
\displaystyle\sf C\overline{A \oplus B} using the XOR operation:


\displaystyle\sf C\overline{(A \cdot \overline{B} + \overline{A} \cdot B)}

Applying De Morgan's law to the expression inside the complement:


\displaystyle\sf C\overline{(A \cdot \overline{B}) \cdot (\overline{A} \cdot B)}

Distributing the complement over the product:


\displaystyle\sf C(\overline{A \cdot \overline{B}} + \overline{\overline{A} \cdot B})

Using De Morgan's law again:


\displaystyle\sf C(\overline{A} + B) \quad (1)

Now, let's simplify the right-hand side of the equation:


\displaystyle\sf BC\overline{A} + AC\overline{B} + AB

Factoring out A from the second term and B from the third term:


\displaystyle\sf (BC + AC)\overline{A} + AB

Applying the distributive law:


\displaystyle\sf (B + A)C\overline{A} + AB

Using the absorption law, which states that
\displaystyle\sf X + XY = X:


\displaystyle\sf BC\overline{A} + AB

Comparing the right-hand side expression with the expression (1) derived from the left-hand side, we can see that they are equal.

Therefore, we have proven that
\displaystyle\sf Co = AB + C\overline{A \oplus B} = BC\overline{A} + AC\overline{B} + AB.


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