Q15Digital Electronics

Boolean algebra relies on fundamental axioms and theorems to systematically minimize complex digital logic circuits. This mathematical exercise demands the rigorous step-by-step proofs of two specific Boolean identities.

Proof i) Show that: $AB+A^{\prime }C+BC=AB+A^{\prime }C$

This highly specific equation is mathematically formally known as the Consensus Theorem. It proves that if you have two terms containing a variable ($A$) and its exact complement ($A^{\prime }$), and a third term containing the remaining variables from the first two terms ($BC$), that third term is mathematically redundant and can be safely permanently deleted.

Step 1 (Expansion): We begin with the Left Hand Side (LHS) expression: $LHS=AB+A^{\prime }C+BC$. To mathematically manipulate the redundant term $BC$, we aggressively multiply it by the Boolean identity $1$, physically represented by the term $(A+A^{\prime })$. $LHS=AB+A^{\prime }C+BC\cdot (A+A^{\prime })$

Step 2 (Distribution): We mathematically distribute the $BC$ term into the parenthesis. $LHS=AB+A^{\prime }C+ABC+A^{\prime }BC$

Step 3 (Rearrangement and Factoring): We group terms that share common variables to isolate expressions. $LHS=(AB+ABC)+(A^{\prime }C+A^{\prime }BC)$ Now, we extract the common factors out of each group. Extract $AB$ from the first, and $A^{\prime }C$ from the second. $LHS=AB(1+C)+A^{\prime }C(1+B)$

Step 4 (Simplification via Null Elements): According to Boolean axioms, anything ORed with 1 is definitively 1. Therefore, $(1+C)=1$ and $(1+B)=1$. $LHS=AB(1)+A^{\prime }C(1)$ $LHS=AB+A^{\prime }C$ This flawlessly equals the Right Hand Side (RHS). The Consensus Theorem is rigorously proven.

Proof ii) Show that: $AB+A^{\prime }C=(A+C)(A^{\prime }+B)$

This proof demonstrates the translation of a Sum of Products (SOP) into a Product of Sums (POS) structure via distributive laws.

Step 1 (Starting with RHS): It is significantly easier to prove this by aggressively multiplying out (distributing) the Right Hand Side to see if it mathematically collapses back into the Left Hand Side. $RHS=(A+C)(A^{\prime }+B)$

Step 2 (FOIL Method Expansion): Distribute every term mathematically. $RHS=A\cdot A^{\prime }+A\cdot B+C\cdot A^{\prime }+C\cdot B$

Step 3 (Simplification via Axioms): By the Boolean axiom of complementarity, a variable ANDed with its inverse is always zero ($A\cdot A^{\prime }=0$). $RHS=0+AB+A^{\prime }C+BC$

Step 4 (Application of the Consensus Theorem): The resulting expression is exactly $AB+A^{\prime }C+BC$. Based entirely on the exhaustive proof we just executed in part (i), we know with absolute mathematical certainty that the final term $BC$ is structurally redundant. $RHS=AB+A^{\prime }C$ This equals the LHS. The equation is proven valid.