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I'm having trouble with rewriting this expression only with the use of: Brackets, variables, $A$, $B$, $\neg$,$\exists$,$\lor$,$\in$

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I tried various solution but the only one that convinces me a little bit is the following one:

$$\forall a \in A \to \neg \exists b \in B \space (a \in B) \equiv \neg(\forall a \in A) \lor \neg \exists b \in B\space (a\in B) \equiv\exists a \in A \lor \neg \exists b \in B (a \in B)$$

Is this right? Could you otherwise give me a hint on how to translate those expressions only with the use of symbols?

•B=$\emptyset$

•$\bigcup A$

•If A and B are sets than exists $C=\{A,B\}$

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3 Answers

up vote 1 down vote accepted

As answer on your first question: the following statements are equivalent

$A\nsubseteq\wp\left(B\right)$

$\exists a\left[a\in A\wedge\neg a\in\wp\left(B\right)\right]$

$\exists a\left[a\in A\wedge\exists x\left[x\in a\wedge\neg x\in B\right]\right]$

$\exists a\neg\left[\neg a\in A\vee\neg\exists x\left[x\in a\wedge\neg x\in B\right]\right]$

$\exists a\neg\left[\neg a\in A\vee\neg\exists x\neg\left[\neg x\in a\vee x\in B\right]\right]$

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Thank you! Could you explain me why you only need to say that that $\exists a [a \in A \land \neg a \in P(B)]$ and not $\forall a[a \in A \land \neg a \in P(B)]$? My questions comes from the fact that actually every a $\in$ A is not contained in any set of $B$ –  Ale Feb 6 at 12:15
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The existence of one element in a set $A$ that does not belong to another set $C$ justifies the conclusion that $A$ is not a subset of $C$. It does not have to be so that every element of $A$ is not an element of $C$. If that is the case then you have $A\cap C=\emptyset$ wich is not required. –  drhab Feb 6 at 12:23
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Basically, make sure you know the definitions of all the involved symbols, expand them, and see where you end up.

Here is the expanding part: \begin{align} & A \not\subseteq \mathcal P(B) \\ \equiv & \qquad \text{"definition of $\;\not\ \;$"} \\ & \lnot (A \subseteq \mathcal P(B)) \\ \equiv & \qquad \text{"definition of $\;\subseteq\;$"} \\ & \lnot \langle \forall V : V \in A : V \in \mathcal P(B) \rangle \\ \equiv & \qquad \text{"definition of $\;\mathcal P\;$"} \\ & \lnot \langle \forall V : V \in A : V \subseteq B \rangle \\ \equiv & \qquad \text{"definition of $\;\subseteq\;$"} \\ & \lnot \langle \forall V : V \in A : \langle \forall x : x \in V : x \in B \rangle \rangle \\ \equiv & \qquad \text{"..."} \\ \end{align} Now get rid of the $\;\forall\;$ using the rules of predicate logic, and you are done.

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It looks like half your $:$ should be $\to $. –  Hagen von Eitzen Feb 6 at 11:19
    
@HagenvonEitzen That's a nice way to put it. :-) But I'm using a quantifier notation due to Dijkstra and Scholten; see, e.g., Dijkstra's EWD1300 under Quantification. (A variant is Gries-Schneider's notation, which is $\;\langle\forall x\mid R:P\rangle\;$.) EWD1300 says we "don’t need a variable range parameter". "But [it] allows more elegant formulae". –  Marnix Klooster Feb 6 at 11:33
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Hint:

We want to say that there is an $x$ such that $x\in A$ and $x$ is not in the powerset of $B$.

Recall that $P\land Q$ is equivalent to $\lnot(\lnot P\lor \lnot Q)$.

So the "and" can be eliminated.

Now we want to say that $x$ is not in the powerset of $B$. This says that there is a $t$ such that $t\in x$ and $t$ is not in $B$. Eliminate "and" as before, and put the pieces together.

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