My Professor and I are debating about absorption law

Question

So the question is \$(w+y)(wz+wz')wy+y\$ and this is my answer by absorption law where \$A+AB=A\$:

$$ (w+y)(wz+wz!)w y + y\\ B A + A $$

So the answer is \$A=y\$.

My professor said I was wrong on using the identity because the long equation can't be considered as one variable.

I said since the long equation is in multiplication, by the rules of basic algebra, it can be considered as one term.

Help me. The picture is attached below

Answer 1

I like the OP's insight better but here is a step by step solution.

(W +Y)(W Z + W Z')(W Y) + Y

(W + Y) {W (Z + Z')} (W Y) + Y

(W + Y) {W (1)}( W Y) + Y

(W + Y) (W) (W Y) + Y

(W) (W + Y) (W Y) + Y

(W W + W Y)(W Y) + Y

(W + W Y) (W Y ) + Y

{W(1 + Y)} (W Y) + Y

{W (1)}(W Y)+ Y

W (W Y) + Y

W W Y + Y

W Y + Y

Y (W + 1)

Y (1)

Y

Answer 2

You are correct (although it's not 'basic algebra').

You can prove it by exhaustively evaluating for all 8 combinations of W,Y, Z.

Answer 3

You had \$ (w+y)(wz+wz')w y + y \$. Let's group it like \$ [(w+y)(wz+wz')w] y + y \$ and look at the subexpression in the brackets.

If this is boolean algebra, then whatever the values of \$ w, y, z \$ are, the subexpression \$ (w+y)(wz+wz')w \$ must be either true (1) or false (0). Not 123, undefined, a cat, or anything else. It can't turn into something completely different just because the expression has a few parts.

So, it has to play by the usual rules, and we can e.g. write a truth table for the whole thing:

[(w+y)(wz+wz')w]	y	[(w+y)(wz+wz')w] y	[(w+y)(wz+wz')w]y + y
0	0	0	0
0	1	0	1
1	0	0	0
1	1	1	1

That's all the possibilities there are. It's rather clear then that the full expression \$ (w+y)(wz+wz')w y + y \$ is equal to \$ y \$.

For the same reason, we could have just given that subexpression some shorter name and saved a bit of typing there, but I guess it might be easier to digest this way (for the professor, I mean).

Now, it's possible they meant this is an exercise or test on other things too, like the subexpression \$ (wz+wz') \$, which also rather obviously simplifies down to \$ w \$. Nothing wrong in that, but it comes to mind they might be a bit miffed about leaving an opening that made it possible to skip a large part of the task they tried to give.

Answer 4

Slightly shorter than Syed's proof is this

(W +Y) (W Z + W Z') (W Y) + Y

(W + Y) (W (Z + Z')) (W Y) + Y

(W + Y) (W (1)) (W Y) + Y

(W + Y) W (W Y) + Y

W (W Y) + Y

(W Y) + Y

Y

Answer 5

Depending on the level of rigour / appeal to axioms expected of you, your professor may be expecting you to state that logical AND and logical OR are both commutative, so that \$A+AB = BA+A\$, but regardless, you're correct. Let \$A = y\$, \$B = (w+y)(wz+w\overline{z})w\$, then the expression reduces to \$BA+A\$, which equals \$A+AB\$, which equals \$A\$ by the absorption law, which we know is \$y\$.

Answer 6

You don't need to learn so many "laws".

Initial expression

$$(w+y) \cdot (w \cdot z+w \cdot \overline{z}) \cdot w \cdot y + y$$

First, apply identity for AND

$$(w+y) \cdot (w \cdot z+w \cdot \overline{z}) \cdot w \cdot y + 1 \cdot y$$

Now, grouping (anti-distributivity)

$$\left[ (w+y) \cdot (w \cdot z+w \cdot \overline{z}) \cdot w + 1 \right] \cdot y$$

Now, the absorbing-point for OR reduces everything inside the brackets

$$\left[ 1 \right] \cdot y$$

And finally identity for AND once again

$$y$$

Done.

You were right not to attempt simplification of any subexpressions left of the \$+ y\$

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What you have called "absorption law (in two variables)" is a consequence of the absorbing-point for OR, namely $$\forall x, 1 + x = 1$$

This is the only one of the basic identities of Boolean algebra that is not shared with ordinary algebra (in ordinary arithmetic, addition has no absorbing-point).

There is a similar absorbing-point identity for AND (this one applies to ordinary multiplication of course) $$\forall x, 0 \cdot x = 0$$

The great thing about these absorbing points is that they absorb any expression, not just a simple variable.