Tengo la función:
Encontré que su función de complemento es:
Tengo que mostrar que: pero no puedo ver cómo hacerlo.
Parece que no hay nada que se cancele mutuamente.
Editar
Como se sugirió, ahora he usado el teorema de DeMorgan y encontré esto:
Pero todavía me parece que no hay nada que me acerque a la realización de
boolean-algebra
Carl
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Respuestas:
Desde que Carl preguntó amablemente. Punto de partida:f(x,y,z,w)=wx+yz
y
f'(x,y,z,w)=w'y'+w'z'+x'y'+x'z'
Siga los siguientes pasos conf′ :
f'(x,y,z,w)=w'(y'+z')+x'(y'+z')
f'(x,y,z,w)=(w'+x′)(y'+z')
DeMorgan:
f'(x,y,z,w)=(wx)'(yz)′
DeMorgan, again:
f'(x,y,z,w)=(wx+yz)′
So now the right-hand side of f′ is just the simple negation of the right-hand side of f . Which is a little anti-climactic, since now we just rely upon the fact that any expression x+x′=1 , which is what people have been saying all along about f+f′=1 , but at least it provides a little Boolean-algebra explanation for why that is true.
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The point is, it really doesn't matter what the functionf() 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 functionf′() , but that's actually irrelevant to the actual question!
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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.
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Mi respuesta es similar a la de Dave Tweed, lo que significa que lo puse en un nivel más formal. Obviamente respondí más tarde, pero decidí publicarlo, ya que alguien puede encontrar este enfoque interesante.
We have that
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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:
and
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 thatf 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 .
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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.
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By simple definition of+ (OR) and ' (NOT)
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