So you can simplify this expression You will find it is the outcome is one. No matter what a, b or c is. So when you create a truth table, it's very simple. when you create the truth table, the x is always one. That's no matter whether a or b or c is

所以，你可以简化这个表达式。你会发现，它结果等于1。无论a、b或c是什么值。所以，当你制作真值表时，就非常简单了。在真值表中，x始终为1。无论a、b还是c是

also, that's the solution for question two. we have done it together, question two.

什么。这就是第二题的解法。这道题我们已经一起做过了。

yes, question two here.

是的，就是这里的第二题。

and question two. Okay, we need to simplify expressions.

还有第二题。好的，我们需要简化这个表达式。

so this is the original expression. of the logic.

这就是原来的表达式。也就是逻辑表达式。

original expression of the circuit, right? But we can simplify it.

也就是电路的原始表达式，对吧？但是我们可以对它进行简化。

is there any questions for question one? and question two.

关于第一题，大家有什么问题吗？还有第二题。

any questions for question one and question two?

第一题和第二题，大家还有问题吗？

let's continue the exercise. let's go back to question three and question four.

好的，那我们继续做练习。我们回到第三题和第四题。

okay, question three to question six.

好的，从第三题到第六题。

i'll give you maybe 15 minutes. to solve

我给大家大概15分钟的时间。来做这些题。

question three's this

把第三题的这个换成这个。

and then

把C改成F，然后再把这个

this

也换一下。

This is the undergate. The first input is a not. The second input is b.

这是下门。第一个输入，是非门。第二个输入是B。

The third input of this and gate is c.

第三个输入，这个与门的是C。

Okay, we have done. And for the second and gate below, the first input is a not.

好的，我们已经完成了。至于下面的第二个与门，第一个输入也是非门。

the second input is b not.

第二个输入是B非。

and the third input is c.

第三个输入是C。

And finally, we have the or gate. So we just add them.

最后，我们有一个或门。所以我们只需要把它们加起来。

So this is the solution.

这就是解决方案。

This is the solution for question three. And the question is draw the truth table.

这是第三题的答案。题目要求画出真值表。

And we can simplify a little bit. because there's always AC, right?

我们可以稍微简化一下。因为这里总是有AC，对吧？

This is one.

这是一个1。

And you should draw the truth table. Okay, there's three variables. a, b, c.

你应该画出真值表。好的，这里有三个变量。a、b、c。

And you should compute what is the output.

然后你需要计算输出是什么。

I will just skip this step. I think question three. is question three understandable? does everybody can solve question three?

我就跳过这一步了。我想说的是第三题。第三题大家能理解吗？每个人都能解出第三题吗？

does everybody can solve? question three is straightforward. if you remember the meaning, the meaning of the logical gates. you can simply write this expression. and just do one step. to one step simplification. so get the final result. and draw the truth table.

大家都能做出来吗？第三题很简单。只要记住它的含义，逻辑门的意思。就可以直接写出这个表达式。只需要一步就能搞定。一步化简。这样就能得到最终结果了。然后画出真值表。

okay, question four. question four is repeat. repeat the question three.

好的，第四题。第四题是重复的。就是第三题的重复。

with each and gate replaced by nand. the or gate replaced by nor.

只是把每个与门都换成与非门。或门换成或非门。

okay. so here is the nand gate. it's not and gate anymore. it's nand gate.

好的。那么，这里是一个与非门。它不再是与门了。而是与非门。

let's look at this nand gate. the first input is here a not.

我们来看一下这个与非门。第一个输入在这里，是A非。

the second input is b.

第二个输入是B。

the third input is c. because here is the nand gate.

第三个输入是C。因为这里是一个与非门。

so we have a log function at the top.

所以顶部有一个逻辑函数。

it's the first nand. for the second nand,

这是第一个与非门。对于第二个与非门，

the first input is a not.

第一个输入是A非。

the second input is b not. and c.

第二个输入是B非。还有C。

because it's a nand gate, so we have a not. at the top.

因为这是一个与非门，所以这里有一个非门。在顶部。

and finally we also have a nor gate.

最后，我们还有一个或非门。

so you have or them first.

所以，你首先要对它们进行或运算。

and finally, not them. so this will be expression. this will be the expression for question four.

