Q12Digital Electronics

In formal Boolean algebra, a mathematical expression is strictly considered "Canonical" or in "Standard Form" if, and only if, every single distinct term within the expression explicitly contains every single variable present in the entire functional domain (whether in its normal or complemented/inverted state). The function provided is $f(A,B,C)=A+BC$. This expression is currently presented in a heavily minimized state. Our objective is to mathematically expand it into a Canonical Product of Sums (POS) form, where the overarching structure is a multiplication (ANDing) of several addition (ORing) terms (Maxterms).

Step 1: Distributive Law Application

First, we must structurally convert the expression from its mixed form into a basic Product of Sums format by applying the fundamental Boolean distributive law: $X+YZ=(X+Y)(X+Z)$. Applying this to our function: $f=A+BC$ $f=(A+B)\cdot (A+C)$

We now possess two distinct sum terms: $(A+B)$ and $(A+C)$. However, neither of these terms is canonical because they are both missing one specific variable from the complete set {A, B, C}.

Step 2: Algorithmic Expansion of Missing Variables

To expand these terms without mathematically altering their intrinsic value, we utilize the identity rule of addition: adding a boolean zero ($0$) changes nothing. In Boolean algebra, $X\cdot X^{\prime }=0$. Therefore, we will forcefully add the product of the missing variable and its complement to each incomplete term.

Expanding the First Term $(A+B)$: The variable strictly missing is $C$. We add $C\cdot C^{\prime }$: $=(A+B)+(C\cdot C^{\prime })$ Now, we aggressively apply the distributive law again, treating $(A+B)$ as a single monolithic unit: $=(A+B+C)\cdot (A+B+C^{\prime })$

Expanding the Second Term $(A+C)$: The variable strictly missing is $B$. We add $B\cdot B^{\prime }$: $=(A+C)+(B\cdot B^{\prime })$ Applying the distributive law: $=(A+B+C)\cdot (A+B^{\prime }+C)$

Step 3: Final Synthesis

We now recombine all the newly generated canonical maxterms together, meticulously replacing the original incomplete terms: $f=[(A+B+C)\cdot (A+B+C^{\prime })]\cdot [(A+B+C)\cdot (A+B^{\prime }+C)]$

Because in Boolean algebra, multiplying a term by itself yields the identical term ($X\cdot X=X$), we can safely eliminate any mathematically redundant maxterms. The term $(A+B+C)$ appears twice. Removing the duplicate yields the final, absolute Canonical POS expression: $f(A,B,C)=(A+B+C)\cdot (A+B+C^{\prime })\cdot (A+B^{\prime }+C)$ In strict Maxterm shorthand notation, this is accurately represented as $\Pi M(0,1,2)$.