It's an important skill. The complement of a sum. This is the complement of a sum.

这是一项重要的技能。和的补码。这是和的补码。

It's equivalent to the

它相当于

the product of the complement.

补码的积。

Okay, second.

好的，第二个。

the complement of the product. This is the complement of the product.

积的补码。这是积的补码。

equals to the sum.

等于和。

of complement.

的补码。

It's De Morgan's Theorem.

这就是德摩根定理。

Okay, exercise. let's use the De Morgan's Theorem to solve problems.

好的，练习。让我们用德摩根定理来解决问题。

simplify the equation. simplify the expression.

表达式化简。表达式化简。

let's solve it in computer. so we have to simplify.

让我们在计算机上求解。所以我们需要化简。

okay, first we use the first form. of De Morgan's Theorem.

好的，首先我们使用第一种形式。德摩根定理。

the complement of the sum. is equivalent to the product. of the complement.

和的反码。它等同于积。反码的积。

it's exactly the complement of a sum, right?

这正是和的反码，对吧？

so this part, can be converted to keep the top not function.

所以这部分，可以转换为保留顶部的非函数。

so the complement of the sum is equivalent to the

所以，和的补码等于

product of the complement.

反码的乘积。

and here you have two not function. so they will be eliminated. and finally you get

这里有两个非函数。所以它们会被消除。最后，你得到

this part directly.

直接就是这部分。

okay, another exercise.

好的，另一个练习。

and for this one, we use the second form. second form of the de morgan's theorem.

对于这个，我们使用第二种形式。也就是德摩根定理的第二种形式。

you have the complement of a product.

这里有一个乘积的补码。

not the first one.

不是第一个。

it's the second form.

是第二种形式。

here is this part. so you have a sum.

这里是这部分。所以这里是一个和。

you have the sum of the complements, right?

是补码的和，对吧？

just like the right side of this equation.

就像这个等式的右边一样。

the sum of the complements is equivalent to the complement of the product. so this part

补码的和等于补码的乘积。所以这部分

is equal to

等于

this is the complement of the product.

乘积的补码。

and this is the not function. are cancelled.

这是非函数。它们相互抵消了。

or are eliminated.

或者被消除了。

so finally you will get this part.

所以最后，你会得到这一部分。

any questions?

有什么问题吗？

but the last exercise is also the most complicated one. yeah, try to solve this one.

但是，最后一个练习也是最复杂的。是的，试着解决它。

any question?

还有问题吗？

有

有。

okay, let's do it together. so this one is the most complicated one.

好的，我们一起做。所以，这是最复杂的一个。

first we have to change

首先，我们必须改变

okay, it's frozen.

状态。好的，它卡住了。

and for g, you can use two not function. so first we have to solve this part.

对于G，你可以使用两个非函数。所以，首先我们必须解决这一部分。

the complement for product.

乘积的补码。

so this part can be

所以这部分可以是

be written like this.

这样写的。

the complement of the product is equivalent to the complement of the sum.

乘积的补码等于和的补码。

now we have the first part remain the same.

现在，第一部分保持不变。

now we have exactly the complement of a sum.

现在我们正好得到了和的补码。

it's equivalent to the product.

它等于乘积。

so this is the part you know the not function. and if we have another not at the top, so these two not can be

所以，这就是你们所知道的非函数部分。如果我们上面还有一个非门，那么这两个非门可以

any questions?

抵消掉。有什么问题吗？

does anybody follow?

大家都跟上了吗？

when or if hunter use distributable?

亨特什么时候或者为什么会使用分配律？

so here we cannot use the distributive law. because the b is not included in the not function.

所以在这里，我们不能使用分配律。因为b没有包含在非函数中。

if you use two not here, you cannot multiply. together.

如果你在这里使用两个非门，你就不能进行乘法运算。把它们放在一起。

because there is a not at the top.

因为上面有一个非门。

you have to first remove the not.

你必须先去掉这个非门。

so this is the second form. of the de morgan's theorem.

所以这是第二种形式。也就是德摩根定理。

converted it to the sum expression.

把它转换成了积之和的形式。

okay, any questions?

好的，有什么问题吗？

why need all function from and the final output?

为什么需要所有的函数，从与函数到最终的输出？

yeah, i mean why the final output is like

是的，我的意思是，为什么最终输出是这样的？

because from this gate to this gate, right?

因为是从这个门到这个门，对吧？

according to this equation.

根据这个公式。

and for every simplification, you should first try to remove the nots.

对于每一次表达式化简，你应该首先尝试删除非函数。

and the top nots. and you should first try to remove the sub level.

还有顶部的非函数。你应该首先尝试删除子层级。

this is the order of the computation. you cannot remove the top not first. try to simplify the part below the top log function.

这就是计算的顺序。你不能先删除顶部的非函数。尝试化简下面的部分。也就是顶部的非函数部分。

i mean this step is right. so this part and this part

我的意思是，这一步是对的。所以这部分和这部分，

could be transformed to this part.

可以转换成这部分。

not right.

不对。

they are not equivalent.

它们是不等价的。

yeah this is equivalent.

是的，这个是等价的。

okay if there's not any other question i would like to continue.

好的，如果没有其他问题的话，我想继续讲下去。

we can also use the symbols. the logical symbols to

我们也可以使用符号。用逻辑符号来

represent de morgan's theorem.

