blocks|key|1228104|text|是否有任何实例>将更改为<=？
|type|blockquote|depth|inlineStyleRanges|entityRanges|data|1228105|unstyled|1228106|答:每次你否定它。|1228107|考虑一下x+>+1。对此的否定显然是x+<=+1。如果简单地将其否定为x+<+1，那么这两种情况都不包括x+==+1大小写。|offset|length|style|CODE|1228108|1228109|尽管如此，给定的布尔!+(+!+(+a+!=+b)++&&++(+b+>+7+)+)可以分解如下：|1228110|给予：
!+(+!(a+!=+b)++&&++(b+>+7))|ordered-list-item|1228111|否定a+!=+b!+((a+==+b)++&&++(b+>+7))|1228112|分发!!(a+==+b)+%7C%7C+!(b+>+7)|1228113|否定a==b(a+!=+b)+%7C%7C+!(b+>+7)|1228114|否定b>7(a+!=+b)+%7C%7C+(b+<=+7)|1228115|1228116|1228117|因此，答案是B。|1228118|entityMap^0|0|0|0|4|5|I|6|Z|5|1G|6|0|0|A|W|0|4|R|0|2|V|0|2|M|0|2|O|0|2|N|0|0|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@]|A|$]]|$1|B|3|-4|5|C|7|16|8|@]|9|@]|A|$]]|$1|D|3|E|5|C|7|17|8|@]|9|@]|A|$]]|$1|F|3|G|5|C|7|18|8|@$H|19|I|1A|J|K]|$H|1B|I|1C|J|K]|$H|1D|I|1E|J|K]|$H|1F|I|1G|J|K]]|9|@]|A|$]]|$1|L|3|-4|5|C|7|1H|8|@]|9|@]|A|$]]|$1|M|3|N|5|C|7|1I|8|@$H|1J|I|1K|J|K]]|9|@]|A|$]]|$1|O|3|P|5|Q|7|1L|8|@$H|1M|I|1N|J|K]]|9|@]|A|$]]|$1|R|3|S|5|Q|7|1O|8|@$H|1P|I|1Q|J|K]]|9|@]|A|$]]|$1|T|3|U|5|Q|7|1R|8|@$H|1S|I|1T|J|K]]|9|@]|A|$]]|$1|V|3|W|5|Q|7|1U|8|@$H|1V|I|1W|J|K]]|9|@]|A|$]]|$1|X|3|Y|5|Q|7|1X|8|@$H|1Y|I|1Z|J|K]]|9|@]|A|$]]|$1|Z|3|-4|5|C|7|20|8|@]|9|@]|A|$]]|$1|10|3|-4|5|C|7|21|8|@]|9|@]|A|$]]|$1|11|3|12|5|C|7|22|8|@]|9|@]|A|$]]|$1|13|3|-4|5|C|7|23|8|@]|9|@]|A|$]]]|14|$]]

<blockquote>
 Is there any instance > would change to &lt;= ?
</blockquote>

Answer: every time you negate it.

Consider <code>x &gt; 1</code>. The negation of this is clearly <code>x &lt;= 1</code>. If you simply negate it as <code>x &lt; 1</code> then neither case covers the <code>x == 1</code> case.

<hr>

That being said, the given boolean <code>! ( ! ( a != b) &amp;&amp; ( b &gt; 7 ) )</code> can be decomposed as follows:

<ol>
<li>Given:

<code>! ( !(a != b) &amp;&amp; (b &gt; 7))</code></li>
<li>Negate <code>a != b</code>:

<code>! ((a == b) &amp;&amp; (b &gt; 7))</code></li>
<li>Distribute the <code>!</code>:

<code>!(a == b) || !(b &gt; 7)</code></li>
<li>Negate <code>a==b</code>:

<code>(a != b) || !(b &gt; 7)</code></li>
<li>Negate <code>b&gt;7</code>:

<code>(a != b) || (b &lt;= 7)</code></li>
</ol>

The answer is, therefore, B.

blocks|key|731509|text|答案应该是B。这是因为(a+=+b)旁边的否定是先求值的，然后再将外部否定分配给整个命题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|731510|使用DeMorgan的法律，&&将切换到\。类似地，!=变成==，>变成<=。|731511|!(!(a+!=+b)+&&+(b+>+7))
!((a+==+b)+&&+(b+>+7))
+(a+!=+b)+%7C%7C+(b+<=+7)|code-block|syntax|javascript|731512|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The answer should be B. This is because the negation next to the (a != b) is evaluated first, then you distribute the outside negation to the entire proposition.

