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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

enter image description here

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    \$\begingroup\$ 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. \$\endgroup\$
    – Syed
    Dec 22, 2021 at 5:24
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    \$\begingroup\$ "multiplication, by the rules of basic algebra..." - ??? en.wikipedia.org/wiki/Boolean_algebra \$\endgroup\$ Dec 22, 2021 at 6:12
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    \$\begingroup\$ You are right. Aaaand the professor needs training. \$\endgroup\$
    – Mitu Raj
    Dec 22, 2021 at 9:45
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    \$\begingroup\$ I’m voting to close this question because it's not about electricity or electronics. \$\endgroup\$ Dec 23, 2021 at 0:58

6 Answers 6

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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
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    \$\begingroup\$ 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. \$\endgroup\$
    – ilkkachu
    Dec 22, 2021 at 23:32
  • \$\begingroup\$ 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 \$\endgroup\$
    – Syed
    Dec 23, 2021 at 3:37
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    \$\begingroup\$ I seriously doubt the poster or their professor is a computer. \$\endgroup\$
    – ilkkachu
    Dec 23, 2021 at 8:36
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You are correct (although it's not 'basic algebra').

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

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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.

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

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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.

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  • \$\begingroup\$ '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. \$\endgroup\$
    – Jivan Pal
    Dec 23, 2021 at 1:04
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    \$\begingroup\$ @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. \$\endgroup\$
    – Ben Voigt
    Dec 23, 2021 at 6:04
  • \$\begingroup\$ 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. \$\endgroup\$
    – Jivan Pal
    Dec 23, 2021 at 9:15

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