表示德摩根定理。

as we know here is the

正如我们所知，这里是

land symbol.

与函数的符号。

land symbol.

与函数的符号。

it's a land symbol.

这是一个与函数符号。

you add first and and last.

你先做与运算，最后再做与运算。

so this is land.

所以这就是与函数。

let's go back to the six symbols we have. so we can also use the symbols. we can also use logic symbols. to express the de morgan's theorem. the first you use a land. land symbol.

让我们回到我们之前提到的六个符号。所以我们也可以使用这些符号。我们也可以使用逻辑符号。来表达德摩根定理。首先，你使用一个与函数。与函数符号。

the land is you add first and and then not. so this is and.

与函数的意思是，你先做与运算，然后再取反码。所以这就是与函数。

so this you pay attention to, you or them. you not a and b first.

所以，你要注意，你要对它们进行或运算。你先对a和b取反码。

the circle, the small circle. small circle means not.

那个小圆圈。小圆圈表示取反码。

you not them first. and then you apply the or. you apply the or operation.

你先对它们取反码。然后再进行或运算。你应用了或运算。

so this is the first form. of de morgan's theorem. using logical symbols.

这就是第一种形式。就是德摩根定理。使用逻辑符号。

and similarly.

同样地，

here is the log, right? logical symbol. you or them first. and then apply the not. not operation.

这里是“与”运算，对吧？逻辑符号。你先对它们进行或运算。然后再取反码。取反运算。

it's equivalent to you not them first. a not, b not, and then you and them together.

这相当于你先对它们取反码。a取反，b取反，然后再把它们进行或运算。

so just two different another expression of de morgan's theorem. using logic symbols.

所以，这就是德摩根定理的另一种表达形式。使用逻辑符号。

we can also use it as use a circle symbols. i don't understand anybody can use the circle symbols, circle logic symbols. to express a very complex equation. for example here. you have b or c and d. and then and with a. so how can we use the how can we use the circle

我们也可以使用圆圈符号。我不太明白，大家能用这种圆圈逻辑符号吗？用来表达一个非常复杂的方程。比如这里。这里有b或c与d。然后再与a进行与运算。那么，我们如何使用圆圈

logic symbols?

逻辑符号呢？

to express this equation.

来表达这个方程？

first you have you and c and d. using the symbol.

首先，你对c和d进行与运算。用这个符号表示。

you get cd. then you or it. you or it with b. is this symbol. so you can get b or cd. and finally you and it with a.

得到cd。然后对它进行或运算。再与b进行或运算。就是这个符号。这样你就能得到b或cd。最后，再与a进行与运算。

you will get this. you get this logical expression of the equation.

结果就是这个。你得到了这个方程的逻辑表达式。

so every equation can be also expressed by logical symbols.

所以，每一个方程都可以用逻辑符号来表示。

why we have to learn logical symbols? because when you design a circuit, in your laptop, right? there's software. or you can provide this. so you can use logical symbols. you just click them and drag. and drag it to the panel. and bind them together. you can finally create very complex circuits. using these kind of simple logical symbols.

那我们为什么要学习逻辑符号呢？因为当你设计电路时，是在笔记本电脑里设计的，对吧？里面有软件。或者你可以提供这个。所以你可以使用逻辑符号。你只需要点击它们，然后拖拽就可以了。把它拖到面板上。然后把它们连接起来。这样，你最终就能创建出非常复杂的电路。只需要使用这些简单的逻辑符号。

and also this one.

还有这个。

yeah, we can use the third law. distributive laws. so we can get a and b or a and c and b. the same equation here. here we apply the third law. distributive law.

是的，我们可以使用第三条定律。那就是分配律。所以，我们可以得到A和B，或者A、C和B。这里的方程也是一样的。这里我们应用了第三条定律。分配律。

and we can get a and b. so in this case, the logical symbols or the circuit looks a little bit different from the previous one.

这样我们就能得到A和B。所以在这种情况下，逻辑符号或者电路，看起来和之前的有点不同。

because here we have to first for example, and c and d. c and d and a. a is here. and also we input a. to this and function. here you and a and b. and finally you or. them together.

因为在这里，我们必须先对C和D进行“与”运算。也就是C、D和A的“与”运算。A在这里。同时，我们还输入A。到这个“与”函数中。这里是将A和B进行“与”运算。最后，再进行“或”运算。把它们结合起来。

so just in one word, we can also use the logical symbols to express the equations. the logical symbol is important. because when you finally design a circuit, the software only provide you such kind of logical symbols. so you must use the logical symbols.

所以，简单来说，我们也可以使用逻辑符号来表达这些方程。逻辑符号非常重要。因为当你最终设计电路时，软件只会为你提供这类逻辑符号。所以你必须使用逻辑符号。

to express the equation.

来表达方程。

so here we have two different logical or two different circuit.

是的，这里我们有两种不同的逻辑，或者说两种不同的电路。

but they both represent the same equation. they both represent the same expression.

但它们都表示同一个方程。它们都表示同一个表达式。

to express the same equation.

