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I need to write the statement 'there exists exactly one x in a domain such that p(x) is true'. This needs to be done without using the uniqueness quantifier $\exists!$. I've been staring at this problem for a while and I honestly haven't a clue how to approach this.

TooTone
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user125535
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3 Answers3

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Something like:

$$ \exists x \in X,\; (p(x) \wedge (\forall y \in X,\; p(y) \rightarrow y=x)) $$

That is, there is at least one element for which $p$ holds, and any element for which $p$ holds is the same.

Luke Mathieson
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  • so that means that there exists at least one y in the domain X such that y having the property p means that y and x are the same? – user125535 Feb 02 '14 at 01:25
  • Yes, it's all the elements $y$ where $y=x$ (i.e. just the one, $x$ itself). – Luke Mathieson Feb 02 '14 at 01:30
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    Not quite. Luke's answer says that there is a solution, $ x$, and that every solution is equal to $ x$. Meaning that $ x $ is the only solution. – Unwisdom Feb 02 '14 at 01:32
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    Let me rewrite the answer with parenthesis so you can see the two implications: $\left(\exists x \in X,; p(x)\right) \wedge \left(\forall y \in X,; p(y) \rightarrow y=x \right)$ – user44197 Feb 02 '14 at 01:55
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Avoiding the uniqueness quantifier which is $\exists!$ as noted in the comments, you could write for the domain $X$

$$\exists x_0 \in X: \forall x \in X: (x=x_0 \wedge p(x)) \vee (x\neq x_0 \wedge \neg p(x) )$$

This is just saying that when $x$ is your special value $x_0$, $p(x)$ is true; otherwise it isn't.

TooTone
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This can only be done if the language has equality i.e., there's a binary predicate "=" which satisfies the equality axioms.

I don't want to spoil the punchline for you, but if you don't see it immediately, think about how you prove the uniqueness of the identity element in a group.

Unwisdom
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  • I danced around your idea for proofing it for about one hour, but still couldn't figure out of proof can you give a more straight-forward hint?? – FazeL Nov 16 '16 at 15:58
  • In a group, you prove the uniqueness of the identity by supposing that $e$ and $e'$ are both solutions to the identity axioms, i.e., $\forall x, ex=x=xe$. Then consider the term $ee'$. Since both $e$ and $e'$ satisfy the identity axioms, we must have $e=ee'=e'$.

    So what was the structure of this argument? We created an identity predicate $I$ and proved $$\forall e \forall e' , I(e) & I(e') \to e=e'.$$

     – Unwisdom Dec 04 '16 at 04:34