How to simplify using algebra laws

Question

Simplify the following by using algebra laws.

(i) X'Y' + XYZ + X'Y + XY

My attempt:

X'Y' + Y(XYZ + X'Y + X.Y)

X'Y' + (XZ + X' + X)

X'(X'Y' + X') + XZ + X

Y' + X' + XZ + X

Y' + X' + X(XZ + X)

Y' + X' + Z

Correct?

(ii) X.Y.Z + XY'Z' + X'YZ + X'Y'Z

This question I don't know how to start..

My lecturer did not teach and ask us to do research on our own which I did but confused the laws.

Answer 1

Okay. Now that I know it is boolean algebra

X'Y'+XYZ+X'Y+XY=

Put the X' s together

X'Y'+X'Y +XYZ+XY=

and factor

X'(Y'+Y) + XYZ +XY=

Y+Y'=1 so

We have

X' + XYZ+XY =

Notice the remaining two terms have XY in common. Factor them out.

X'+XY (Z+1)

Note 1+x=1 so

X'+XY

That's about it. We could have factored out the Y first and got:

X'Y' + XYZ+X'Y+XY=

X'Y' +Y (XZ + X'+ X) =

X'Y' + Y (XZ+1)=

X'Y' + Y

Interestingly X'+XY = X'Y'+Y.

Hmmm, I actually don't know the algebraic rule to go the last step (XY')'

Answer 2

(i) X’.Y’ + X.Y.Z. + X’.Y + X.Y

My attempt:

X’.Y’ + Y(X.Y.Z + X'Y + X.Y)

X’.Y’ + (X.Z + X' + X)

That doesn't make a lick of sense. You've tried to distribute $Y$ out of the conjunctions (okay) then cancelled it with... itself? That won't work.

Study your Laws of Boolean Algebra some more.   (Note that that wiki page uses $\wedge$ for 'and' (conjunction) and $\vee$ for 'or' (disjunction), rather than $\cdot, +$)

$$\begin{align}&\rm X'\cdot Y'+ (X\cdot Z+X'+X)\cdot Y& \text{Distribution} \\& \rm X'\cdot Y'+(X\cdot Z+1)\cdot Y & \text{Disjunctive Complement} \\& \rm X'\cdot Y'+1\cdot Y & \text{Disjunctive Annihilation} \\& \rm X'\cdot Y'+Y & \text{Conjunctive Identity} \\& \rm \underline{\qquad\qquad} & \text{Distribution} \\& \rm \underline{\qquad\quad} & \text{Disjunctive Complement} \\& \rm \underline{\qquad} & \text{Conjunctive Identity} \end{align}$$

Can you fill in the blanks?

(ii) X.Y.Z + X.Y’.Z’ + X’.Y.Z + X’.Y’.Z

This question I don't know how to start..

$$\begin{align} &\rm X\cdot Y'\cdot Z'+(X\cdot Y+X'\cdot Y+X'\cdot Y')\cdot Z& \text{Commutation & Distribution} \\ & \rm X\cdot Y'\cdot Z'+(X\cdot Y+X'\cdot Y+X'\cdot Y+X'\cdot Y')\cdot Z& \text{Idempotence} \\ & \rm X\cdot Y'\cdot Z'+((X+X')\cdot Y+X'\cdot(Y+Y'))\cdot Z& \text{Distribution} \\ & \vdots \end{align}$$

...and so forth. Can you keep going now?

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Remark: Idempotence is a powerful but counterintuitive 'trick'.   Sometimes when simplifying it actually makes sense to copy a term, via $[x = x+x]$ or $[x=x\cdot x]$ , when you can see ways for it to interact usefully with two different parts of the sequence.

Always remember, the laws of simplification are equalities, and sometimes using them in reverse can supply you with two terms needed to simplify different areas.   However, it is a bit of an art to realising when to do this; pattern recognition comes with practice.

Answer 3

$\overline X\ \overline Y + XYZ + \overline XY + XY$

$\overline X\ (\overline Y + Y) + XY (1 + Z)$ Distributive Law.

$\overline X\ + XY$ Complement Law: $\overline A + A = 1$ and Annulment Law: $1 + A = 1$

$\overline X\ + Y$ Redundancy Law: $A + \overline A B = A + B$

Second one: Look for common terms (Distributive). Duplicate terms as needed (Idempotent). Then apply Complement.

$XYZ + X\ \overline Y\ \overline Z + \overline XYZ + \overline X\ \overline YZ + \color {red} {\overline XYZ}$

$X\ \overline Y\ \overline Z + \color {blue} {\overline XYZ + \overline X\ \overline YZ} + \color {red} {\overline XYZ + XYZ}$

This will get you started. Edit your question to show your workings if you are still unclear but sometimes boolean expressions do not simplify to simple expressions. ii is one of those.

Laws and Theorems of Boolean Algebra