最后，再取非。这就是表达式。这就是第四题的表达式。

and we should simplify it. let's do it. We are simplify.

我们应该对它进行化简。我们来化简一下。

and there are many ways. to simplify the expression.

方法有很多。来化简这个表达式。

first, okay, according to De Morgan's theorem. the complement of a sum. This is complement. The complement of sum.

首先，根据德·摩根定律。取反，一个和的取反。这就是取反。和的取反。

is equivalent to the

等同于

product, right?

各个部分的乘积，对吧？

and these two nots is eliminated. so can be removed.

这两个非号就被消掉了。所以可以去掉。

so finally you have this.

最后，你就得到了这个结果。

and here is a 2c.

这里有一个2C。

so you get a b. and a not. okay. b and b not. it's always zero, right? so the final outcome is also zero.

所以你得到了一个B。和一个非A。好的。B和非B。这总是等于零，对吧？所以最终的结果也是零。

so there is another ac. don't forget them.

还有另一个AC。别忘了它们。

okay.

好的。

question five.

第五题。

do you have any questions for question five?

关于第五题，大家有什么问题吗？

do you know how solve the problem?

你知道怎么解这道题吗？

okay, let's do question five.

好的，那我们来做第五题。

question five is use and gate. or gate. and inverters.

第五题是使用与门。或门。还有反相器。

to implement the following logical expressions.

来实现下面的逻辑表达式。

i think for this one is quite simple, right? this is and. and this is the and again. but you have to use a lot of functions first. and this is or.

我觉得这个挺简单的，对吧？这是与门。这也是与门。但是，你需要使用很多逻辑门来实现。而这个是或门。

so for the simple, you have a,

所以，简单来说，你有A，

you have two input, three input, a, b, c. first you apply the and gate.

还有三个输入：A、B和C。首先，你使用与门。

and for this one, you

而对于这个，你

apply the not function first.

要先使用非门。

it's a not.

这是一个非门。

i don't know if i'm wrong.

我不知道我有没有弄错。

this is b not, and you and c.

这是B的非，然后和C进行与运算。

let me answer the question.

让我来回答这个问题。

and finally you have a or.

最后，你得到一个或门。

right? okay, this is the solution.

对吧？好的，这就是解决方案。

i think a is very simple.

我觉得A非常简单。

any question for a?

关于A，有什么问题吗？

这个答案是负一

这个答案是负一。

这个答案是不是不是特定的一个

这个答案是不是不是特定的一个？

我是化简成别的然后 也没化简成这个格式

我是把它化简成了别的，然后也没有化简成这个格式。

那第一个是

那第一个是

第一个

第一个。

有点麻烦吧这个

这个有点麻烦吧。

ok第一个的话你换成这样的

好的，第一个的话，你可以这样写。

这是或吧 或的话没有这个点 没有这个圈的

这是“或”吧？如果是“或”的话，就没有这个点。没有这个圈。

i i wrote on the on the on the screen so this is the solution for question 5a

我在屏幕上已经写出来了。这就是第5题A问的解答。

you have an led and you apply a signal and then and we see

你有一个LED，然后施加一个信号。然后我们就能看到。

and to question b

接下来是B问。

what is question b? maybe let's do the the most complex one let's do the d or e together okay because of time i cannot do every task d seem to be a little bit complex

B问是什么？也许我们先做最复杂的那个。我们一起做D或者E吧。好的，因为时间关系，我不能把每个任务都做一遍。D看起来稍微有点复杂。

because here is we use an and an or this is and, right? this is or. so very straightforward. just you have a, a, b, c, d, e, f, g. you have six input. you have a, b, c,

因为这里用到了“与”和“或”门。这是“与”门，对吧？这是“或”门。所以，非常简单明了。你只需要有A、B、C、D、E、F、G。你有六个输入。还有A、B、C，

you and this. and then you apply

然后把它们“与”起来。然后，你再应用

this the first term, right? second term is e f. you have e f.

这个第一项，对吧？第二项是E和F。你有E和F。

you and them first.

先把它们“与”起来。

you or it.

然后再“或”上它。

or it with g.

最后再与G“或”起来。

this is g not, right? you or it.

