Boolean Algebra

[sxy]

Hi
Why the folllowing equation is correct:
L'C+I'C+L'I=L'C+L'I
thanks

 

[andre_luis]

Did you try put these terms into the K-map ?
There you can find out what terms can be simplified.

 

[std_match]

Hi
Why the folllowing equation is correct:
L'C+I'C+L'I=L'C+L'I
thanks

It is obviously not correct. The right side always depends on L. The left side can be true regardless of L.

 

[kaz1]

Hi
Why the folllowing equation is correct:
L'C+I'C+L'I=L'C+L'I
thanks

Using consensus theorem (XY + X'Z + YZ = XY + X'Z)
let L' = X
let C = Y
let I = Z

Hence your expression becomes:
XY + Z'Y + XZ
rearrange as XZ + YZ' + XY

Hence based on above theory XY is redundant:
= XZ + YZ'
i.e. = I'C + L'I

So either I or you may have typing error.

 

[sxy]

Using consensus theorem (XY + X'Z + YZ = XY + X'Z)
let L' = X
let C = Y
let I = Z

Hence your expression becomes:
XY + Z'Y + XZ
rearrange as XZ + YZ' + XY

Hence based on above theory XY is redundant:
= XZ + YZ'
i.e. = I'C + L'I

So either I or you may have typing error.

Hi
If you do the truth table for every side of the equation;
you will see that they are equal to each other; but I didn't succed to prove in in the terms of Boolean Algebra.
thank you very much for your answer.

 

[andre_luis]

Have you even tried the approach proposed in post #2?

 

[albbg]

As already said by std_match in post #2 the equation can't be correct. If you need an example simply let: I=0, C=1 and L=1 then LHS we have

0*1 + 1*1 + 0*0 = 1

RHS, intstead

0*1 + 0*0 = 0

Thus RHS <> LHS

Possibly your equation contains a typo, please check it.