来表达同一个方程。

this is just three examples. of sum of products. you have product first. and then you sum. you have the second example. you have product first. and then you sum them together. for the third example,

好的，这只是三个例子。是关于“积之和”的。首先有乘积项。然后把它们相加。我们来看第二个例子。同样，首先有乘积项。然后把它们相加。对于第三个例子，

you have product first. and then you sum them. so to summarize is a sum of products. product expressions consist of a number of anded functions. right? the product is the and function.

也是先有乘积项。然后相加。所以，总结一下就是，“积之和”。乘积表达式由若干个“与函数”组成。对吧？这里的“乘积”就是“与函数”。

and also or. or functions together. so a sum of products expression consists of a number of anded functions. together.

还有“或”。“或函数”组合在一起。所以，“积之和”表达式由若干个“与函数”组成。它们通过“或”运算连接起来。

any logical expressions can be converted into a sum of a products.

任何逻辑表达式都可以转换为“积之和”形式。

using the rules of boolean algebra.

这利用了布尔代数的规则。

so any logical expression can be converted to the sum of products.

所以，任何逻辑表达式都可以转换为“积之和”形式。

that means we can only use anded functions and or function to express any logical expression. you can use and function. Here is and. and or function. to express any logical expression.

这意味着，我们只需要使用“与函数”，和“或函数”，就能表示任何逻辑表达式。你可以使用“与函数”。这里就是“与”。还有“或函数”。用来表示任何逻辑表达式。

because any logical expression can be converted into the sum of products. but the sum of product only include and function. and or function.

因为任何逻辑表达式都可以转换为“积之和”形式。但是，“积之和”形式只包含“与函数”。以及“或函数”。

so this is an example. so a sum of products expression. a sum of products expression can be implemented as two stage gate network.

这是一个例子。这是一个“积之和”表达式。“积之和”表达式可以实现为两级门网络。

the maximum number of gates for a single possible input is up to 18.

门网络的最大门数对于单个可能的输入，最多可达18个。

here you have the and function first. these three they are anded together, right? and e f g. also anded together. and finally you or them.

这里首先有“与函数”。这三个是“与”在一起的，对吧？还有e、f、g。它们也是“与”在一起的。最后，把它们“或”起来。

it's like this example. the third one.

就像这个例子一样。第三个。

you and b c d. you and e f g. and then you or them.

你把b、c、d“与”起来。然后把e、f、g“与”起来。然后把它们“或”起来。

you can use this and function. and logical symbol. use the or logical symbol. you can use only you just use two logical symbols. and and or. to express the sum of product.

你可以使用这个“与函数”。还有“与”逻辑符号。使用“或”逻辑符号。你只需要使用两种逻辑符号。一个是“与”，一个是“或”。用来表达“积之和”。

and also the second example. product of sum.

还有第二个例子。“和之积”。

product of sum means you first sum them together. you sum b c with d. you sum e f with g. and then you make the multiplication.

“和之积”的意思是，你首先把它们“或”起来。比如把b、c和d“或”起来。再把e、f和g“或”起来。然后，再把它们“与”起来。

product of sum. the first one is sum of product, right? and now is products of the sums. you first sum them together and then make the multiplication.

这就是“和之积”。第一个是“积之和”，对吧？现在是“和之积”。你先把它们“或”起来，然后再“与”起来。

so the order is different.

所以顺序是不同的。

similarly or identically. a product of sums also consists a number of or.

同样地，或者说是完全一样的。“和之积”也由若干个“或”组成。

or functions and and it together, right?

或函数。然后把它们“与”起来，对吧？

you first apply or. and then times. so any logic expression can be converted into a product of sums. using rules of boolean algebra.

你先执行“或”运算。然后再执行“与”运算。所以，任何逻辑表达式都可以转换为“和之积”的形式。这是利用布尔代数的规则实现的。

just two different forms. you can either use a sum of product or a product of sums. but both forms just use and and or function.

只是两种不同的形式。你可以使用“积之和”，或者“和之积”。但这两种形式都只使用了“与”和“或”函数。

for example, for this one, we or. we or b, c, d first. and we or e, f, g. and finally we and. this is the case here. we or b, c, d, and we or e, f, g, and then we and them together. it's three terms.

举个例子，对于这个，我们先执行“或”运算。先把b、c、d“或”起来。然后把e、f、g“或”起来。最后再把它们“与”起来。这里就是这种情况。对b、c、d进行“或”运算，对e、f、g进行“或”运算，然后将结果“与”起来。这就是三项。

and here you just use the or and and. and or and and logical symbols.

这里你只使用了“或”和“与”函数。也就是“或”和“与”的逻辑符号。

okay, another example to simplify the expressions.

好的，再举一个例子来演示表达式化简。

let's just do it together. we just apply the distributive law. so we can this equation can be written as can be written as c

我们一起来做。我们应用分配律。所以，我们可以把这个式子写成c

right? we get this one.

的形式，对吧？就得到了这个结果。

and this part is zero. because b not b and b. is zero.

而这一部分等于零。因为b和b的反码相与。结果为零。

so we have get a

所以，我们得到了a。

let's use the distributive law again.

让我们再次使用分配律。

now we will have a

现在我们得到a

plus c.

加上c。

two c c is just can be replaced by one, right?

两个c相或，就等于1，对吧？

yeah. now, similarly, we are use the third distributive law.

是的。那么，同样地，我们使用第三个分配律。

and we reach at

最终得到

and this part is one.

这一部分等于1。

right?

