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I am teaching someone about statements, the very basis of mathematical thinking, but now I am a bit confused myself. For

$ \forall x \in \mathbb{R}, x > 2 \Rightarrow x > 3 $

to be a statement, does it need the $ \in \mathbb{R} $ part? And the quantifier?

So is $ x > 2 \Rightarrow x > 3 $ a statement or not?

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    $\begingroup$ I think $\forall x \in \mathbb{R}$ $x > 2 \implies x > 3$ is a shorthand for $\forall x$ $(x \in \mathbb{R} \wedge x > 2) \implies x > 3$. But I am not familiar with mathematical logic. $\endgroup$ – Alex Vong Apr 24 '17 at 14:26
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    $\begingroup$ This is incorrect, see Clive Newstead's answer. $\endgroup$ – Alex Vong Apr 24 '17 at 14:53
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    $\begingroup$ @AlexVong: It's less incorrect than you might think, since in general $(p \wedge q) \Rightarrow r$ is equivalent to $p \Rightarrow (q \Rightarrow r)$, so what you write is equivalent to the thing it's shorthand for. (The statement itself is false, though.) $\endgroup$ – Clive Newstead Apr 24 '17 at 16:00
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    $\begingroup$ @user21820 Yes, I get it now. An alternative way to see it would be that both $(p \wedge q) \implies r$ and $p \implies (q \implies r)$ expands to $\neg p \vee \neg q \vee r$. $\endgroup$ – Alex Vong Apr 25 '17 at 6:12
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    $\begingroup$ @AlexVong: Correct. Though note that the equivalence "$A ⇒ B \equiv \neg A \lor B$" is only valid for classical logic, but that's what you should focus on for now. $\endgroup$ – user21820 Apr 25 '17 at 6:34
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If it's understood that the variable $x$ refers to a real number, then it can be omitted; otherwise, it cannot.

This practice is known as bounded quantification. In general, when you write $\forall x$, the variable $x$ is taken to range over the entire universe of discourse, whatever that may be. If the universe of discourse is not specified, then it is typically understood by context (e.g. the von Neumann universe in set theory).

If $p(x)$ is some statement with a free variable $x$, then the expression $\forall x \in X,\, p(x)$ is shorthand for $\forall x,\, (x \in X \Rightarrow p(x))$. It then doesn't matter what the universe of discourse is, because in order for the hypothesis $x \in X$ to be specified, you've instantly restricted yourself to elements of $X$.

Thus the statement $\forall x \in \mathbb{R},\, x > 2 \Rightarrow x > 3$ is shorthand for $$\forall x,\, (x \in \mathbb{R} \Rightarrow (x > 2 \Rightarrow x > 3))$$ If it were understood from context that the variable $x$ refers to a real number, then you could omit the "$\in \mathbb{R}$" part so that the statement becomes $\forall x,\, (x > 2 \Rightarrow x > 3)$; in fact, in this case, you could shorten this even further to become just $\forall x > 2,\, x > 3$.

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  • $\begingroup$ Thanks! So the quantifier is not optional? $\endgroup$ – Tempestas Ludi Apr 24 '17 at 14:44
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    $\begingroup$ You need the quantifier, otherwise the expression has a free variable and so cannot be assigned a truth value. $\endgroup$ – Clive Newstead Apr 24 '17 at 14:53
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    $\begingroup$ In addition, $\exists x\in X, p(x)$ is commonly the shorthand for $\exists x\left(x\in X \land p(x)\right)$. Note that upon negating $\exists$ and $\forall$ generally switch roles, but the shorthand form remains, but switches its meaning: $\lnot\left(\exists x\left(x\in X \land p(x)\right)\right) = \lnot\left(\exists x\in X, p(x)\right) \equiv \forall x\in X, \lnot p(x) = \forall x\left(x\in X \rightarrow \lnot p(x)\right)$. $\endgroup$ – ComFreek Apr 24 '17 at 16:12
  • $\begingroup$ There is a very very slight problem with your claimed short-hand of "$∀x,(x∈R⇒(x>2⇒x>3))$", because at least in standard first-order logic the constituents of a compound proposition are to be interpreted separately. So "$x>2$" must make sense for every object $x$, which is not the case since "$<$" is only defined for real numbers. You could argue that "$<$" is just a relation on all objects that happens to be a set of pairs of reals. But what about more complex things like "$\forall f : S \to S\ ( \exists x \in S\ ( f(x) = x ) )$"? $\endgroup$ – user21820 Apr 24 '17 at 16:13
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    $\begingroup$ @user21820: Sure, and I agree. These issues magically disappear when working within type theory, rather than set theory+FOL... unfortunately type theory has its own shortcomings. $\endgroup$ – Clive Newstead Apr 24 '17 at 18:13
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the very basis of mathematical thinking

is about communication as much as it is about formal statements.

Whether or not there's ambiguity in your example depends on the context. If it's part of a discussion about the real numbers you don't need the extra specificity. If it's a standalone example about quantifiers, you do.

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If you don't specify it, there is a possibility of ambiguity. For example, consider the example $\forall x, x>3 \Rightarrow x>2$. If $x$ is real, it's obviously true. If however we take it over the lattice of divisibility, $x=9$ contradicts the statement, as $9$ is an odd number.

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  • $\begingroup$ Nice counterexample, for a statement that, at first sight, seems to be universally true! $\endgroup$ – Tempestas Ludi Apr 24 '17 at 14:40
  • $\begingroup$ Note that the OP's example is in fact a false statement (which is OK since the question is about formulating a statement). You quite naturally reversed it. $\endgroup$ – Ethan Bolker Apr 24 '17 at 14:43
  • $\begingroup$ @EthanBolker yes, I did it deliberately, but I hope you understand the purpose $\endgroup$ – enedil Apr 24 '17 at 14:51
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Better to specify $\forall x \in \mathbb R$, because it could be possibly that your statement is about $\forall x \in \mathbb Q$, or $\forall x \in \mathbb Z$, etc.

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  • $\begingroup$ To down-voter: please leave reason why you down-voted, any suggestion for edit to improve. Is it something fundamentally wrong, or are there minor issues that could be fixed/corrected?For soft question like this, I'm not expecting a down-vote without any reason/suggestion. $\endgroup$ – Yujie Zha Apr 25 '17 at 12:10
  • $\begingroup$ Also, this is the first post to the question, so if you down-vote because you think it mimic/simplifies other answers, you are wrong. $\endgroup$ – Yujie Zha Apr 25 '17 at 12:27

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