这是G的非，对吧？然后把它“或”上。

oh sorry.

哦，对不起。

this is or.

这是“或”运算。

and then you and it with b, okay? so we have to plot this. let's note here. a branch.

然后，再把它和B“与”起来，好吗？所以我们得把这个画出来。我们记在这里。一个分支。

you and.

然后“与”。

here is a b, we make a note here. so this another branch.

这里是B，我们在这里做个标记。所以，这是另一个分支。

and then you or it.

然后“或”上它。

and for the last one,

最后一个是，

last e is

最后一个E是

we can simplify it a little bit.

我们可以简化一下。

so we have a, b c a.

所以，我们有A，B、C、A。

or b b c is b c, right?

或者BBC就是BC，对吧？

and bc is bc.

而BC就是BC。

and we have two a. and just keep one a.

我们有两个A。我们只保留一个A。

okay, bc and bc. they are the same. so just keep one.

好的，BC和BC。它们是一样的。所以只保留一个。

that's no.

没有了。

okay, this is the same, just keep one.

好的，这个也是一样的，只保留一个。

that's abc.

那就是ABC。

it's just abc.

就是ABC。

so just use the and gate.

所以，只用与门。

any questions?

有什么问题吗？

the solution provide this because they didn't simplify the expression. right?

提供这个解法是因为他们没有简化表达式。对吧？

the task is we or abcd. just use the or gate. you get abcd or.

任务是，我们要对ABCD进行或运算。所以，只用或门就可以了。这样就能得到ABCD的或运算结果。

and then you and with abc.

然后，再与ABC进行与运算。

so they didn't simplify the expression. it's also correct. but sometimes if you don't simplify, the logical circuit will be very complex. if you simplify it, that could be much simpler.

所以他们没有简化表达式。不过，这也是正确的。但有时候，如果不进行简化，逻辑电路会变得非常复杂。如果简化一下，就会简单得多。

yeah, depend on

是的，这取决于

whether that task is complex.

任务本身的复杂程度。

okay, so both solution are correct. this one is non-simplified solution.

好的，所以这两个解都是正确的。这个是没有简化的解。

so any questions?

那么，有什么问题吗？

no.

没有。

okay, there's no questions. that's end.

好的，没有问题。就到这里。

let's solve question six. and lock gate.

我们来解答第六题。还有锁存器。

implement the following logic expressions.

实现以下逻辑表达式。

we know that is like this, right?

我们知道，与门是这样的，对吧？

so a, b and not. so basically we have to convert. we have to convert the expressions. to include the form like this.

所以是A、B和非门。所以，基本上我们需要进行转换。我们需要转换这些表达式。使其符合这种形式。

so if we can successfully convert, each expressions to this form.

所以，如果我们能成功转换，把每个表达式都转换成这种形式，

we can use the land gate. directly.

就可以使用与门了。直接实现。

for the first one, okay, the first one looks very complex, right? it's the longest one compared to b and c. And let me look at the form. The first thing is, okay, this is okay because it's just like very similar to this form, right? we can use the land gate directly. and also it's very looks friendly. we can use the land gate directly.

对于第一个，好的，第一个看起来非常复杂，对吧？和B和C相比，它是最长的。让我来看一下它的形式。首先，好的，这样是可以的，因为它和这种形式非常相似，对吧？我们可以直接使用与门。而且这样看起来也很简洁。可以直接使用与门。

this also okay. but here is not.

这样也可以。但是这里不行。

we have an or. and also we have or here. so what we have to do is try to convert

这里有一个或门。这里也有一个或门。所以，我们需要做的是尝试转换

this term and this to something like this.

这一项和这一项。变成类似这样的形式。

and if you think about the de morgan's theory. the first term can be convert to

如果你回想一下德摩根定律。第一项可以转换成

a or b.

A或B。

and now we can use the distributive law.

现在，我们可以利用分配律。

because a not and a a not and a is zero. this is the reason that i want to apply distributive law.

对吧？因为非A与A，等于零。这就是我想应用分配律的原因。

here.

就是这里。

this is the reason i want to apply distributive law here. a not and a is zero. so this term is zero.

这就是我想在这里应用分配律的原因。非A与A，等于零。所以这一项等于零。

so finally we are get

所以最后我们得到

this one.