对吗？

so finally you will get

所以最后，你会得到

this.

这个结果。

so now i will introduce to

现在，我将介绍

we have already learned the NAND gate. the that's logic symbol, right? that is not and. so you first and, and then apply the not function. the symbol is like this.

我们已经学过的与非门。这是它的逻辑符号，对吧？它表示“非与”。所以，你先进行“与”运算，然后再应用“非函数”。符号就是这样。

and we can why do we call it universal? that means

我们可以为什么称它为“通用”门呢？意思是，

because we can use the NAND logic symbol. to express all the logic gate we have learned before. for example here, we can use the NAND gate to express the NOT gate. so everything we learn how to use, how to express with the logic gate, right? but we can also use the NAND gate. to replace the logic gate. and also we can use the NAND gate to replace the AND gate. so it's called universal. we can use the NAND gate to express

因为我们可以用与非门的逻辑符号来实现其他逻辑门。表达我们之前学过的所有逻辑门。比如这里，我们可以用与非门来表示非函数。所以我们学的所有内容，都是关于如何使用逻辑门来表达的，对吧？但我们也可以用与非门来实现。用来替代其他逻辑门。同样，我们也可以用与非门来替代与函数。所以它被称为“通用”门。我们可以用与非门来表达

all remaining gates.

所有其他的逻辑门。

here we can also use the NAND gate to replace the OR gate. and replace the NOR gate.

在这里，我们也可以用与非门来替代或函数。以及或非函数。

but how can we achieve this?

但是，我们该如何实现这一点呢？

a NAND a, you will get not a.

A与非A，得到的就是A的反码。

let's have a look what we have already learned at the beginning.

让我们回顾一下，一开始我们已经学过的内容。

is we first and. and not bit.

首先，是与函数。还有非函数。

sorry, this is or.

抱歉，这里是或函数。

so this is the expression.

所以这就是表达式。

of not.

表示非函数。

and we can use the land gate to replace the not. if we have two a's, two a as input.

我们可以用与门来代替非门。如果有两个A，作为输入。

so you have a, a and not.

那么就是A、A和非A。

this is exactly equals to a not, right?

这正好等于A的非，对吧？

so we have successfully used the land gate to replace the not gate. similarly, we can also use the land gate to replace the and gate.

所以我们成功地使用与门来替代了非门。同样地，我们也可以使用与门来替代与函数。

here we have to use two land gates.

这里我们需要使用两个与门。

first we have the first one is a and b. then we will get this form.

首先，第一个是A和B。这样我们就会得到这个形式。

and then we use the second land, right?

然后我们使用第二个与门，对吧？

second land is b and c.

第二个与门应该是B和C。

and then we use the second land. let's first try to remove. according to the de morgan's theorem.

然后我们使用第二个与门。我们先试着把它去掉。根据德摩根定理。

it can be expressed like this.

它可以这样表示。

and these two not can be removed. so finally you will get a b or a b c.

这两个非门可以被去掉。所以最后你会得到A和B的或，再加上C。

a b or a b c is exactly a b c, right?

A或B和C，正好就是A、B、C，对吧？

just like you have a or a. it's exactly a.

就像你有A或A一样。结果就是A。

so here we successfully use just two not gate.

所以在这里，我们成功地只用了两个非门。

to replace the and gate. and similarly, you can also use the and gate to replace the or gate. but it becomes a little bit complex.

来替代与函数。同样地，你也可以用与函数来替代或函数。但是这样会变得稍微复杂一些。

here we have to use three. three not gate.

这里我们需要用三个。三个非门。

the first just use not gate for one

首先，只用一个非门来处理一个

and you get enough.

输入，这样就足够了。

and a not b not

还有A非B非，

you use the third not gate.

用第三个非门。

you use the third not gate, you use a not b not, third not gate, you and them together.

用第三个非门，把A非和B非用与函数连接起来。

and using the de morgan's theorem.

然后利用德摩根定理。

and the two not can be removed. and you get a or b.

这样两个非号就可以消掉了。最后得到A或B。

so this case is another example. to describe that we can use three. three logic gates. to represent one or gate.

所以，这个例子是另一个示例。我们可以用三个来描述。三个逻辑门。来表示一个或函数。

and similarly we can use four. four logic gates. to represent the or gate.

同样地，我们也可以使用四个。四个逻辑门。来表示或函数。

so this reason we call it universal. it just use one logic symbol.

所以，这就是我们称之为“通用”的原因。它只需要一个逻辑符号。

and successfully replace all the remaining gates. just use the nand gates.

就能成功地替代所有其他的逻辑门。只需要使用与非门。所以它是通用的。

we have several minutes left, so let's just do this exercise. that we can end the class.

我们还有几分钟时间，那就来做一下这个练习吧。这样我们就可以结束这节课了。

just only use the nand gates. to express or to represent this expression.

同样，只能使用与非门。来表达或表示这个布尔表达式。

because we have just one minute left, we are close to the answer.

因为只剩一分钟了，我们快得出答案了。

because the nand gate is this form, right? not and. You should not do the and first and not. so the equation should be converted. to this form. because this form only includes and. and not.

因为与非门的符号是这样的，对吧？它是“非与”，而不是先做“与”再做“非”。所以这个表达式需要转换一下。变成这种形式。因为这种形式只包含“与”函数。和“非”函数。

and the two being are and not.

