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Must the $\forall$ statement be used in front of the consequent statement?

For example, must we write $\forall x\in\mathbb{Z}^{+}, x>0$, or can we also write $x>0, \forall x\in\mathbb{Z}^{+}$

In proofs, the 1st one is most common, but I have a personal preference to write the 2nd one. I just wanted to ask which one (or both) is accepted, and which one is preferred. Thanks in advance.

IraeVid
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    It's up to you, as soon as it is understandable and unambiguous. Remember that such a sentence is just a translation in formal language of a real-life spoken sentence, and that "All positive integers are positive" and "A number is positive as soon as it is a positive integer" are both valid. – Didier May 20 '23 at 12:06
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    Usually the quantificator has to be on the left of the statement (1st one). Using them in a "natural language" way is bad practice (my teacher would give penalties for such use ahah... ). Use english words if you want a formulation like the second statement, not logical quantificator that have to obey rules. Also, if you have multiple quantificators, the order is critical and it can be even more misleading if you do not respect this. – Lelouch May 20 '23 at 12:25
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    Besides the issue of hanging quantifiers (do a site search), I also suggest omitting that comma; both issues relate to your quantifiers' scope ambiguity. By the way, there is no consequent in your displayed statement; consequent means the right side of a conditional statement. Finally: "I have a personal preference to write the 2nd one": I feel that peppering mathematical prose with symbols decreases readability; however, when writing in symbols is actually helpful (e.g., to clarify meaning), there's no point doing so ambiguously, right? – ryang May 20 '23 at 13:35

4 Answers4

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In every formal system I have ever encountered, only $\forall x \in \mathbb{Z} \ , \ x > 0$ makes sense. The reason is that quantifiers bind what is on their right - if the statement comes first and the quantifier second, the variable in the statement would be free.

Assuming that your math is written in a context where reading happens from left to right, it is also confusing if the variable gets introduced only later. As such, even in informal writing, I'd consider the second very bad style.

You are of course free to introduce your own idiosyncratic notation.

Arno
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The scope of a quantifier should be very clear. This is especially true with statements that have nested quantifiers of different types. If you get in a habit of mixing the positions of quantifiers around, sometimes before and sometimes after, your reader will be unable to correctly detect the scope of the quantifier, and may misread the statement, and the sh** hits the fan.

This issue of making the scope clear, by the way, is independent of whether you use symbols or words for your quantifiers. I'll use words, but feel free to substitute $\forall$ for my words "for all" and $\exists$ for my words "there exists" in what follows.

Let's take, for example, the definition of continuity of a function $f : \mathbb R \to \mathbb R$ at $a \in \mathbb R$, correctly written at first, with quantifiers preceding their scope:

For every $\epsilon > 0$ there exists $\delta > 0$ such that for every $x \in \mathbb R$, if $|x-a|<\delta$ then $|f(x)-f(a)|<\epsilon$.

Suppose now you that you decide you like to put your first universal quantifier at the end.

There exists $\delta > 0$ such that for every $x \in \mathbb R$ if $|x-a| < \delta$ then $|f(x)-f(a)| < \epsilon$, for every $\epsilon > 0$.

Big trouble. What's the scope of that $\epsilon$ quantifier? As written, it could easily be interpreted to be the same as the scope of the $x$ quantifier. Which, if that intepretation were correct, would then be equivalently the the following statement with quantifiers in their correct position preceding their scope:

There exists $\delta > 0$ such that for every $\epsilon > 0$ and every $x \in \mathbb R$, if $|x-a| < \delta$ then $|f(x)-f(a)| < \epsilon$.

BIG BIG BIG trouble. This statement is NOT equivalent to the definition of continuity.

Lee Mosher
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We need to distinguish between the formal syntax of the predicate calculus, and prose exposition.

Predicate calculus has precise syntax rules for good reasons, and these dictate that the quantifiers come in front. (Examples below.) But in prose exposition, people often put them afterwards. On the other hand, in textbooks and such, you won't usually find the symbol $\forall$ used as a shorthand for the words "for all". (In handwritten notes or on the blackboard, sure.)

The formal syntax handles alternating universal and existential quantifiers without ambiguity. Consider the distinction between uniform and pointwise convergence of a sequence of functions, $f_n\to f$, say on the interval $(0,1)$. First, uniform:

$$\forall \epsilon\; \exists N\; \forall n\!\!>\!\!N\;\forall x\!\!\in\!\!(0,1)\; |f_n(x)-f(x)|<\epsilon$$

and then pointwise:

$$\forall \epsilon\;\forall x\!\!\in\!\!(0,1)\; \exists N\; \forall n\!\!>\!\!N\; |f_n(x)-f(x)|<\epsilon$$

The distinction is subtle enough that Cauchy initially got it wrong. (Slight oversimplification of the history; see Jeremy Gray's The Real and the Complex: A History of Analysis in the 19th Century for the full story.)

Of course, you can make any chain of quantifiers clear enough in regular prose, but authors don't always do a good job of this.

By the way, "consequent" has a technical meaning in logic. It's the $B$ in an implication $A\Rightarrow B$, with $A$ be the antecedent.

Finallly, a personal opinion: I think of Aristotle’s logic as “always, sometimes, never” logic. The premises and conclusions of his syllogisms can always be cast in the form $\forall x P(x)$ or $\exists x P(x)$, where $P(x)$ is a boolean combination of some sort. Formal logic first advanced decisively beyond Aristotle when it learned to handle the alternation of quantifiers.

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If $x>0$ is understood to mean $x\in\{z\in\mathbb R : z>0\}$, then $x\in\mathbb Z^+$, $x>0$ and $x>0$, $x\in\mathbb Z^+$ virtually mean the same: $x\in\mathbb Z^+$, $x>0$ $\iff$ $x\in(\mathbb Z^+\cap \{z\in\mathbb R : z > 0\})$. Since the set intersection is commutative, it holds that $\mathbb Z^+\cap \{z\in\mathbb R : z > 0\} = \{z\in\mathbb R : z > 0\} \cap \mathbb Z^+$, and $x>0$, $x\in\mathbb Z^+$ $\iff$ $x\in(\{z\in\mathbb R : z > 0\} \cap \mathbb Z^+)$.