Mengapa F + F '= 1?

15

Saya memiliki fungsi: $f(x,y,z,w) = wx + yz$

Saya menemukan fungsi komplemennya adalah: $f '(x,y,z,w) = w'y' + w'z' + x'y' + x'z'$

Saya harus menunjukkan bahwa: $f + f '=1$ tetapi saya tidak bisa melihat bagaimana melakukannya.

Sepertinya tidak ada yang membatalkan satu sama lain.

Edit

Seperti yang disarankan, saya sekarang menggunakan teorema DeMorgan dan menemukan ini:

$f + f' = wx+yz+(w+y)'+(w+z)'+(x+y)'+(y+z)'$

Tapi sepertinya masih ada yang membuat saya lebih dekat dengan realisasi $f+f' = 1$

boolean-algebra

— Carl
sumber

6

Petunjuk: Gunakan Hukum DeMorgan

— Spehro Pefhany

11

Baik f atau f 'harus 1

— Chu

4

Anda hanya memiliki 4 input. Jika tidak ada yang lain, Anda bisa menulis tabel kebenaran.

— The Photon

2

Spehro benar dalam hal uang, tetapi ya menerapkan DeMorgan sebagai langkah pertama tidak membantu. Jadi untuk memperluas sedikit petunjuk Spehro: solusinya melibatkan melakukan beberapa aljabar dasar, yang mencakup DeMorgan sebagai langkah. Menggunakan aljabar + DeMorgan sederhana, Anda dapat mengubah fungsi f 'menjadi negasi yang jelas-jelas dari f. Menuliskannya di selembar kertas, hanya butuh 4 langkah untuk melakukannya.

— Tn. Snrub

1

@ Mr.Snrub langkah pertama "Saya menemukan fungsi komplemennya" seharusnya (wx + yz) ′

— OrangeDog

4

Karena Carl bertanya dengan baik. Titik awal:

f (x, y, z, w) = w x + y z

$f(x,y,z,w)=wx+yz$ dan

f' (x, y, z, w) = w' y' + w' z' + x' y' + x' z'

$f′(x,y,z,w)=w′y′+w′z′+x′y′+x′z′$

Ambil langkah-langkah berikut dengan $f'$ :

f' (x, y, z, w) = w' (y' + z') + x' (y' + z')

$f′(x,y,z,w)=w′(y′+z′)+x′(y′+z′)$

f' (x, y, z, w) = (w' + x^{'}) (y' + z')

$f′(x,y,z,w)=(w′+x')(y′+z′)$ DeMorgan:

f' (x, y, z, w) = (w x)' (y z)^{'}

$f′(x,y,z,w)=(wx)′(yz)'$ DeMorgan, sekali lagi:

f' (x, y, z, w) = (w x + y z)^{'}

$f′(x,y,z,w)=(wx + yz)'$ Jadi sekarang sisi kanan

f^{'}

$f'$ hanyalah negasi sederhana dari sisi kanan

f

$f$ . Yang sedikit anti-klimaks, karena sekarang kita hanya mengandalkan fakta bahwa setiap ekspresi

x + x^{'} = 1

$x + x' = 1$ , yang telah dikatakan orang selama ini tentang

f + f^{'} = 1

$f+f'=1$ , tapi setidaknya itu memberikan sedikit Penjelasan aljabar Boolean mengapa itu benar.

— Tuan Snrub
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Saya tidak mengerti bagaimana Anda sampai ke baris kedua tanpa melewati jawaban akhir Anda. Jawaban akhir Anda adalah langkah pertama saya: itu hanya negasi dari kedua belah pihak.

— C. Lange

The first two lines are the formulas given by OP. They are the starting point, by definition. I fully agree that the stuff later on may have been part of OP's derivation of those first two formulas. But we don't have that information; we just cannot confirm.

— Mr. Snrub

Understood -- on the assumption that

f

$f$ and

f^{'}

$f'$ were given in the question like OP has written them out. My understanding was that OP had already tried to expand

f^{'}

$f'$ and didn't know where to go from there.

— C. Lange

41

The point is, it really doesn't matter what the function $f()$ actually is. The key fact is that its output is a single binary value.