这个结果。

and we can also use the de morgan's law again.

我们还可以再次使用德摩根定律。

and here we have two d not, so just get one.

这里有两个非D，所以合并成一个。

great. and now

很好。那么现在，

all the four terms, are very similar to this one, right? so this is exactly what we want. but the problem here is we have the or.

所有的四项，都和这一项非常相似，对吧？这正是我们想要的。但这里的问题是，有一个“或”运算。

okay, we have the or.

好的，有一个“或”。

a not b or a not b not c 等于 a not.

A非B或A非B非C，等于A非。

oh yes, so for these two terms, we can simplify it further.

哦，是的。所以对于这两项，我们可以进一步简化。

just a not. because b or b not is one.

就是A非。因为B或B非等于1。

and there's a a not here.

而且这里还有一个A非。

so finally it should be

所以最后应该是

just a not or cd, right?

A非或CD，对吧？

然后变成那个and

然后变成那个与。

is this step understandable?

这一步能理解吗？

no. okay.

不能。好的。

when you compare

好的，当你比较

this one and this one, they both have the a not.

这个和这个的时候，它们都有A非。

so if you a not, if you

所以，如果有A非，那么

it's one.

它就等于1。

so this is the final form. if you simplify the expression.

所以，这就是最终形式。如果你简化这个表达式的话。

but a little bit different than the language.

不过，这和语言有点不同。

so let's apply the

那么，让我们应用

you have a not not, right?

双重否定。对吧？

yeah.

是的。

so finally you get

所以最后，你会得到

a complex expression. just using two language.

一个复杂的表达式。只用了两个逻辑门。

let's draw the circuit. you have a.

我们来画一下电路图。这里有一个A。

you have c.

还有一个C。

do you have any questions?

大家有什么问题吗？

let's just use this de morgan's law.

好的，那我们直接用德摩根定律。

the first turn is you have c.

第一步是，我们有C。

so you can obtain this form. the form like this. and let me use the de morgan's theory.

所以你可以得到这个形式。就是这样的形式。接下来，让我用一下德摩根定理。

this is right.

这样对吗？

and here we can use the de morgan's theory again.

这里我们可以再次使用德摩根定理。

yeah. now we can do some simplification. for example, this one is just a c not, right?

是的。现在我们可以进行一些化简了。比如，这个就是非C，对吧？

this one is b not.

这个就是非B。

and this one, this one.

还有这个，这个。

it's also b not.

也是非B。

okay, maybe we don't have time.

好的，可能时间不够了。

完蛋了

完了。

You can simplify these expressions. The outcome is 1 no matter what A, B, or C is. When you create the truth table, it is very simple: X is always 1. It does not matter whether A or B or C is what. Also, the solution for question two—we have done together question two.

你可以简化这些表达式。无论A、B还是C取何值，结果都是1。当你创建真值表时，非常简单：X始终为1。不管A、B或C是什么都不影响结果。另外，关于第二题的解法——我们已经一起完成了第二题。

Question two here. We need to simplify expressions. This is the original expression of the circuit, but we can simplify it.

这里是第二题。我们需要简化表达式。这是电路的原始表达式，但我们可以对其进行简化。

Is there any question for Question 1 and Question 2? Let us continue on the next slide. Let us go back to Question 3 and Question 4.

关于第一题和第二题，有没有问题？我们继续看下一张幻灯片，回到第三题和第四题。

[This segment contains no substantive lecture content. The audio consists of ambient classroom noise including paper shuffling, coughing, and indistinct background murmuring with no coherent instructional speech.]

[该片段不包含实质性讲课内容。音频由课堂环境噪音组成，包括翻纸声、咳嗽声和模糊的背景低语，没有连贯的教学语音。]

This is an AND gate. The first input is A₀, the second input is B, and the third input is C. We have completed the first AND gate. For the second AND gate below, the first input is A₀.

这是一个与门。第一个输入是A₀，第二个输入是B，第三个输入是C。第一个与门已经完成。对于下方第二个与门，第一个输入是A₀。

We can simplify a little bit because there is always AC, right? A AND C. This is one. And you should draw the truth table. There are three variables: A, B, C.

