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		<title>23666: HW 1 - Revision history</title>
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		<updated>2026-05-04T16:36:58Z</updated>
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		<id>http://helmut.knaust.info/mediawiki/index.php?title=23666:_HW_1&amp;diff=2595&amp;oldid=prev</id>
		<title>HelmutKnaust: Created page with &quot;'''Problem 1.''' Show that the following two statements are equivalent: $(P\wedge Q)\Rightarrow R$ and $(P\wedge \neg R)\Rightarrow \neg Q$.  '''Problem 2.''' You have seen ho...&quot;</title>
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				<updated>2019-01-29T23:48:20Z</updated>
		
		<summary type="html">&lt;p&gt;Created page with &amp;quot;&amp;#039;&amp;#039;&amp;#039;Problem 1.&amp;#039;&amp;#039;&amp;#039; Show that the following two statements are equivalent: $(P\wedge Q)\Rightarrow R$ and $(P\wedge \neg R)\Rightarrow \neg Q$.  &amp;#039;&amp;#039;&amp;#039;Problem 2.&amp;#039;&amp;#039;&amp;#039; You have seen ho...&amp;quot;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;'''Problem 1.''' Show that the following two statements are equivalent: $(P\wedge Q)\Rightarrow R$ and $(P\wedge \neg R)\Rightarrow \neg Q$.&lt;br /&gt;
&lt;br /&gt;
'''Problem 2.''' You have seen how to generate compound statements using the four connectives $\neg, \vee, \wedge$ and $\Rightarrow$. This problem addresses the question whether all four connectives are necessary. &lt;br /&gt;
#Use a truth table to show that $A\Rightarrow B$ is equivalent to $\neg(A \wedge \neg B)$.&lt;br /&gt;
#Show that $A\vee B$ can be written using only the connectives $\neg$ and $\wedge$. &lt;br /&gt;
#Thus the two connectives $\neg$ and $\wedge$ suffice to generate all compound statements. It is possible to further reduce to only one connective, albeit a different one: Let us define the new connective $\mbox{NOR}$ by setting $A \mbox{ NOR } B \Leftrightarrow \neg(A\vee B).$ Show that the four compound statements $\neg A,\ A\vee B,\ A\wedge B$ and $A\Rightarrow B$ can be written using only $\mbox{NOR}$-connectives.&lt;br /&gt;
&lt;br /&gt;
'''Problem 3.''' In each case, give an example, or explain why such an example&lt;br /&gt;
cannot exist:&lt;br /&gt;
#Is there a predicate $A(x,y)$ such that the statement $\forall x\&lt;br /&gt;
\exists y:\ A(x,y)$ is true, while the statement $\exists y\&lt;br /&gt;
\forall x:\ A(x,y)$ is false?&lt;br /&gt;
#Is there a predicate $A(x,y)$ such that the statement $\exists y\&lt;br /&gt;
\forall x:\ A(x,y)$ is true, while the statement $\forall x\&lt;br /&gt;
\exists y:\ A(x,y)$ is false?&lt;br /&gt;
&lt;br /&gt;
'''Problem 4.''' Prove Theorem 1.3.2: If $A(x)$ is an open statement with variable $x$, then&lt;br /&gt;
#$(\exists ! x)A(x)\Rightarrow(\exists x) A(x)$.&lt;br /&gt;
#$(\exists ! x) A(x)$ is equivalent to $(\exists x)A(x)\wedge(\forall y)(\forall z)(A(y)\wedge A(z)\Rightarrow  y=z)$.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
'''Problem 5.''' A clothing store advertises: ''For every customer we have a rack of clothes that fit.''&lt;br /&gt;
#Write the statement above using quantifier(s) and predicate(s). &lt;br /&gt;
#Negate the sentence using quantifier(s) and predicate(s).&lt;br /&gt;
#Write the negation in the form of an English sentence. (Don't just write ''It is not true that...'')&lt;/div&gt;</summary>
		<author><name>HelmutKnaust</name></author>	</entry>

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