It is a fundamental fact in Boolean algebra that the complement of a binary value is true whenever the value itself is false. This is known as the law of excluded middle. So ORing a value with its complement is always true, and ANDing a value with its complement is always false.

It's nice that you were able to derive the specific function $f'()$ , but that's actually irrelevant to the actual question!

— Dave Tweed
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1

This is known as the law of excluded middle.

— BallpointBen

@BallpointBen: Thanks! I added it to my answer.

— Dave Tweed

13

All previous answers are correct, and very much in depth. But a simpler way to approach this might be to remember that in boolean algebra, all values must be either 0 or 1.

So... either F is 1, then F' is 0, or the other way around: F is 0 and F' is 1. If you then apply the boolean OR-function: F + F', you will always have one of both terms 1, so the result will always be 1.

— Opifex
sumber

11

My answer is similar to the one of Dave Tweed, meaning that I put it on a more formal level. I obviously answered later, but I decided to nevertheless post it since someone may find this approach interesting.

The relation you are trying to prove is independent from the structure of the function $f$ since it is, as a matter of fact, a tautology. To explain what I mean, I propose a demonstration for a general, correctly formed, Boolean expression $P$ in an arbitrary number of Boolean variables, say $n\in\Bbb N$ , $y_1,\ldots,y_n$ , where $y_i\in\{0,1\}$ for all $i=1,\ldots,n$ .
We have that $P(y_1,\ldots,y_n)\in\{0,1\}$ and consider the following two sets of Boolean values for the $n$ -dimensional Boolean vector $(y_1,\ldots,y_n)$

\begin{aligned} Y & = {(y_{1}, \dots, y_{n}) \in {0, 1}^{n} | P (y_{1}, \dots, y_{n}) = 1} \\ \bar{Y} & = {(y_{1}, \dots, y_{n}) \in {0, 1}^{n} | P (y_{1}, \dots, y_{n}) = 0} \end{aligned}

$\begin{align} Y&=\{(y_1,\ldots,y_n)\in\{0,1\}^n|P(y_1,\ldots,y_n)=1\}\\ \bar{Y}&= \{(y_1,\ldots,y_n)\in\{0,1\}^n|P(y_1,\ldots,y_n)=0\} \end{align}$ These set are a partition of the full set of values the input Boolean vector can assume, i.e.

Y \cup \bar{Y} = {0, 1}^{n}

$Y\cup\bar{Y}=\{0,1\}^n$ and

Y \cap \bar{Y} = \emptyset

$Y\cap\bar{Y}=\emptyset$ (the empty set), thus

\begin{aligned} P (y_{1}, \dots, y_{n}) & = {\begin{cases} 0 & if (y_{1}, \dots, y_{n}) \in \bar{Y} \\ 1 & if (y_{1}, \dots, y_{n}) \in Y \end{cases} \\ ⇕ \\ P^{'} (y_{1}, \dots, y_{n}) & = {\begin{cases} 1 & if (y_{1}, \dots, y_{n}) \in \bar{Y} \\ 0 & if (y_{1}, \dots, y_{n}) \in Y \end{cases} \end{aligned}

$\begin{align} P(y_1,\ldots,y_n)&= \begin{cases} 0&\text{if }(y_1,\ldots,y_n)\in \bar{Y}\\ 1&\text{if }(y_1,\ldots,y_n)\in Y\\ \end{cases}\\ &\Updownarrow\\ P'(y_1,\ldots,y_n)&= \begin{cases} 1&\text{if }(y_1,\ldots,y_n)\in \bar{Y}\\ 0&\text{if }(y_1,\ldots,y_n)\in Y\\ \end{cases} \end{align}$ therefore we always have

P + P^{'} = 1 \forall (y_{1}, \dots, y_{n}) \in {0, 1}^{n}

$P+P'=1\quad\forall(y_1,\ldots,y_n)\in\{0,1\}^n$

— Daniele Tampieri
sumber

11

All good answers that provide the necessary justification in one way or the other. Since it is a tautology, it's hard to create a proof that doesn't just result in "it is what it is!". Perhaps this method help tackle it from yet another, broader angle:

Expand both statements to include their redundant cases, and the remove the repeated cases:

