# My Professor and I are debating about absorption law

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

• Verifying using Mathematica: BooleanMinimize[((w || y) && (w && z || w && ! z) && (w && y)) || y] gives y. You are right. Please edit and write title in lower-case. And make it about the question, not the prof.
– Syed
Dec 22, 2021 at 5:24
• "multiplication, by the rules of basic algebra..." - ??? en.wikipedia.org/wiki/Boolean_algebra Dec 22, 2021 at 6:12
• You are right. Aaaand the professor needs training. Dec 22, 2021 at 9:45
• I’m voting to close this question because it's not about electricity or electronics. Dec 23, 2021 at 0:58

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

• Well, isn't the point there that all of this is totally unnecessary, since we can immediately see that whatever $(w+y)(wz+wz')w$ is, it has no effect on the value of the expression? Now, if the expression turned into $(w+y)(wz+xz')wy + y$ instead (I changed one $w$ to $x$), you'd have to do the step-by-step simplification again, while the insight they've made just gives the result far more simply and quickly. Dec 22, 2021 at 23:32
• Computers don't have insight. They need to do a step by step solution. Hence it is important to know how it is done. To emphasize, the only way out for a computer (perhaps running a synthesis tool) would be a step by step solution even for the simplest of cases. @ilkkachu
– Syed
Dec 23, 2021 at 3:37
• I seriously doubt the poster or their professor is a computer. Dec 23, 2021 at 8:36

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.

• No, you are right. I missed the extra y. Dec 23, 2021 at 0:20
• @Math, ok, thanks. I tried to clarify the post a bit. Dec 23, 2021 at 0:33

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.

• 'You don't need to learn so many "laws".' — Depending on who you ask, the absorption law is a defining property of Boolean algebra, not merely an emergent theorem, so it is reasonable to expect it be known by those studying it. Dec 23, 2021 at 1:04
• @JivanPal: But it can be proved using simpler properties, and it's quite perverse to choose a complicated basis set when a simpler one works just as well. Particularly because the simpler properties appear elsewhere in abstract algebras, such as groups and fields (two operators denoted $A+B$ and $A \cdot B$, with commutivity, distributivity, and identity elements for each operator), sets (negation $\overline A$, exclusivity of $A \cup \overline A = {\bf 0}$, coverage of $A \cap \overline A = \Omega$), and then theory of Markov chains gives us absorbing states. Dec 23, 2021 at 6:04
• I said "depending on who you ask", because you can also formulate Boolean algebra as just a bounded lattice with distributivity of × over +, and the absorption law is a defining property of a (bounded) lattice. "Simpler" is subjective. If you use "simpler" to mean "having fewer axioms", then see here. Dec 23, 2021 at 9:15