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
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
You are correct (although it's not 'basic algebra').
You can prove it by exhaustively evaluating for all 8 combinations of W,Y, Z.
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.
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
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\$.
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\$
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.
