Re: Algebraic Manipulation

Hi Sham,

Proving Boolean algebra equations is just like proving other equations. You should start with one side, and, by a chain of equations, reach the other side.

(x+y)(x+z) = xx+xz+xy+yz = x + xz +xy +yz = x(1 + z +y) + yz = x1+yz = x + yz

(x+y)(x+y!) = xx +xy +xy! +yy! = x + xy +xy! + 0 = x(1+y+y!) = x1 = x

Re: Algebraic Manipulation

John, thanks so much! But what is **(1 +y + y!)** = to? I know y*y! = 0 but what about **(1+y+y!)**?

Re: Algebraic Manipulation

y OR y! is 1. If y is false then not y is true

Also 1 OR [anything]=1

Do you understand the OR opperation?

X OR Y is asking if either X is true or Y is true. Therefore 1 OR X is asking if either 1 is true or X is true. 1 is always true so regardless of the value of X: 1 OR X is always true

Re: Algebraic Manipulation

Hi Shamieh,

In Boolean algebra 1 + anything = 1; i.e. true or anything is true no matter what the truth value of anything.

Re: Algebraic Manipulation

Quote:

Hi Shamieh,

In Boolean algebra 1 + anything = 1; i.e. true or anything is true no matter what the truth value of anything.

So for example, x(1 + x +y +x +y +x +z +x +y) you are saying that that will evaluate to 1? Getting x AND 1 which evaluates to x for example? And I'm not sure what shakari is talking about because y AND y! definitely evaluates to **0**. Unless my book is lying...