我们可以稍微简化一下，因为始终存在AC，即A与C。这就是1。你们应该画出真值表。共有三个变量：A、B、C。

The second input is B naught, and the third input is C. Finally, we have the OR gate, so we just add them. This is the solution for question three. The question asks us to draw the truth table.

第二个输入是B₀，第三个输入是C。最后，我们有一个或门，所以将它们相加。这就是第三题的解法。题目要求我们画出真值表。

Question 4 is a repeat of Question 3. In this question, each AND gate is replaced by a NAND gate, and each OR gate is replaced by a NOR gate. So here we have a NAND gate—it is no longer an AND gate; it is a NAND gate. Let us look at this NAND gate. The first input is A and NOT...

第4题是第3题的重复。在这道题中，每个与门被替换为与非门，每个或门被替换为或非门。所以这里得到的是一个与非门——它不再是与门，而是与非门。让我们来看这个与非门。第一个输入是A和非……

Let's skip this step. Question three is straightforward. If you remember the meaning of the logical gates, you can simply write these expressions and do one-step simplification to get the final result, then draw the truth table. Now, question four...

我们跳过这一步。第三题很直接。只要记住逻辑门的含义，就可以直接写出这些表达式，做一步化简得到最终结果，然后画出真值表。接下来是第四题……

We have a NOT and a NOT. Finally, we also have a logit. You have all of them first, and then NOT them. This is the expression for Question 4. We can simplify it. There are many ways to do so.

我们有一个NOT和一个NOT。最后，我们还有一个logit。先将它们全部组合在一起，然后对整个结果取NOT。这就是第四题的表达式。我们可以对其进行化简，化简的方法有很多种。

According to De Morgan's theorem, the complement of a sum is equivalent to the product of the complements. So the complement of A plus B plus C equals A dot B dot C.

根据德摩根定理，和的补等于补的积。因此，A加B加C的补等于A点乘B点乘C。

There is another AC. Don't forget them.

还有另一个交流电（AC）。不要忘记它们。

Question five. Do you have any questions for question five? Do you know how to solve the problem, question five? Let's do question five. Question five is—

第五题。大家对第五题有什么问题吗？你们知道怎么解第五题吗？我们来做第五题。第五题是——

This term T₀ is eliminated, so it can be removed. We are left with B and B̄. B and B̄ is always zero. Therefore, the final outcome is zero.

这一项 T₀ 被消去，因此可以移除。剩下的结果是 B 与 B̄。B 与 B̄ 恒等于零。所以最终结果为零。

Use AND gates, OR gates, and inverters to implement the following logical expressions. I think for this one it is quite simple. This is AND, and this is AND again, but you have to use a lot of functions first.

使用与门、或门和非门来实现以下逻辑表达式。我认为这一题相当简单。这里是与运算，这里也是与运算，但你首先需要用到很多函数。

You have to use the NOT function first, and this is OR. For the same expression, you have three inputs: A, B, C. First you apply the AND gate. And for this one, you apply the NOT function first.

你必须先使用非函数，这是一个或门。对于同一个表达式，你有三个输入：A、B、C。首先应用与门。而对于这一个，你先应用非函数。

This is not wrong. This is really not—I'm seeing. Maybe I should try.

这没有错。这确实不是——我看到了。也许我应该换个方式试试。

He lost the answer C, and finally you have earned more. Right? This is the solution for A. I think A is very simple.

他失去了答案C，最终你赚得更多。对吧？这就是A的解法。我认为A非常简单。

Any questions for part A? [Student] Is the answer unique, or is it not a specific single value? [Instructor] Mm-hmm.

关于A部分，大家有什么问题吗？ 【学生】这个答案是唯一的吗，还是说它不是一个特定的值？ 【教师】嗯。

Student: I simplified it and then redrew it in a different format. Instructor: The first one — you draw it like this. For "货" (goods), there is no dot; it uses a circle instead.

学生：我先把它化简了，然后换了一种格式重新画出来。 教师：第一个——你这样画。"货"字没有点，用的是圈。

As I wrote on the screen, this is the solution for Question 5a. You had AB, applied B dot, and then multiplied by BC. Now let's proceed to Question D. Due to time constraints, let's tackle the most complex one—D or E—together. I cannot cover every task individually. D seems to be a little bit complex.