这两个也是“与”和“非”。

so you only include the and and not.

所以你只用了“与”和“非”。

if he can convert this expression to this form, then he can successfully implement only the nand gate. to express this equation.

如果他能转换这个表达式为这种形式，那么他就能成功只用与非门来实现。表达这个方程。

so first we have to convert this equation to this form. and then you can use the nand gates.

所以首先，我们必须把这个表达式转换为这种形式。然后你就可以使用与非门了。

so conversion process is very first you add two nots. and for this not, the same level not. you use the de morgan's theorem.

所以，转换过程非常简单。首先，你要加上两个非门。对于这个非门，是同一级别的非门。你用的是德摩根定理。

and you convert it into the product.

然后你把它转换成积之和的形式。

a not is the, has the a not, this is a not. if the input is 2a, you get a not.

A非就是，这里有一个A非，这也是一个A非。如果输入是两个A，你就会得到A非。

so finally you use three, four. to implement this implementation. okay, next one. Thank you.

所以最后，你用了三个，四个。来实现这个功能。好的，下一个。谢谢。

In more concise form, the complement of a sum is equivalent to the product of the complements. Second form.

更简洁地说，和的补码等于补码的积。第二种形式。

The complement of a product equals the sum of the complements. This is De Morgan's Theorem.

一个乘积的反码等于各反码之和。这就是德摩根定理。

Next, let's use De Morgan's Theorem to simplify the expression.

接下来，让我们使用德摩根定理来化简该表达式。

Let's solve it together. We have to simplify.

我们一起来解这道题。我们需要进行化简。

First, we use the first form of De Morgan's Theorem: the complement of a sum is equivalent to the product of the complements. This is exactly the complement of a sum.

首先，我们使用德摩根定理的第一种形式：和的补码等价于补码的乘积。这正是和的补码。

This part can be converted to a NOT function. So the complement of a sum is equivalent to the product of the complements. And here you have two NOT functions, so they will be eliminated.

这部分可以转换为一个非函数。因此，和的补码等价于补码的积。这里有两个非函数，它们将被消去。

And finally, you obtain this part. Now, let us proceed to another exercise.

最后，你得到这一部分。现在，我们进入另一个练习。

For this one, we use the second form of De Morgan's theorem.

对于这个，我们使用德摩根定理的第二形式。

You have the complement of the whole product of the first term, which is the second form of De Morgan's Theorem. Here is this part—so you have the sum of the complements. You have the sum of the complements, just like the…

你得到第一项整个乘积的反码，这就是德摩根定理的第二种形式。这一部分——所以你得到的是反码之和。你得到反码之和，就像……

The right side of this equation shows that the sum of the complements is equivalent to the complement of the product. So this part is equal to the complement of the product, and these NOT functions are cancelled or eliminated.

这个方程的右边表明，补码之和等价于积的反码。因此这一部分等于积的反码，而这些非函数被抵消或消除了。

Finally, you will get this part. Any questions? But the most expensive is also the most complicated one—it has the most components.

最终，你将得到这部分结果。有什么问题吗？但最昂贵的那个也是最复杂的——它拥有最多的组件。

No discernible lecture content is present in this segment. The audio from 576.0 to 636.0 seconds contains only background noise, indistinct ambient sounds, and no intelligible speech that can be transcribed.

该片段无可辨识的课堂讲授内容。576.0至636.0秒的音频仅包含背景噪音、模糊的环境声响，没有可转写的清晰语音。

Now let's do it together. This one is the most complicated expression. First, we have to change the form of D, because D is not the complement of a sum. The other terms are complements of sums, but D is not included in the sum part. So we should solve the problem with D first. For D, if I use two-input functions, right?

现在我们一起做。这个是最复杂的表达式。首先，我们必须改变D的形式，因为D不是一个和的补码。其他项都是和的补码，但D不包含在和的部分中。所以我们应该先解决D的问题。对于D，如果我使用两输入函数，对吧？

For G, we can use two NOT functions. First, we have to solve this part.

对于G，我们可以使用两个非函数。首先，我们需要求解这一部分。

The complement of a product can be written as the complement of a sum. Right? So this part can be...

一个乘积的反码可以写成一个和的反码。对吧？所以这一部分可以……

The complement of a product is equivalent to the complement of a sum. Now the first part remains the same, and we have exactly the complement of a sum.

一个乘积的反码等价于一个和的反码。现在第一部分保持不变，我们得到的恰好是一个和的反码。

It is implemented with a block. This is the part below the logic function, and we have another block at the top. These two blocks can be removed.

它由一个模块来实现。这是逻辑函数下方的部分，而我们在顶部还有另一个模块。这两个模块可以被移除。

Finally, we are going to get to this one. Any questions? Does anybody have any follow-up? Student: Teacher, when will a counter use the distributive law?

最后，我们要进入这个部分。有什么问题吗？有没有人需要进一步提问？ 学生：老师，计数器什么时候会用到分配律？

Here we cannot use the distributive law because the variable is not included in the NOT function. If you use two NOT functions here, you cannot multiply them together because there is a NOT at the top.

这里我们不能使用分配律，因为该变量没有被包含在非函数中。如果你在这里使用两个非函数，你不能将它们相乘，因为顶部有一个非运算。

You have to first remove the NOT at the top. This is the second form of De Morgan's Theorem. It will lead to the sum expression. Any questions?

