Proving Boolean Expressions Using Algebraic Identities and Consensus Theorem

Boolean algebra is a fundamental part of digital electronics and computer science, particularly in the design of digital circuits and systems. Understanding how to simplify Boolean expressions is crucial for creating efficient and reliable digital systems. In this article, we will explore the proof of the Boolean expression xz xy yz xz y using algebraic identities and the Consensus Theorem.

Step-by-Step Proof Using Boolean Algebra Identities

Step 1: Start with the Original Expression

Let us start with the expression:

xz xy yz

Step 2: Apply the Consensus Theorem

The Consensus Theorem in Boolean algebra states that:

ab ac bc ab ac

In our case, we apply the theorem to the terms xy yz with:

a y b z c x

This allows us to rewrite xy yz as:

yx z

Substituting this back into the original expression, we get:

xz yx z

Step 3: Simplify Using Absorption Law

Next, we simplify yx z. According to the Absorption Law, which states that:

ab ac ab if c ab

We can use this to simplify xz y xz

Let us simplify xz y xz:

If y 1, then xz xz If y 0, then xz 0, which depends on the value of xz

Thus, we can simplify this to:

xz y

Conclusion

We now have simplified the left-hand side:

xz xy yz xz y

This completes the proof demonstrating that:

xz xy yz xz y

By using Boolean algebra identities and the Consensus Theorem, we have successfully simplified and proven the given Boolean expression.

Related Proofs:

Using De Morgan's Laws

Let's also explore the use of De Morgan's Laws in a related expression:

xz (overline{X} overline{Y} overline{Y}) overline{Z} xz (overline{XY} overline{YZ})

Using AB AC ABC from the distributive law, we can simplify:

xz overline{YXZ} xz overline{Y} overline{XZ}

And further simplify to:

xz overline{Y}

Again using the distributive law:

xy overline{x} x overline{x}xy xy

Q.E.D.

Using Karnaugh Maps

Let's verify a similar expression using a Karnaugh Map:

xz x' y' y' z' xz y'

A Karnaugh Map will help you see:

x' y' x' y' 1 x' y' z z' x' y' z x' y' z' z z' 1 (since either z or z' is always true) y' z' y' z' 1 y' z' x x' y' xz' x'z' x x' 1

Thus, we get:

xz x' y' z' xz y' xz y' x'z x'z' y' xz'

Simplifying further:

xz y'x[zz'] x' [z z'] xz y'x 1 x'1 xz y'x x'

Finally:

xz y' 1 xz y'

LHS: xz xy yz

xz yx z

xz y overline{xz}

Let P xz, then:

LHS PyP

PPyP

1Py

xzy RHS Q.E.D.