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$$\exists! x : A(x) \Rightarrow \exists x : A(x)$$ Assuming that $A(x)$ is an open sentence. I'm new to abstract mathematics and proofs, so I came here to ask for some simplification. Thanks

AlexR
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user2628815
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  • In words this means "If there is exaclty one object satisfying a property, then there is at lease one." – AlexR Apr 08 '14 at 22:46

3 Answers3

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If there is a unique $x$ such that $A(x)$ is true,

then there must be some $x$ for which $A(x)$ is true.

gt6989b
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Notation $\exists! x. P(x)$ means "there is a unique $x$ such that $P(x)$". The common way to express this is by adding "$P(y)$ implies $x = y$", that is:

$$\exists x :P(x) \land \Big(\forall y : P(y) \implies (x = y)\Big).$$

However, please check, if this actually is the definition of $\exists!$ your book/professor uses. If yes, then your problem could be rewritten to

$$\Bigg(\exists x :A(x) \land \Big(\forall y : A(y) \implies (x = y)\Big)\Bigg) \implies \Bigg(\exists x : A(x)\Bigg),$$

which you can see is of the form $\alpha \land \beta \implies \alpha$ modulo the quantifier. If this isn't your definition, use the one from your course and proceed accordingly.

I hope this helps $\ddot\smile$

dtldarek
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The problem states that:

If there exists exactly one $x$ such that $A(x)$ is true, then there exists at least one $x$ such that $A(x)$ is true.

The proof should be trivial.