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DeMorgan's Second Law:&nbsp; <math class = "center">{(\bigcap^{n}{S_n})}^{c} = \bigcup^{n}{(S_n)}^c</math>&nbsp;  
DeMorgan's Second Law:&nbsp; <math>{(\bigcap^{n}{S_n})}^{c} = \bigcup^{n}{(S_n)}^c</math>&nbsp;  
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Proof:  
 
Proof:  
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#<math class = "center">
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\begin{align}
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x \notin {(\bigcap^{n}{S_n})}^{c} & \Rightarrow x \in {\bigcap^{n}{S_n}} \\
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& \Rightarrow \forall{n}, x \in {S_n} \\
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& \Rightarrow \forall{n}, x \notin {(S_n)}^{c} \\
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& \Rightarrow x \notin {\bigcup^{n}{(S_n)}^{c}}
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\end{align}
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</math>&nbsp;
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#<math class = "center">{(\bigcap^{n}{S_n})}^{c} \subseteq \bigcup^{n}{(S_n)}^c</math>
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#<math class = "center">
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\begin{align}
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x \notin {\bigcup^{n}{(S_n)}^{c}} & \Rightarrow \forall{n}, x \notin {(S_n)}^{c} \\
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& \Rightarrow \forall{n}, x \in {S_n} \\
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& \Rightarrow x \in {\bigcup^{n}{S_n}} \\
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& \Rightarrow x \notin {(\bigcup^{n}{S_n})}^{c}
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\end{align}
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</math>&nbsp;
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#<math class = "center>{(\bigcap^{n}{S_n})}^{c} \supseteq \bigcup^{n}{(S_n)}^c</math>
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#By lines&nbsp;2 and 4:  &nbsp;<math class = "center">{(\bigcap^{n}{S_n})}^{c} = \bigcup^{n}{(S_n)}^c</math>&nbsp;<span class="texhtml">&nbsp;</span>&nbsp;
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----
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Formatting help:
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<math>
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\begin{align}
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x \notin {(\bigcap^{n}{S_n})}^{c} & \Rightarrow  x \in {\bigcap^{n}{S_n}}\\
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&\Rightarrow \forall{n}, x \in {S_n}
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\end{align}
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</math>
  
#If&nbsp;&nbsp; &nbsp;<math>x \notin {(\bigcap^{n}{S_n})}^{c}</math>
 
#<math>\Rightarrow x \in {\bigcap^{n}{S_n}}</math>&nbsp;
 
#<math>\Rightarrow \forall{n}, x \in {S_n}</math>
 
#<math>\Rightarrow \forall{n}, x \notin {(S_n)}^{c}</math>&nbsp;
 
#<math>\Rightarrow x \notin {\bigcup^{n}{(S_n)}^{c}}</math>&nbsp;
 
#By lines 1 through&nbsp;5: <math>x \notin {(\bigcap^{n}{S_n})}^{c} \Rightarrow x \notin {\bigcup^{n}{(S_n)}^{c}}</math>
 
#By line 6; <math>{(\bigcap^{n}{S_n})}^{c} \subseteq \bigcup^{n}{(S_n)}^c</math>
 
#If <math>x \notin {\bigcup^{n}{(S_n)}^{c}}</math>
 
#<math>\Rightarrow \forall{n}, x \notin {(S_n)}^{c}</math>&nbsp;
 
#<math>\Rightarrow \forall{n}, x \in {S_n}</math>&nbsp;
 
#<math>\Rightarrow x \in {\bigcup^{n}{S_n}}</math>&nbsp;
 
#<math>\Rightarrow x \notin {(\bigcup^{n}{S_n})}^{c}</math>&nbsp;
 
#By lines&nbsp;8 through 12:<math>x \notin {\bigcup^{n}{(S_n)}^{c}} \Rightarrow x \notin {(\bigcup^{n}{S_n})}^{c}</math>
 
#By line 13:<math>{(\bigcap^{n}{S_n})}^{c} \supseteq \bigcup^{n}{(S_n)}^c</math>
 
#By lines&nbsp;7 and 14&nbsp;<math>{(\bigcap^{n}{S_n})}^{c} = \bigcup^{n}{(S_n)}^c</math>&nbsp;<span class="texhtml">&nbsp;</span>&nbsp;
 
  
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Example of alignments: <math>x+y</math> <math>x_3^7+y </math> <math class="inline">x_3^7+y</math>
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----
 
<br>[[2013 Spring ECE 302 Boutin|Back to 2013 Spring ECE 302 Boutin]]  
 
<br>[[2013 Spring ECE 302 Boutin|Back to 2013 Spring ECE 302 Boutin]]  
  
 
[[Category:ECE302Spring2013Boutin]] [[Category:ECE]] [[Category:ECE302]] [[Category:Probability]] [[Category:Proofs]] [[Category:DeMorgan's_Second_Law]]
 
[[Category:ECE302Spring2013Boutin]] [[Category:ECE]] [[Category:ECE302]] [[Category:Probability]] [[Category:Proofs]] [[Category:DeMorgan's_Second_Law]]

Latest revision as of 12:01, 19 March 2013



Alternative Proof for DeMorgan's Second Law

By Oluwatola Adeola

ECE 302, Spring 2013, Professor Boutin


During lecture, a proof of DeMorgan’s second law was given as a possible solution to the quiz which was based on showing that both sets are subsets of each other and are therefore equivalent. Here’s is an alternative method of proving the law that relies on determining a subset based on the exclusion of an element rather than inclusion.

  DeMorgan's Second Law:  $ {(\bigcap^{n}{S_n})}^{c} = \bigcup^{n}{(S_n)}^c $ 

Proof:

  1. $ \begin{align} x \notin {(\bigcap^{n}{S_n})}^{c} & \Rightarrow x \in {\bigcap^{n}{S_n}} \\ & \Rightarrow \forall{n}, x \in {S_n} \\ & \Rightarrow \forall{n}, x \notin {(S_n)}^{c} \\ & \Rightarrow x \notin {\bigcup^{n}{(S_n)}^{c}} \end{align} $ 
  2. $ {(\bigcap^{n}{S_n})}^{c} \subseteq \bigcup^{n}{(S_n)}^c $
  3. $ \begin{align} x \notin {\bigcup^{n}{(S_n)}^{c}} & \Rightarrow \forall{n}, x \notin {(S_n)}^{c} \\ & \Rightarrow \forall{n}, x \in {S_n} \\ & \Rightarrow x \in {\bigcup^{n}{S_n}} \\ & \Rightarrow x \notin {(\bigcup^{n}{S_n})}^{c} \end{align} $ 
  4. $ {(\bigcap^{n}{S_n})}^{c} \supseteq \bigcup^{n}{(S_n)}^c $
  5. By lines 2 and 4:  $ {(\bigcap^{n}{S_n})}^{c} = \bigcup^{n}{(S_n)}^c $   



Formatting help: $ \begin{align} x \notin {(\bigcap^{n}{S_n})}^{c} & \Rightarrow x \in {\bigcap^{n}{S_n}}\\ &\Rightarrow \forall{n}, x \in {S_n} \end{align} $


Example of alignments: $ x+y $ $ x_3^7+y $ $ x_3^7+y $



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