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The Following Is an Outline of a Proof That (AB)cAcBc( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c }

question 18

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The following is an outline of a proof that (AB)cAcBc( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks.
 The following is an outline of a proof that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks.  Given sets  A  and  B , to prove that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose  x \in    then we show that  x \in \ldots    . So suppose that   Then by definition of complement,   So by definition of union, it is not the case that (  x  is in  A  or  x  is in  B  ). Consequently,  x  is not in  A    x  is not in  B  because of De Morgan's law of logic. In symbols, this says that  x \notin A  and  x \notin B . So by definition of complement,  x \in \ldots    and  x \in  Thus, by definition of intersection,  x \in   [as was to be shown].  Given sets AA and BB , to prove that (AB)cAcBc( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose xx \in  The following is an outline of a proof that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks.  Given sets  A  and  B , to prove that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose  x \in    then we show that  x \in \ldots    . So suppose that   Then by definition of complement,   So by definition of union, it is not the case that (  x  is in  A  or  x  is in  B  ). Consequently,  x  is not in  A    x  is not in  B  because of De Morgan's law of logic. In symbols, this says that  x \notin A  and  x \notin B . So by definition of complement,  x \in \ldots    and  x \in  Thus, by definition of intersection,  x \in   [as was to be shown].  then we show that xx \in \ldots  The following is an outline of a proof that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks.  Given sets  A  and  B , to prove that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose  x \in    then we show that  x \in \ldots    . So suppose that   Then by definition of complement,   So by definition of union, it is not the case that (  x  is in  A  or  x  is in  B  ). Consequently,  x  is not in  A    x  is not in  B  because of De Morgan's law of logic. In symbols, this says that  x \notin A  and  x \notin B . So by definition of complement,  x \in \ldots    and  x \in  Thus, by definition of intersection,  x \in   [as was to be shown].  . So suppose that  The following is an outline of a proof that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks.  Given sets  A  and  B , to prove that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose  x \in    then we show that  x \in \ldots    . So suppose that   Then by definition of complement,   So by definition of union, it is not the case that (  x  is in  A  or  x  is in  B  ). Consequently,  x  is not in  A    x  is not in  B  because of De Morgan's law of logic. In symbols, this says that  x \notin A  and  x \notin B . So by definition of complement,  x \in \ldots    and  x \in  Thus, by definition of intersection,  x \in   [as was to be shown].  Then by definition of complement,  The following is an outline of a proof that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks.  Given sets  A  and  B , to prove that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose  x \in    then we show that  x \in \ldots    . So suppose that   Then by definition of complement,   So by definition of union, it is not the case that (  x  is in  A  or  x  is in  B  ). Consequently,  x  is not in  A    x  is not in  B  because of De Morgan's law of logic. In symbols, this says that  x \notin A  and  x \notin B . So by definition of complement,  x \in \ldots    and  x \in  Thus, by definition of intersection,  x \in   [as was to be shown].  So by definition of union, it is not the case that ( xx is in AA or xx is in BB ). Consequently, xx is not in AA  The following is an outline of a proof that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks.  Given sets  A  and  B , to prove that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose  x \in    then we show that  x \in \ldots    . So suppose that   Then by definition of complement,   So by definition of union, it is not the case that (  x  is in  A  or  x  is in  B  ). Consequently,  x  is not in  A    x  is not in  B  because of De Morgan's law of logic. In symbols, this says that  x \notin A  and  x \notin B . So by definition of complement,  x \in \ldots    and  x \in  Thus, by definition of intersection,  x \in   [as was to be shown].  xx is not in BB because of De Morgan's law of logic. In symbols, this says that xAx \notin A and xBx \notin B . So by definition of complement, xx \in \ldots  The following is an outline of a proof that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks.  Given sets  A  and  B , to prove that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose  x \in    then we show that  x \in \ldots    . So suppose that   Then by definition of complement,   So by definition of union, it is not the case that (  x  is in  A  or  x  is in  B  ). Consequently,  x  is not in  A    x  is not in  B  because of De Morgan's law of logic. In symbols, this says that  x \notin A  and  x \notin B . So by definition of complement,  x \in \ldots    and  x \in  Thus, by definition of intersection,  x \in   [as was to be shown].  and xx \in  The following is an outline of a proof that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks.  Given sets  A  and  B , to prove that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose  x \in    then we show that  x \in \ldots    . So suppose that   Then by definition of complement,   So by definition of union, it is not the case that (  x  is in  A  or  x  is in  B  ). Consequently,  x  is not in  A    x  is not in  B  because of De Morgan's law of logic. In symbols, this says that  x \notin A  and  x \notin B . So by definition of complement,  x \in \ldots    and  x \in  Thus, by definition of intersection,  x \in   [as was to be shown].  Thus, by definition of intersection, xx \in  The following is an outline of a proof that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } . Fill in the blanks.  Given sets  A  and  B , to prove that  ( A \cup B ) ^ { c } \subseteq A ^ { c } \cap B ^ { c } , we suppose  x \in    then we show that  x \in \ldots    . So suppose that   Then by definition of complement,   So by definition of union, it is not the case that (  x  is in  A  or  x  is in  B  ). Consequently,  x  is not in  A    x  is not in  B  because of De Morgan's law of logic. In symbols, this says that  x \notin A  and  x \notin B . So by definition of complement,  x \in \ldots    and  x \in  Thus, by definition of intersection,  x \in   [as was to be shown].  [as was to be shown].


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