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  1. According to the definition 3.6 of A.G. Hamilton's "Logic for Mathematicians", it looks like the following is a well-formed formula of a first order language: $$ (\forall x_1)(\exists x_1)(x_1<x_1) $$ What is the interpretation of this? Since the last quantifier to be employed is the existential one, is this the same as $$ (\forall x_1)(\exists x_2)(x_2<x_2) $$ which is also the same as $$ (\exists x_2)(x_2<x_2) $$
  2. In the same book, the Axiom Scheme of Replacement (one of the axioms for ZF) is written $$ (\forall x_1)(\exists! x_2)\mathscr{A}(x_1,x_2)\to(\forall x_3)(\exists x_4)(\forall x_5)(x_5\in x_4\leftrightarrow(\exists x_6)(x_6\in x_3\wedge\mathscr{A}(x_6,x_5))) $$ and the author says that "$\mathscr{A}(x_1,x_2)$ can be any well-formed formula in which $x_1$ and $x_2$ occur free (and in which, we may suppose without loss of generality, the quantifiers $(\forall x_5)$ and $(\forall x_6)$ do not appear)". Are these restrictions on the well-formed formula $\mathscr{A}(x_1,x_2)$ really necessary? Why do we want such restrictions? Does it have something to do with the kind of example in the part 1. above of this question?
Guest
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  • $(\forall x_1)(\exists x_1)(x_1<x_1)$ looks like a typo.You should probably avoid such constructs in mathematics even if it might be allowed in some flavours of FOL. – Dan Christensen Sep 16 '15 at 16:05
  • @DanChristensen No, it wasn't a typo. My question came from worrying about such constructions – Guest Sep 16 '15 at 16:07
  • If it is not technically a typo, it certainly looks like one. Avoid it if you can. I can't imagine a circumstance in mathematics in which $(\forall x_1)(\exists x_1)$ would be unavoidable. – Dan Christensen Sep 16 '15 at 16:28
  • I don't think you would have any less expressive power if you disallowed such constructs. – Dan Christensen Sep 16 '15 at 16:48
  • @DanChristensen I agree with you, but for some reason these constructs are not disallowed. – Guest Sep 16 '15 at 16:50

1 Answers1

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To 1. : YES. A quantifier $\forall x$ acting on a formula $\varphi$ without $x$ free has "no effect"; i.e., in this case, $\forall x \varphi$ is equivalent to $\varphi$.

Thus, as you say, $(∀x_1)(∃x_1)(x_1 < x_1)$ is equivalent to $(∃x_1)(x_1 < x_1)$.

For 2. : the antecedent of the axiom $(∀x_1)(∃!x_2)A(x_1,x_2)$ simply states that the formula $A(x_1,x_2)$ "defines" a functional mapping : i.e. a binary relation $R(x,y)$ such that for any value of $x$ there exists exactly one value of $y$ satisfying the relation [e.g. "$x$ is son of $y$" satisfy the functionality condition, while "$x$ is father of $y$" does not].

This has nothing to do with the fact discussed in 1.

In principle, there is no reason to forbid an occurrence of $\forall x_5$ into $A(x_1,x_2)$; if so, this quantifier has as scope a sub-formula of $A(x_1,x_2)$ and thus has no effect outside it, and specifically it does not "interact" with the consequent of the axiom : $(∀x_3)(∃x_4)(∀x_5)(\ldots)$.

It's only a good "hygienical" practice to use distinct quantified variables in the same formula.

For the axiom specifically, we will "apply" it to formulae $A(x_1,x_2)$ satisfying the "functionality" condition expressed by the antecedent (if not, we do not care of them...).

For these formulae, the consequent licenses us to assert, for every set $x_3$, the existence of a set $x_4$ "collecting" all those sets $x_5$ that are "picked up" from $x_3$ by the formula $A$, i.e. such that $A(x_6,x_5)$ holds for $x_6 \in x_3$.

In order to "pick up" the $x_5$, we need that the functionality condition apply.

Mauro ALLEGRANZA
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  • Thank you. For 2., if I understand correctly, there is also no need to assume that $x_1$ and $x_2$ are free in $\mathscr{A}$ and the author assumes this for "hygienical" reasons too? – Guest Sep 15 '15 at 20:47
  • @Guest - Wrong... If we say : $\forall x \exists ! y (y=x+1)$ this formula corerctly define a function : $(n,m)$ such that $m$ is the successor of $n$. If we write e.g. $\forall x \exists ! y (0=0)$ this is a true formula but it does not define a "function". Thus, it is necessary, in order for the axiom "to work", that $x_1$ and $x_2$ are free. – Mauro ALLEGRANZA Sep 15 '15 at 20:56
  • Yes but isn't the antecedent justly there to verify that it defines a function? I mean, $\forall x\exists!y(0=0)$ does not define a function, but in that case the antecedent is just false and we will never be able to use that instance of the axiom schema, so it's no big deal. Even if $x_1$ and $x_2$ do both appear free, the formula won't necessarily define a function... e.g. $\forall x\exists!y(x=x\wedge y=y)$. No? – Guest Sep 15 '15 at 21:04
  • @Guest - Correct... in general, it does not define a function because, in every "world" with more than one object, the formula is false. – Mauro ALLEGRANZA Sep 15 '15 at 21:10
  • Okay so in the end, since the "world" we have in mind when constructing ZF has much more than one object, I understand that it must be the case that $x_1$ and $x_2$ must be free in $\mathscr{A}(x_1,x_2)$. My objection was that to explicitly mention this is not necessary since this is an implicit condition in order for the antecedent to be true and the corresponding instance of the axiom schema to be useful. Anyway, thank you for your time. – Guest Sep 15 '15 at 21:17