Dlaczego F + F '= 1?

15

Mam funkcję: $f(x,y,z,w) = wx + yz$

Odkryłem, że jego funkcją dopełniacza jest: $f '(x,y,z,w) = w'y' + w'z' + x'y' + x'z'$

Muszę pokazać, że: $f + f '=1$ ale nie widzę, jak to zrobić.

Wygląda na to, że po prostu nie ma nic, co by się wzajemnie anulowało.

Edytować

Jak sugerowałem, użyłem teraz twierdzenia DeMorgan i znalazłem to:

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

Ale nadal wydaje mi się, że nic nie przybliża mnie do realizacji $f+f' = 1$

boolean-algebra Carl
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6

Wskazówka: skorzystaj z prawa

DeMorgan

11

Albo f albo f 'musi wynosić 1

Chu

4

Masz tylko 4 wejścia. Jeśli nic więcej, możesz po prostu napisać tabelę prawdy.

Photon

2

Spehro ma rację co do pieniędzy, ale tak, zastosowanie DeMorgan jako pierwszego kroku nie pomaga. Aby więc nieco rozwinąć wskazówkę Spehro: rozwiązanie wymaga wykonania podstawowej algebry, która zawiera DeMorgan jako krok. Za pomocą prostej algebry + DeMorgan możesz zmienić funkcję f 'w wyraźnie oczywistą negację f. Pisząc to na kartce papieru, zajęło mi to tylko 4 kroki.

Pan Snrub,

1

@ Mr.Snrub pierwszym krokiem „Znalazłem jego funkcję uzupełniającą” powinno być (wx + yz) ′

OrangeDog

4

Since Carl asked nicely. Starting point:

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

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

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

$f′(x,y,z,w)=w′y′+w′z′+x′y′+x′z′$

Take the following steps with $f'$ :

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, again:

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

$f′(x,y,z,w)=(wx + yz)'$ So now the right-hand side of

f^{'}

$f'$ is just the simple negation of the right-hand side of

f

$f$ . Which is a little anti-climactic, since now we just rely upon the fact that any expression

x + x^{'} = 1

$x + x' = 1$ , which is what people have been saying all along about

f + f^{'} = 1

$f+f'=1$ , but at least it provides a little Boolean-algebra explanation for why that is true.

Mr. Snrub
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I don't understand how you got to the second line without passing by your final answer. Your final answer was my first step: it's just the negation of both sides.

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
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11

Moja odpowiedź jest podobna do Dave'a Tweeda, co oznacza, że postawiłem ją na bardziej formalnym poziomie. Oczywiście odpowiedziałem później, ale postanowiłem jednak opublikować to, ponieważ ktoś może uznać to podejście za interesujące.

$f$ $P$ $n\in\Bbb N$ $y_1,\ldots,y_n$ , 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
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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
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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
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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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Dlaczego F + F '= 1?

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