Using DeMorgan's Laws, the &amp;&amp; will switch to ||. Similarly, != becomes ==, and > becomes &lt;=.

<pre><code>!(!(a != b) &amp;&amp; (b &gt; 7))
!((a == b) &amp;&amp; (b &gt; 7))
 (a != b) || (b &lt;= 7)
</code></pre>

blocks|key|1228134|text|!+(+!+(+a+!=+b)++&&++(+b+>+7+)+)

=+!+(+(a+=+b)+&&+(b+>+7))

=+(a+!=+b)+%7C%7C+(b+<=+7)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1228135|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

blocks|key|1228162|text|答案是B：((a=+b)+&(b>7))|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1228163|让我们把它分成几个部分。部分虚拟：(a!=b)第X部分：！虚拟部分Y：(b>7)|1228164|现在!X+=虚拟=>虚拟=>的双负(a!=b)|1228165|!Y+=+!(b>7)+=>+b不应大于7+=>+b应小于或等于7+=>+(b<=7)|1228166|现在剩下的问题是，如何&+is+=|1228167|原来的问题是：(X+&+Y)+=>不应该是(X和Y+)+=>，它不应该是X的否定，也不应该是Y的否定，因为如果它不是X，则条件(X和Y)变成假的，从而(X和Y)变为真，从而得到原来的条件。Y.也是如此。|1228168|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|P|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|Q|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|R|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|S|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|M|$]]

answer is B.
to understand this :
! ( ! ( a != b) &amp;&amp; ( b > 7 ) )

Lets break it into parts. 
Part dummy: (a!=b)
Part X: !dummy
Part Y: (b>7)

Now !X = double negate of dummy => dummy => (a!=b)

!Y = !(b>7) => b should not be greater than 7 => b should be less than or equal to 7 => (b&lt;=7)

Now problem left is how &amp;&amp; becomes ||

So original question is: !( X &amp;&amp; Y ) => should not be (X and Y) => it should be either negate of X or it should be negate of Y, because if instead of X it is ~X, the condition (X and Y) becomes false and hence !(X and Y) becomes true and hence original condition is achieved. Similarly for Y.