正如我在屏幕上所写的，这是第5a题的解答。你已有AB，对B施加点积运算，再乘以BC。接下来我们进入D题。由于时间有限，我们把最复杂的那道——D题或E题——一起处理。我无法逐一讲解每道题。D题看起来稍微复杂一些。

Here we use an AND and an OR. This is AND, this is OR, this is AND, this is OR. It is very straightforward. You have a, b, c, d, e, f, g. You have six inputs. You have a, b, c.

这里我们使用一个 AND 和一个 OR。这是 AND，这是 OR，这是 AND，这是 OR。非常直接。你有 a、b、c、d、e、f、g，共六个输入。你有 a、b、c。

You have A, B, C. You undate, then you apply. This is the first term. The second term is E, F. You have E, F. You undate first, before each.

你有 A、B、C。先 undate，然后 apply。这是第一项。第二项是 E、F。你有 E、F。先 undate，在每一项之前。

This is OR. And then you add here this B. So we have two clauses. Let the NOT here, a branch, branch here.

这是OR门。然后在这里加入B。所以我们有两个子句。把NOT放在这里，一个分支，分支在这里。

Here is B. We draw a node here, so this is another branch, B. And then the OR — same procedure. On the last one...

这里是B。我们在这里画一个节点，所以这是另一个分支，B。然后是OR——同样的操作。最后一个……

We can simplify it a little bit. We have a...

我们可以把它稍微简化一下。我们有……

BCA. B, B, C is B, C, right? And BC is BC. This is C is BC. And we have two A's, so we just keep one A. BC and BC, they're the same, so just keep one.

BCA。B、B、C就是B、C，对吧？BC还是BC。这里C是BC。我们有两个A，所以只保留一个A。BC和BC是一样的，所以只保留一个。

Just the K one. Oh, that's no. A, B, C, O, H, G, C, D. This is the same—just the K one.

只是K那个。哦，不对。A、B、C、O、H、G、C、D。这是相同的——只是K那个。

That gives us ABC. This is one, so the final result is simply ABC. Use the AND gate to draw ABC. Any questions?

结果为ABC。这是一，所以最终结果就是ABC。用与门来画ABC。有问题吗？

The solution provided this form because they did not simplify the expression. When you look at the task...

解答给出这种形式是因为他们没有简化表达式。当你看这道题时……

You can use OR with A, B, C, D. Just use OR here, and you get ABCD OR. Then you AND with ABC. So they did not simplify the expression. It is also correct if you do not simplify. But sometimes if you do not simplify, those logic circuits could be very complex. If you simplify it, that could be much simpler. It depends on whether the task...

你可以对 A、B、C、D 使用 OR 运算。这里直接用 OR，得到 ABCD OR。然后与 ABC 进行 AND 运算。所以他们没有简化这个表达式。不简化也是正确的。但有时候如果不简化，那些逻辑电路可能会非常复杂。如果进行简化，就会简单得多。这取决于任务……

The false solution is not correct. That one is a non-simplified solution. Our solution is the simplified one.

错误解法是不正确的。那个是非简化解。我们的解法是简化后的解。

Any questions? No. There are no questions, so we'll continue. Let's move to question six. Question six is a NAND and NOT gate. Implement the following…

有问题吗？没有。没有问题，我们继续。来看第六题。第六题是NAND门和NOT门。实现以下内容……

We know that 'not' is like this: A, B, and not. Basically, we have to convert all the expressions to include the form like this. If we can successfully convert each expression to this form, we can use the...

我们知道，'非'的形式是这样的：A、B 和非。基本上，我们必须将所有表达式转换为包含这种形式。如果我们能够成功地将每个表达式转换为这种形式，我们就可以使用……

For the first one, it looks very complex. It is the longest one compared to B and C. When we look at the form, the first part is fine because it is very similar to this form. We can use Lanczos directly.

第一个表达式看起来非常复杂，与B和C相比它是最长的。当我们观察其形式时，第一部分是没有问题的，因为它与这个形式非常相似。我们可以直接使用Lanczos方法。

This variable looks friendly; we can use lambda directly. But here, this is also okay. However, here it is not. We have an "or" here, and also "or" here. So what we have to do is try to convert this term and this term to something like this. If you think about De Morgan's theory, the first term can be converted to A dot H.