$𝑓=𝑤𝑥+𝑦𝑧\\ \ \ =wx\cdot(y'z'+y'z+yz'+yz)\ +\ yz\cdot(x'w'+x'w+xw'+xw)\\ \ \ =wxy'z'+wxy'z+wxyz'+wxyz\ +\ yzx'w'+yzx'w+yzxw'+yzxw\\ \ \ =wxy'z'+wxy'z+wxyz'+wxyz\ +\ yzx'w'+yzx'w+yzxw'$

and

$𝑓′=𝑤′𝑦′+𝑤′𝑧′+𝑥′𝑦′+𝑥′𝑧′\\ \ \ \ = w'y'\cdot(x'z'+x'z+xz'+xz)\ +\ 𝑤′𝑧′\cdot(x'y'+x'y+xy'+xy)\ +\\ \ \ \ \ \ \ \ \ \ x′y′\cdot(w'z'+w'z+wz'+wz)\ +\ x′𝑧′\cdot(w'y'+w'y+wy'+wy)\\ \ \ \ = w'y'x'z'+w'y'x'z+w'y'xz'+w'y'xz\ +\ 𝑤′𝑧′x'y'+𝑤′𝑧′x'y+𝑤′𝑧′xy'+𝑤′𝑧′xy\ +\\ \ \ \ \ \ \ \ \ \ x′y′w'z'+x′y′w'z+x′y′wz'+x′y′wz\ +\ x′𝑧′w'y'+x′𝑧′w'y+x′𝑧′wy'+x′𝑧′wy\\ \ \ \ = w'y'x'z'+w'y'x'z+w'y'xz'+w'y'xz\ +\ 𝑤′𝑧′x'y+𝑤′𝑧′xy\ +\\ \ \ \ \ \ \ \ \ \ x′y′wz'+x′y′wz\ +\ x′𝑧′wy$

I've kept the terms in consistent order to make the derivation more obvious, but they could be written alphabetically to be clearer. In any case, the point is that $f$ ORs seven 4-bit cases, and $f'$ ORs nine, distinct 4-bit cases. Together they OR all sixteen 4-bit cases, so reduce to $1$ .

— Heath Raftery
sumber

4

+1 this is the only answer that is answering the true intention of the OPs question, which is to do some Boolean algebra rather than making theoretical arguments. But per my comment on the OP, note that a more elegant solution does exist; this problem can be solved without needing to add in the redundant cases.

— Mr. Snrub

I would very much like to see that as well. That is, if you have the time and the generosity to do it.t

— Carl

8

F + F' = 1 means that you have to show that no matter the state of the 4 inputs, OR'ing the result of those 2 always result in 1,

A few minutes in excel shows it is indeed the case. You can use "NOT()" to invert between 0 and 1 in excel.

F = W * X + Y * Z

F' = W' * Y' + W' * Z' + X' * Y' + X' * Z'

As to why this is the case, If you want F to be false, e.g. setting W and Y low, you just made F' true. If you make X and Z low, you also made F" true, same for swapping there pairs.

— Reroute
sumber

2

"F + F' = 1 means that you have to show that no matter the state of the 4 inputs, OR'ing the result of those 2 always result in 1" . No, it doesn't. It merely means that you have to show that regardless of the output (which can only have two possibilities) and the corresponding output of its complement, the relation holds. The inputs are irrelevant, as is the function. The only truth table needed is the one showing the relationship between the output of the function and the output of anything qualifying as its complement.

— Chris Stratton

@ChrisStratton, that depends if the question is to show that the OR of a function and its complement is always 1 (which is trivial by definition of the complement) or to show that the proposed function F' is actually the complement of F. From OP's wording, I think they had a 2 part problem. Part A: find the complement function. Part B: show that it actually is the complement.

— The Photon

0

By simple definition of $+$ (OR) and $′$ (NOT)

 A | B | A + B
---------------
 0 | 0 |   0
 1 | 0 |   1
 0 | 1 |   1
 1 | 1 |   1

 A | A′| A + A′
----------------
 0 | 1 |   1
 1 | 0 |   1

$∴ ∀f. f + f′ = 1$

— OrangeDog
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