The Boolean function AB + AC is equivalent to ______.
ABC + ABC' + AB'C
Understanding the equivalence of Boolean functions is a fundamental concept in digital electronics and Boolean algebra. We are asked to find the equivalent expression for the Boolean function \(AB + AC\).
To determine the equivalent expression from the given options, we can expand the original function \(AB + AC\) into a Sum of Products (SOP) form that includes all three variables (A, B, C) in each term, similar to the format of the options.
We will expand each term in the original expression \(AB + AC\) by introducing the missing variables. This is done using the Boolean identity \(X = X \cdot 1\) and \(1 = Y + Y'\), where \(Y\) is the missing variable. So, we use \(X = X(Y + Y')\).
The term \(AB\) is missing the variable \(C\). We can introduce \(C\) by multiplying the term with \((C + C')\):
\(AB = AB(C + C')\)
\(AB = ABC + ABC'\)
The term \(AC\) is missing the variable \(B\). We can introduce \(B\) by multiplying the term with \((B + B')\):
\(AC = AC(B + B')\)
\(AC = ABC + AB'C\)
Now, we combine the expanded forms of both terms:
\(AB + AC = (ABC + ABC') + (ABC + AB'C)\)
We notice that the term \(ABC\) appears twice in the expression. According to the Boolean algebra idempotence law, \(X + X = X\). Therefore, \(ABC + ABC\) simplifies to just \(ABC\).
Applying the idempotence law, our combined expression becomes:
\(ABC + ABC' + AB'C\)
This is the simplified Sum of Products form for the given Boolean function \(AB + AC\).
Let's compare our derived expression with the provided options:
Our derived equivalent expression \(ABC + ABC' + AB'C\) exactly matches Option 4. This demonstrates how the original Boolean function \(AB + AC\) can be expressed in an expanded form by including all variables in each product term and applying basic Boolean algebra identities.
How many different Boolean functions of degree n are there?
Consider a Boolean function of ‘n’ variables. The order of an algorithm that determines whether the Boolean function produces a output 1 is:
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