Efficient Boolean Minimization Techniques: Simplifying Y = \bar{X}_1+\bar{X}_0

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SUMMARY

The discussion focuses on Boolean minimization techniques, specifically simplifying the expression Y = \bar{X}_1\bar{X}_0 + \bar{X}_1X_0 + X_1\bar{X}_0 to Y = \bar{X}_1 + \bar{X}_0. The user successfully factored \bar{X}_1 from the first two terms and applied the reduction of \bar{X}_0 + X_0 to 1. The query revolves around understanding how the last term simplifies to \bar{X}_0, emphasizing the distributive properties of Boolean algebra.

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CentreShifter
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This is just a general question regarding Boolean minimization.

Expression:
[tex]Y=\bar{X}_1\bar{X}_0+\bar{X}_1X_0+X_1\bar{X}_0[/tex]

Minimized expression:
[tex]Y=\bar{X}_1+\bar{X}_0[/tex]

My first attempt was to minimize it algebraically. I factored [tex]\bar{X}_1[/tex] from the first two terms, then the [tex]\bar{X}_0+X_0[/tex] reduce to 1. So I end up with [tex]\bar{X}_1+X_1\bar{X}_0[/tex].

My question then is, how does the second term (last term from the original expression) reduce to [tex]\bar{X}_0[/tex] to end up with the known correct expression?
 
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Multiply the first term by

[tex]\bar{X}_0+1[/tex]
 
Try factoring out [tex]\bar{X_1}[/tex] from the first two terms using the distributive property.

Also, remember that in boolean algebra not only does multiplication (AND) distribute over addition (OR), but addition also distributes over multiplication. That is,

[tex]A(B + C) = AB + AC[/tex]

and

[tex]A + (B C) = (A + B)(A + C)[/tex]
 
Last edited:

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