Question

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

Answer 1

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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