Logical negation of statements

[mathmari]

Hey! :unsure:

It is given that the negation of "$\forall a\in A: \alpha (a)$" is "$\exists a\in A: \neg \alpha (a)$" and the negation of "$\exists b\in B: \beta (b)$" is "$\forall b\in B: \neg \beta (b)$".

I want to show that the negation of "$\forall a \in A \ \exists b \in B \ : \ \alpha (a, b)$" is "$\exists a \in A \ \forall b \in B \ : \ \neg \alpha (a, b)$".

Do we show that as follows? \begin{align*}\neg \left (\forall a \in A \ \exists b \in B \ : \ \alpha (a, b)\right )&\equiv \exists a \in A \ \neg \left ( \exists b \in B \ : \ \alpha (a, b)\right ) \\ & \equiv \exists a \in A \ \forall b \in B \ : \ \neg \alpha (a, b)\end{align*} Let $A\subset \mathbb{R}$ and let $f, g, h:A\rightarrow \mathbb{R}$ be functions.

I want to determine the negation of "$\forall \epsilon>0 \ \exists \alpha ,\beta >0 \ : \ [x\in A, f(x)\leq \alpha , g(x)\geq \beta \Rightarrow h(x)\leq \epsilon]$".

I have done the following:
\begin{align*}&\neg \left (\forall \epsilon>0 \ \exists \alpha ,\beta >0 \ : \ [x\in A, f(x)\leq \alpha , g(x)\geq \beta \Rightarrow h(x)\leq \epsilon]\right ) \\ & \equiv \exists \epsilon>0 \ \forall \alpha ,\beta >0 \ : \ \neg [x\in A, f(x)\leq \alpha , g(x)\geq \beta \Rightarrow h(x)\leq \epsilon] \\ & \equiv \exists \epsilon>0 \ \forall \alpha ,\beta >0 \ : \ \neg [\neg \left (x\in A, f(x)\leq \alpha , g(x)\geq \beta\right ) \lor h(x)\leq \epsilon] \\ & \equiv \exists \epsilon>0 \ \forall \alpha ,\beta >0 \ : \ [\left (x\in A, f(x)\leq \alpha , g(x)\geq \beta\right ) \land h(x)> \epsilon]\end{align*}

Is everything correct?

:unsure:

 

[Evgeny.Makarov]

Yes, everything is correct. It makes sense to write either only commas or only $\land$ in the last formula.

 

[solakis1]

Hey! :unsure:

It is given that the negation of "$\forall a\in A: \alpha (a)$" is "$\exists a\in A: \neg \alpha (a)$" and the negation of "$\exists b\in B: \beta (b)$" is "$\forall b\in B: \neg \beta (b)$".

I want to show that the negation of "$\forall a \in A \ \exists b \in B \ : \ \alpha (a, b)$" is "$\exists a \in A \ \forall b \in B \ : \ \neg \alpha (a, b)$".

Do we show that as follows? \begin{align*}\neg \left (\forall a \in A \ \exists b \in B \ : \ \alpha (a, b)\right )&\equiv \exists a \in A \ \neg \left ( \exists b \in B \ : \ \alpha (a, b)\right ) \\ & \equiv \exists a \in A \ \forall b \in B \ : \ \neg \alpha (a, b)\end{align*} Let $A\subset \mathbb{R}$ and let $f, g, h:A\rightarrow \mathbb{R}$ be functions.

I want to determine the negation of "$\forall \epsilon>0 \ \exists \alpha ,\beta >0 \ : \ [x\in A, f(x)\leq \alpha , g(x)\geq \beta \Rightarrow h(x)\leq \epsilon]$".

I have done the following:
\begin{align*}&\neg \left (\forall \epsilon>0 \ \exists \alpha ,\beta >0 \ : \ [x\in A, f(x)\leq \alpha , g(x)\geq \beta \Rightarrow h(x)\leq \epsilon]\right ) \\ & \equiv \exists \epsilon>0 \ \forall \alpha ,\beta >0 \ : \ \neg [x\in A, f(x)\leq \alpha , g(x)\geq \beta \Rightarrow h(x)\leq \epsilon] \\ & \equiv \exists \epsilon>0 \ \forall \alpha ,\beta >0 \ : \ \neg [\neg \left (x\in A, f(x)\leq \alpha , g(x)\geq \beta\right ) \lor h(x)\leq \epsilon] \\ & \equiv \exists \epsilon>0 \ \forall \alpha ,\beta >0 \ : \ [\left (x\in A, f(x)\leq \alpha , g(x)\geq \beta\right ) \land h(x)> \epsilon]\end{align*}

Is everything correct?

:unsure:

Does the above hold for some x or for all x
What do the symbols (:) (,) mean

 

[Evgeny.Makarov]

It's a bad idea to overquote. The first part of the quote has nothing to do with $x$.

Does the above hold for some x or for all x

The equivalence holds for each particular $x$, so for all $x$.

What do the symbols (,) mean

I am not sure about the smiley, but comma means conjunction in this context.

 

[solakis1]

Sorry i mean ... :

 

[Evgeny.Makarov]

In this case colon is just a part of formula syntax: $\forall x:\,A$. There are many variations of this syntax.

 

[solakis1]

oh yes $\forall x$