这个变量看起来是友好的，我们可以直接使用 lambda。这里也是可以的。但这里不行。我们这里有一个"或"，这里也有一个"或"。所以我们要做的是尝试将这个项和这个项转换为类似这样的形式。如果你考虑德摩根定律，第一个项可以转换为 A dot H。

Now we can apply Kirchhoff's law. Because A and B are not connected, the current between them is zero. This is the reason I want to apply Kirchhoff's law here.

现在我们可以应用基尔霍夫定律。由于A和B之间没有连接，它们之间的电流为零。这就是我想要在此处应用基尔霍夫定律的原因。

This is the reason I want to apply distributed load here. A naught is zero, so this term is zero. Finally, we get this one.

这就是为什么我要在此处施加分布载荷。A零等于零，因此这一项为零。最终我们得到这个结果。

We can also use the same symbol again. Here we have two denotes, so just keep one. Great. And now, all the four terms.

我们也可以再次使用相同的符号。这里有两个 denote，所以只保留一个。很好。现在，所有四项。

All four terms are very similar to this one. This is exactly what we want. But the problem here is we have the OR. We have the OR. So for these two terms, we can simplify further.

这四项与这一项非常相似。这正是我们想要的。但这里的问题是我们有或运算。我们有或运算。所以对于这两项，我们可以进一步化简。

Just A not, because B or B not is one. And there is A not here. So finally it should be just A not or C D, right?

只剩A非，因为B或B非等于1。这里还有一个A非。所以最终结果应该就是A非或CD，对吗？

When you compare this one and this one, they both have A₀. If you use the VDF law, P₀V₀ = 1.

当你比较这一个和这一个时，它们都有A₀。如果使用VDF定律，则P₀V₀ = 1。

One or anything is one. So this is the final form if you simplify the expression. But it is a little bit different than the NAND gate. So let us apply De Morgan's theorem again, because we cannot just apply it to NOTs that have been eliminated. Let us think about the term below.

一或任何值都等于一。所以这就是化简表达式后的最终形式。但它与与非门略有不同。因此让我们再次应用德摩根定理，因为我们不能直接对已经被消除的非门应用它。让我们考虑下面的项。

You have A, not, not, right? So finally you get A, C, D. This is exactly two NAND gates. This is one NAND gate, and the total part is another NAND gate. So you simplified a very complex expression just using two NAND gates.

你有A，非，非，对吧？所以最终你得到A、C、D。这恰好是两个与非门。这是一个与非门，而整体部分是另一个与非门。所以你仅用两个与非门就简化了一个非常复杂的表达式。

In this circuit, you have A, C, and D. First, you apply a NAND gate to C and D. Then, you apply A with the output of that NAND gate. Do you have any questions?

在这个电路中，有A、C和D。首先，对C和D应用一个NAND门。然后，将A与该NAND门的输出一起应用。大家有问题吗？

Let's use the distributive law. First, try...

我们来用分配律。首先，试着……

You have C, D, E. So you can obtain this form, the form like this. And then you use De Morgan's theorem.

你有C、D、E。这样你就可以得到这种形式，形如这样的表达式。然后你使用德摩根定理。

And here we can use Tiboga's theory again. Now we can do some simplification. For example, this one is just C naught, right?

这里我们可以再次使用Tiboga理论。现在我们可以做一些简化。例如，这一项就是C零，对吧？

And this one is also V₀. Perhaps we do not have time.

这一项也等于 V₀。也许我们没有时间了。

[No discernible English lecture content identified in this segment. The audio consists primarily of background noise, overlapping indistinct chatter, and ambient sounds without clear instructional speech.]

[该片段中未识别出可辨识的英文授课内容。音频主要由背景噪音、重叠的模糊人声和环境音组成，无清晰的教学语音。]

No coherent English lecture content is present in this segment. The audio consists of ambient classroom noise, shuffling, and indistinct background chatter without identifiable instructional speech.

该时间段内无可辨识的英文授课内容。音频仅包含教室环境噪音、翻动声及模糊的背景交谈声，未检测到可整理的教学语句。