blocks|key|731638|text|首先应用内支架Logical+NOT+(!)**:**+|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|CODE|entityRanges|data|731639|!(!(a+!=+b)+&&+(b+>+7))+becomes+!((a+==+b)+&&+(b+>+7))|code-block|syntax|javascript|731640|使用De+Morgan's+Law，我们将operators反转为对应的。|731641|我们将>更改为<=，因为>+operator不包含7本身或更少的内容，因此<=是唯一满足该条件的。|731642|​|731643|📷|atomic|731644|731645|现在的!**:**是，现在是外部|731646|在查看truth+tables+(上图)时，您会注意到Logical+AND+(&&)和Logical+OR+(%7C%7C)在比较两个不同的boolean+expressions+(即真假、假真)时有相反的结果，因此当我们应用!时，我们用%7C%7C反转&&。最后，我们需要再次将==切换到!=。|731647|全部启动，这会产生|731648|((a+!=+b)+%7C%7C+(b+<=+7))|731649|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://developer.qcloudimg.com/http-save/yehe-900000/50321dc91e8ab73bd202521c4b894ab8.png|imageAlt^0|7|H|7|F|0|0|2|F|L|9|0|3|1|7|2|C|1|E|8|11|2|0|0|0|1|0|0|0|3|3|A|6|3|1|0|3|C|R|G|18|F|1V|J|32|1|38|2|3C|2|3P|2|3U|2|0|0|9|0|0|M|0^^$0|@$1|2|3|4|5|6|7|1C|8|@$9|1D|A|1E|B|C]|$9|1F|A|1G|B|D]]|E|@]|F|$]]|$1|G|3|H|5|I|7|1H|8|@]|E|@]|F|$J|K]]|$1|L|3|M|5|6|7|1I|8|@$9|1J|A|1K|B|D]|$9|1L|A|1M|B|D]]|E|@]|F|$]]|$1|N|3|O|5|6|7|1N|8|@$9|1O|A|1P|B|D]|$9|1Q|A|1R|B|D]|$9|1S|A|1T|B|D]|$9|1U|A|1V|B|D]|$9|1W|A|1X|B|D]]|E|@]|F|$]]|$1|P|3|Q|5|6|7|1Y|8|@]|E|@]|F|$]]|$1|R|3|S|5|T|7|1Z|8|@]|E|@$9|20|A|21|1|22]]|F|$]]|$1|U|3|Q|5|6|7|23|8|@]|E|@]|F|$]]|$1|V|3|W|5|6|7|24|8|@$9|25|A|26|B|C]|$9|27|A|28|B|C]|$9|29|A|2A|B|D]]|E|@]|F|$]]|$1|X|3|Y|5|6|7|2B|8|@$9|2C|A|2D|B|D]|$9|2E|A|2F|B|D]|$9|2G|A|2H|B|D]|$9|2I|A|2J|B|D]|$9|2K|A|2L|B|D]|$9|2M|A|2N|B|D]|$9|2O|A|2P|B|D]|$9|2Q|A|2R|B|D]|$9|2S|A|2T|B|D]]|E|@]|F|$]]|$1|Z|3|10|5|6|7|2U|8|@$9|2V|A|2W|B|C]]|E|@]|F|$]]|$1|11|3|12|5|6|7|2X|8|@$9|2Y|A|2Z|B|D]]|E|@]|F|$]]|$1|13|3|-4|5|6|7|30|8|@]|E|@]|F|$]]]|14|$15|$5|16|17|18|F|$19|1A|1B|-4]]]]

Firt apply the inner bracket <code>Logical NOT (!)</code>:

<pre><code>!(!(a != b) &amp;&amp; (b &gt; 7)) becomes !((a == b) &amp;&amp; (b &gt; 7))
</code></pre>

With <code>De Morgan's Law</code> we reverse the <code>operators</code> to their counterpart.

We change <code>&gt;</code> to <code>&lt;=</code> because the <code>&gt;</code> <code>operator</code> does not include 7 itself or anything less, hence the <code>&lt;=</code> is the only one that satisfies that condition.

<img src="https://i.stack.imgur.com/VxEE3.png" alt="Truth Table">

Now the outer <code>!</code>:

When looking at <code>truth tables</code> (picture above), you'll notice that <code>Logical AND (&amp;&amp;)</code> and <code>Logical OR (||)</code> have opposite results when comparing 2 different <code>boolean expressions</code> (i.e. truth false, false true), hence when we apply the <code>!</code>, we reverse the <code>&amp;&amp;</code> with <code>||</code>. Finally we need to switch the <code>==</code> to <code>!=</code> again.

All up, this produces

<code>((a != b) || (b &lt;= 7))</code>

One of my exam questions reads:

<pre><code>! ( ! ( a != b) &amp;&amp; ( b &gt; 7 ) )
</code></pre>

The choices:

<pre><code>a) (a != b) || (b &lt; 7)
b) (a != b) || (b &lt;= 7)
c) (a == b) || (b &lt;= 7)
d) (a != b) &amp;&amp; (b &lt;= 7)
e) (a == b) &amp;&amp; (b &gt; 7)
</code></pre>

Initially, I thought it would be D. This is incorrect, and I realize why. I don't understand how the logical negation operator reverses &amp;&amp; and greater than/less than. I believe I have narrowed it down to the first two. Is there any instance > would change to &lt;= ?

Boolean negation

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我的一个试题是：! ( ! ( a != b) && ( b > 7 ) )选择：a) (a != b) || (b < 7)b) (a != b) || (b <= 7)c) (a == b) || (b <= 7)d) (a != b) && (b <= 7)e) (a == b) && (b > 7)一开始，我以为是D。这是不正确的，我知道原因。我不明白逻辑否定操作符是如何反转的&而且大

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