Mathematical Logic: Doubt on meaning of statement

Question

The notation $\exists ! x P(x)$ denotes the proposition “there exists a unique $x$ such that $P(x)$ is true”. Give the truth values of the following statements : 

I)${\color{Red} {\exists ! x P(x)}} \rightarrow \exists x P(x)$

II)${\color{Red} {\exists ! x\sim P(x)}} \rightarrow \neg \forall x P(x)$

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What will be answer here?? Is the assumption only for left hand side and not right hand side??

Comments

sorry, not getting what you are asking.

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check below comment.

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$\exists!$ is a symbol. it is not $\exists$ and then negation.

You can check the interpretation of $\exists!$ here https://math.stackexchange.com/questions/3245560/write-out-exists-xpx-where-the-domain-consists-of-the-integers-1-2-a

In (1) Try to put P(1) or P(2) or P(3) true..either one and remaining false. and then check right side whether you are getting true or false.

In (2) do opposite...

Answer 1

Let $P(x):(x-1=0)$ defined over $\mathbb{Z}^{+}$

then $x=1$ is the only unique solution for which $P(x)$ is true.

 

  $\exists !xP(x)$ means " there exists a unique $x$ such that $P(x)$ is true” and here the unique value of $x=1$

$\Rightarrow$ We can say that there exist a value of $x$ for which proposition holds true.

$\Rightarrow$ $\exists xP(x)$

So if  there exists a unique $x$ such that $P(x)$ is true then there will exist atleast $1$ value of $x$ for which proposition holds true.

$\therefore$ $\exists !xP(x)\rightarrow\exists xP(x)$ is true.

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Let $P(x):(5/x$ is defined $)$ defined over $\mathbb{R}^{+}$

then $x=0$ is the only unique solution for which $P(x)$ is false since $5/0$ is undefined.

 

$\exists !x\sim P(x)$ means there exists a unique $x$ such that $P(x)$ is not true i.e. $P(x)$ is false and here the unique value is $x=0$

$\sim \forall xP(x)$ can also be written as $\exists x\sim P(x)$ which means there exists at least one $x$ such that $P(x)$ is not true i.e. $P(x)$ is false and here we are having x=0 which is satisfying.

So if there exists a unique $x$ such that $P(x)$ is not true i.e. $P(x)$ is false then there will definitely  exists at least one $x$ such that $P(x)$ is not true i.e. $P(x)$ is false

$\Rightarrow$  $\exists !x\sim P(x)\rightarrow\sim \forall xP(x)$ is true.

Comments

means when r u getting !then only we r using unique

right??

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Say in. 2nd statement LHS, we took both not outside

Then what will be LHS??

${\color{Red} {\exists ! x\sim P(x)}}$=${\color{Red} {\exists \sim x\sim P(x)}}$

=${\sim\forall x\sim P(x)}$

=${\forall x P(x)}$

Isnot it??

Then $LHS\neq RHS$

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It is a quantifier also known as uniqueness quantifier. Please read about it.

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${\color{Red} {\exists ! x\sim P(x)}}$=${\color{Red} {\exists \sim x\sim P(x)}}$

@srestha You can't do this.

! and ~ are not the same.

$\exists !$ is a single symbol which represents uniqueness quantifier and ~ represents not operation