# Help with Boolean Algebra Simplification

#### Emile_Uk

Joined Nov 17, 2003
1
Hi,

If anyone can help me simplify the following two boolean expressions whilst listing the laws used at each stage then that would be hugely appreciated!

The two expressions are:

([P]*Q*[X]*Y)+([P]*Q*X*[Y])+(P*[Q]*X*[Y])+(P*[Q]*X*Y)+(P*Q*[X]*Y)+(P*Q*X*Y)

and

(P*[Q]*[X]*Y)+(P*Q*[X]*Y)+(P*Q*X*[Y])

Where * is AND and [] is an inverted input.

Any help would be greatly appreciated!

Cheers,

Emile

#### Dave

Joined Nov 17, 2003
6,969
Firstly the second one (because its easier):

(P*[Q]*[X]*Y)+(P*Q*[X]*Y)+(P*Q*X*[Y])

Deal with the first two brackets and take out P*[X]*Y as common factors

Therefore: P*[X]*Y([Q]+Q) where [Q]+Q = 1 (Complementary Law)

So that gives (P*[X]*Y) + (P*Q*X*[Y]) (with the third bracket reintroduced)

Now take out P as a common factor:

So you get P*{([X]*Y) + (Q*X*[Y])}

Thats it in its simplest form.

The first one is considerably more difficult:

([P]*Q*[X]*Y)+([P]*Q*X*[Y])+(P*[Q]*X*[Y])+(P*[Q]*X*Y)+(P*Q*[X]*Y)+(P*Q*X*Y)

Look at brackets 3 and 4 and take out P*[Q]*X as common factors

So P*[Q]*X([Y]+Y) where [Y]+Y = 1

Do the same with brackets 5 and 6 taking P*Q*Y out as common factors

So P*Q*Y([X]+X] where [X]+X = 1

Brackets 1 and 2 are just take out [P]*Q as common factors and get:

[P]*Q*{([X]*Y)+(X*[Y])}

I haven't had chance to look at this with another method but will post back if the answer comes out any different. I'd also appreciate comments on any errors I have made or if you'd like any further help 