首先，你需要去掉顶部的非运算。这是德摩根定理的第二种形式，它会导出和的表达式。有什么问题吗？

Student: Why do all functions need to have a final output? Teacher: All functions have a final output.

学生：为什么所有函数都需要有一个最终输出？ 教师：所有函数都有最终输出。

We call these… From this state to this state, right? Because from that to this state…

我们把这些称为……从这个状态到这个状态，对吧？因为从那个到这个状态……

For every simplification, you should first try to remove the NOT functions at the top level.

对于每一次表达式化简，你首先应该尝试移除顶层的非函数。

This is the top log block. You should first try to remove the sub-level log block. This is the order of the computation. You cannot remove the top log first. Try to simplify the part below the top log function. This step is right. So this part, that part.

这是顶层对数块。你应该首先尝试去除子层对数块。这就是计算的顺序。你不能先去除顶层对数。尝试简化顶层对数函数以下的部分。这一步是正确的。所以这部分，那部分。

Can this be transformed into this point? No. They are not equivalent. This point is correct. This cannot be transformed — it is not equivalent.

这个能否转换为这一点？不能。它们不等价。这一点是正确的。这个无法转换——它们不等价。

If there are no further questions, I would like to continue. We can also use logical symbols to represent De Morgan's Theorem. As we know, here is the...

如果没有其他问题，我想继续讲下去。我们也可以使用逻辑符号来表示德摩根定理。如我们所知，这里是……

This is the NAND symbol. You AND first, then invert at the last stage. So this is NAND. Let's go back to the second symbol we have drawn here. We can also use logical symbols to express De Morgan's Theorem. The first step uses the NAND symbol.

这是与非符号。先进行与运算，然后在最后一级取反。所以这就是与非。我们回到这里画出的第二个符号。我们也可以使用逻辑符号来表达德摩根定理。第一步使用与非符号。

The NAND function: you perform AND first and then NOT. This is equivalent to NOT-ing A and B first, and then applying the OR operation. The small circle on the gate symbol means NOT.

与非函数：先执行与运算，再执行非运算。这等同于先对A和B分别取非，再执行或运算。门符号上的小圆圈表示非。

This is the first form of De Morgan's theorem, expressed using logical symbols. Similarly, here is the logical symbol for the second form: you OR them first, and then apply the NOT operation. This is equivalent to...

这是德摩根定理的第一种形式，用逻辑符号表示。类似地，这里是第二种形式的逻辑符号：先对它们进行或运算，然后应用非运算。这等价于……

De Morgan's Theorem can be expressed using logical symbols in two different forms: first negate each variable individually, then combine them with the opposite operator. We can also use set symbols, which function identically to logical symbols, to express very complex equations. For example, consider B OR C AND D, then AND with A. How can we use set symbols to represent this?

德摩根定理可以用逻辑符号以两种不同形式表达：先对每个变量分别取反，再用相反的运算符将它们组合起来。我们也可以使用集合符号，其功能与逻辑符号完全相同，来表达非常复杂的方程。例如，考虑 B OR C AND D，再与 A 进行 AND 运算。我们如何用集合符号来表示这个表达式？

We use logical symbols to express this equation. First, you AND C and D using the AND symbol. You get C and D. Then you OR it with B using the OR symbol. So you get B OR C and D. Finally, you AND it with A.

我们使用逻辑符号来表达这个方程。首先，用与符号将C和D进行与运算，得到C与D。然后，用或符号将其与B进行或运算，得到B或C与D。最后，再与A进行与运算。

You get this logical expression of the equation. Every equation can also be expressed by logic symbols. Why do we have to learn logic symbols? Because when you design a circuit on your laptop, there is software that already provides you with logic symbols. You just click them and drag them to the panel, then combine them together. You can finally create very complex circuits using these kinds of simple logic symbols.

你可以得到方程的逻辑表达式。每个方程都可以用逻辑符号来表示。我们为什么要学习逻辑符号？因为当你在笔记本电脑上设计电路时，软件已经为你提供了逻辑符号。你只需点击它们并将它们拖到面板上，然后将它们组合在一起。你最终可以使用这些简单的逻辑符号来创建非常复杂的电路。

We can use the third law, the distributive law. Applying it, we get AB + ACB. Here we apply the distributive law again, and we can get AB.

我们可以使用第三条定律，即分配律。应用它，我们得到 AB + ACB。这里我们再次应用分配律，可以得到 AB。

In this case, the logic symbol or circuit looks different from the previous one. Here, we first AND C and D, then input A into this AND function, and finally OR them together.

在这种情况下，逻辑符号或电路与之前的有所不同。这里，我们首先对 C 和 D 执行与运算，然后将 A 输入到该与函数中，最后将它们进行或运算。

Just as in one word, we can also use the logical symbols to express the equations. The logical symbol is important because when you finally design a circuit, the software only provides you certain kinds of logical symbols. So you must use the logical symbols to express the equation. Here we have two different circuits, but they both represent...

除了用文字之外，我们还可以使用逻辑符号来表达方程。逻辑符号非常重要，因为当你最终设计电路时，软件只提供特定类型的逻辑符号。因此你必须使用逻辑符号来表达方程。这里我们有两个不同的电路，但它们都表示……

They both represent the same equation. They express the same expression. Right? Same importance.

