# Using existential quantifier within implication

I had used an existential quantifier within an implication, as below:

$$\exists ~ \sigma_{opt}~~~ \wedge~~~ n \in R^{+}~ ,~0<n<1 \rightarrow ~~ \sigma_{opt} = n$$

I've, recently, found that it is illegal to do that and just universal quantifier could be used in this situation. Could anybody explain the reason?

• This doesn't make sense... If I write it out, it says, "The existence of a $\sigma_{opt}$ and an $n$ within the positive real number range between 0 and 1 non-inclusive implies that $\sigma_{opt}$ is $n$." I don't see anything wrong with that. Oct 9 '16 at 20:23
• @VermillionAzure: Actually, a professor in mathematics just stated the incorrectness of such statement, but didn't explain much about the underlying reason.
– User
Oct 9 '16 at 20:26
• You should include the proposed answer to replace this. Oct 9 '16 at 20:28
• @VermillionAzure It's completely meaningless as a statement of first-order logic. First, there is no comma operator in FO. Second, $\land$ can only be placed between syntactically legal formulas: $\exists\,\sigma_{\mathrm{opt}}$ is not syntactically legal. Third, it's unclear how much of the stuff of the implication operator is supposed to be part of that implication. Oct 15 '16 at 10:19

Your statement attempts to be express in formal logic the following sentence:

There exist $\sigma_{opt}$ and $n$ in $R^+$ such that if $0 < n < 1$ then $\sigma_{opt} = n$.

The attempt at translation goes about by replacing English words by mathematical symbols, while keeping English syntax. Unfortunately, that is not how this sort of translation works. Just as when translating between human languages the syntax needs to be changed, here as well you have to conform to the syntax of statements in predicate logic. Your statement is, unfortunately, ungrammatical (more formally, it is not well-formed, that is, it is not a wff).

The syntax of sentences in predicate logic is described in many sources, so let me just mention that the following sentence is valid: $$\exists \sigma_{opt} \in R^+ \exists n \in R^+ \; (0 < n \land n < 1) \to \sigma_{opt} = n.$$ We often allow shortcuts to be used, and so consider the following statement to be valid as well (standing for the former statement): $$\exists \sigma_{opt},n \in R^+ \; 0 < n < 1 \to \sigma_{opt} = n.$$ You are using a different shortcut for folding the two existential quantifiers, but your convention is not standard; it's better to use standard conventions, unless there's a good reason to deviate from standard practice.

Finally, I'm not sure what your statement is supposed to express, but it probably doesn't express what you were hoping for: it doesn't depend on any parameter (this is why it's a statement rather than just a formula), which was probably not your original intention.

• To be a bit more pedantic by adding brackets: $\exists \sigma_{opt} \in R^+, \exists n \in R^+, [(0 < n < 1) \rightarrow (\sigma_{opt} = n)]$ Oct 10 '16 at 5:03
• One could also write "$\exists (\sigma_\text{opt}, n) \in (R^{+})^2 \dots$". This more clearly indicates that $n$ need not depend on $\sigma_\text{opt}$ (if that is the OP's intent, which is not clear to me). Oct 10 '16 at 7:05
• When you say "the following sentence is valid", I'm not sure if you mean "it's a legitimate expression" or "it's a tautology". But it's a tautology. It would usually be parsed as "there exist $\sigma_{\mathrm{opt}}$ and $n$ such that, if $n\in(0,1)$, then $n=\sigma_{\mathrm{opt}}$." Well, I'll make that true by choosing $n=2$ and $\sigma_{\mathrm{opt}}=\pi$. Oct 15 '16 at 10:15
• @DavidRicherby Validity was intended only on the syntactic level. Oct 15 '16 at 10:31