它们都代表同一个方程。它们表达的是同一个表达式。对吧？同等重要。

These are three examples of Sum of Products. For the first example, you have products first, and then you sum them together. For the second example, you have product, product, product, and then you sum them together. For the third example, you have product, product, product, and then you sum them together. To summarize, a Sum of Products expression consists of a number of AND functions OR'd together.

这是积之和的三个例子。第一个例子中，先有乘积项，然后将它们相加。第二个例子中，有乘积、乘积、乘积，然后将它们相加。第三个例子中，同样有乘积、乘积、乘积，然后将它们相加。总结来说，积之和表达式由若干个与函数通过或运算组合而成。

AND function or OR function. Any logical expression can be converted into a sum of products form using the rules of Boolean algebra. That means

与函数或或函数。任何逻辑表达式都可以利用布尔代数的规则转换为积之和形式。也就是说

That means we can only use the AND Function and the OR Function to express any Boolean expression. Because any Boolean expression can be converted into Sum of Products, and Sum of Products only includes the AND Function and the OR Function.

这意味着我们只需要使用与函数和或函数就能表达任意布尔表达式。因为任意布尔表达式都可以转换为积之和，而积之和只包含与函数和或函数。

AND function, AND/OR function. A sum of products expression can be implemented as a two-stage gate network. The maximum number of gates for a single pass through the input.

与函数、与/或函数。积之和表达式可以实现为两级门网络。单次通过输入的最大门数量。

Here you have the AND function first. B, C, D are ANDed together, and E, F, G are also ANDed together. Finally, you OR them. This is like the third example: you AND B, C, D, you AND E, F, G, and then you OR them.

这里首先使用与函数。B、C、D 先进行与运算，E、F、G 也先进行与运算，最后将它们进行或运算。这就像第三个例子：先将 B、C、D 做与运算，再将 E、F、G 做与运算，然后对两者做或运算。

This is the AND logical symbol, and this is the OR logical symbol. You can use just these two logical symbols—AND and OR—to express the Sum of Products. And also the second form, Products of Sums.

这是与逻辑符号，这是或逻辑符号。你只需使用这两个逻辑符号——与和或——就可以表达积之和。还有第二种形式，和之积。

Product of Sums means you first sum them together—for example, you sum B, C with D; you sum E, F with G. You sum them together, and then you make the multiplication. That is the Product of Sums. The first form we discussed was Sum of Products. Now it is Products of Sums: you first sum them together, and then take the product. So the order is reversed.

和之积的意思是，你先将它们相加——例如，将B、C与D相加；将E、F与G相加。先把它们加在一起，然后再做乘法运算。这就是和之积。我们之前讨论的第一种形式是积之和。现在是和之积：先求和，再求积。所以顺序是相反的。

The order is different. Similarly, a Product of Sums also consists of a number of OR functions ANDed together. You first apply OR, and then AND the function. Any logical expression can be converted into a Product of Sums form using rules of Boolean algebra.

顺序是不同的。类似地，和之积也由若干个或函数通过与运算组合而成。先进行或运算，再进行与运算。任何逻辑表达式都可以利用布尔代数的规则转换为和之积形式。

There are two different forms: sum of products or product of sums. Both forms just use AND and OR functions. For example, in this one, we OR BCD first, then we OR EFG, and finally we AND them together. This is the case here: we OR BCD and we OR EFG, and then we AND these three terms together.

有两种不同的形式：积之和或和之积。这两种形式都只使用与函数和或函数。例如，在这个例子中，我们先对BCD进行或运算，然后对EFG进行或运算，最后将它们进行与运算。这就是这里的情况：我们对BCD进行或运算，对EFG进行或运算，然后将这三项进行与运算。

Here you simply use the OR and AND logical symbols. Another example: simplifying the expression.

这里只需使用或（OR）和与（AND）逻辑符号。再举一个例子：对表达式进行化简。

First, we apply the distributive law. This equation can be written as C.

首先，我们应用分配律。该方程可以写为C。

We get this one. This part is zero, because NOT B AND B is zero.

我们得到这一项。这部分为零，因为 B 的非与 B 相与的结果为零。

Let's read it again. It does not even have a…

让我们再读一遍。它甚至没有一个……

C can be replaced by 1. Now, similarly, we use the third variable.

C可以用1来替代。现在，类似地，我们使用第三个变量。

This part is one. And this part is one, right? Finally, you will get one output. You can see it has four, five, three—it still has many inputs, maybe a total of five. So now I want to introduce to...

这一部分等于1。这一部分也等于1，对吧？最终你会得到一个输出。你可以看到它有四个、五个、三个——它仍然有很多输入，总共可能有五个。所以现在我想介绍……

We have already learned the NAND logic symbol. It is NOT-AND: first an AND function, then apply the NOT function. The symbol looks like this. Why do we call it a universal logic gate? Because we can use the NAND logic symbol to express all the logic gates we have learned before. For example, we can use the NAND gate to express the NOT gate. Previously we learned how to use the NOT logic gate, but we can also use the NAND gate to replace the NOT gate. And also we can use the NAND gate to replace—

我们已经学习了NAND逻辑符号。它是NOT-AND：先执行AND函数，再施加NOT函数。符号如下所示。为什么我们称之为通用逻辑门？因为我们可以使用NAND逻辑符号来表达之前学过的所有逻辑门。例如，我们可以用NAND门来表达NOT门。之前我们学习了如何使用NOT逻辑门，但我们也可以用NAND门来替代NOT门。而且我们还可以用NAND门来替代——

NAND gates are called universal. They can be used to express other logic gates. You can also use NAND gates to replace OR gates and NOT gates. But how can we achieve that? A NAND A gives NOT A.

与非门被称为通用门。它们可以用来实现其他逻辑门的功能。你也可以用与非门来替代或门和非门。但我们如何实现这一点呢？A 与非 A 得到非 A。

We have already learned at the beginning that NAND is the first AND and then NOT. So this is the expression of NAND.

我们在开头已经学过，NAND（与非）就是先做 AND（与），再做 NOT（非）。这就是 NAND 的表达式。

We can use the NAND gate to replace the NOT gate. If we have two A's as inputs, we have A NAND A. This is exactly equal to NOT A. So we have successfully used the NAND gate to replace the NOT gate. Similarly, we can also use the NAND gate to replace the AND gate.

我们可以用与非门来替代非门。如果将两个A作为输入，就得到A NAND A，这恰好等于NOT A。因此，我们成功地用与非门替代了非门。类似地，我们也可以使用与非门来替代与门。

To replace the AND gate, we have to use two NAND gates. The first one takes A and B as inputs, giving us this form. Then we use the second NAND gate.

要用与非门替代与门，我们需要使用两个与非门。第一个与非门的输入为A和B，得到这个形式。然后我们使用第二个与非门。

The second line is lacking this. Let's first try to remove this line according to De Morgan's Theorem.

第二行缺少这一项。我们先尝试根据德摩根定理去掉这条线。

It can be expressed like this. And this whole term can be removed, so that means you just get AB or AB. AB or AB is exactly AB. Just like if you have A or A, it is exactly A.

它可以这样表示。这一整项可以被去掉，所以结果就是AB或AB。AB或AB恰好等于AB。就像A或A恰好等于A一样。

Here we successfully use just two NAND gates to replace the AND gate. Similarly, we can also use NAND gates to replace the OR gate, but it becomes a little bit complex. Here we have to use three NAND gates. The first NAND gate is used for A. If you just use a NAND gate for A, you will get A-bar, just like here. If you just use a NAND gate for one variable, you will get its complement.

这里我们成功地仅用两个与非门替换了与门。类似地，我们也可以使用与非门来替换或门，但会稍微复杂一些。这里我们需要使用三个与非门。第一个与非门用于A。如果仅对一个变量使用一个与非门，你将得到A的反码，就像这里所示。如果仅对一个变量使用一个与非门，你将得到它的补码。

For the NOT function, just use one NAND gate here. You get A bar, and you get B bar. Then you use the third NAND gate to AND them together. Using De Morgan's Theorem, this equals to—

对于非函数，只需在此处使用一个与非门。你得到A的非，也得到B的非。然后使用第三个与非门将它们进行与运算。利用德摩根定理，这等于——

An XOR function can be removed and replaced by an OR gate. This case is another example to describe that we can use three NAND gates to represent one OR gate. And similarly, we can use four NOR gates to represent an AND gate.

一个异或函数可以被移除并用一个或门替代。这个案例是另一个例子，说明我们可以使用三个与非门来表示一个或门。类似地，我们可以使用四个或非门来表示一个与门。

This is the reason we call it universal. You just use one Logic Symbol and successfully replace all the remaining gates. Just use the NAND gate. So it is universal. We have several minutes left, so let's do some exercises, and then we can end the class.

这就是我们称之为通用的原因。你只需使用一个逻辑符号，就能成功地替代所有其余的逻辑门。只需使用与非门即可。所以它是通用的。我们还有几分钟时间，来做一些练习，然后我们就可以下课了。

Only use brackets for expressing or to represent in an expression. Only use brackets.

仅在表达式中需要表示分组时才使用括号。只使用括号。

We have just one minute left. If you have questions, please go ahead and ask.

我们还剩一分钟。如果有任何问题，请现在提出。

The NAND gate performs the AND operation first, followed by the NOT operation. Therefore, the equation should be converted to this form, which only includes ANDs and NOTs.

与非门先执行与运算，再执行非运算。因此，该方程应转换为这种形式，其中仅包含与运算和非运算。

If you can convert this expression to this form, then you can successfully implement it using only NAND gates. So first, you have to convert this equation to this form, and then you can use the NAND gates. The conversion process is as follows: first, you add two NOTs—double negation. For these NOTs at the sum level, you use De Morgan's theorem, and you convert them to NAND form.

如果你能将这个表达式转换为这种形式，那么你就可以仅使用与非门来实现它。所以首先，你必须将这个方程转换为这种形式，然后就可以使用与非门了。转换过程如下：首先，你添加两个非运算——即双重否定。对于求和层面的这些非运算，你使用德摩根定理，将它们转换为与非形式。

A naught is the output. The input is two A's. So finally, you use three or four NAND gates to implement this function. Next one. Thank you.

A₀ 是输出，输入是两个 A。因此最终需要三到四个与非门来实现该函数。下一位，谢谢。

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