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Let's consider language with predicate $P$ and following derivation

$${{{[P[a/x]]^1} \over {P[a/x] \rightarrow P[a/x]}}\rightarrow I^1 \over {\forall_x (P(x) \rightarrow P(x))}}\forall I$$

Doesn't this first order derivation proves second order sentence $\forall_P \forall_x (P(x) \rightarrow P(x))$? Of course interpretation would assign some concrete predicate to symbol $P$ but I can always create model where $P$ means whatever I want because there is no any assumption about $P$. So for me it looks like I have proved property of all predicates but it would mean that FOL has same power as second order logic, which I know is not true. Please correct my reasoning.

Trismegistos
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    I know what was my mistake. I thought that I can switch model and if I do that I can take $P$ to be any predicate I want. It is not the case when I fix the model. Then in FOL $P$ is fixed to some concrete predicate or if I allow any formula in place of $P$ than I can represent only definable predicates/set but in SOL I can take any set I want in place of $P$ event when model is fixed. – Trismegistos Oct 05 '14 at 16:10

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You have proved $∀x(P(x)→P(x))$ with a derivation in first-order logic.

You have not proved $∀P∀x(P(x)→P(x))$, simply because this is not a well-formed formula of f-o logic; in f-o logic, the quantifier $\forall P$ is not allowed.

But the above derivation is schematic, i.e. you can "recycle" it for whatever predicate you want, and it is still valid.

If we "apply" our first-order logic to the language of elementary arithmetic, with constant symbol $0$ for the number "Zero", (binary), function symbols $+$ and $\times$ for the operations of "addition" and "multiplication" respectively, and (unary) function symbol $S$ for the "successor", (binary) predicate symbol $<$ for the relation "less then", we can apply the above schematic derivation to the (unary) predicate $x \ge 0$ [which is an abbreviation for : $\lnot (x < 0)$] in place of $P(x)$ to have :

$\vdash \forall x((x \ge 0) \rightarrow (x \ge 0))$.

The sentence above is a theorem of first-order arithmetic because it is a (well-formed) formula in the language of first-order arithmetic which is provable, due to the fact that it is an instance of a "law" of first-order logic (i.e. a valid f-o formula).

In second-order logic the quantifier $\forall P$ is allowed; thus the formula $∀P∀x(P(x)→P(x))$ is well-formed.

In SOL we have also the $\forall^2$-I rule, which allows us to add to the above derivation the final step concluding with : $∀P∀x(P(x)→P(x))$.

What is the difference ?

In SOL, if we agree on the "intuitive" semantics, let $D$ be the domain of the interpretation (where the individual varibale $x$ range); then the predicate variable $P$ range over all the subsets of $D$.

If we are interpreting an arithmetic formula, the domain $D$ is the set $\mathbb N$ of natural numbers, which has countable infinite ($\aleph_0$) objects .

Thus the predicate $P$ range over the power-set of $\mathbb N$, which has $2^{\aleph_0}$ objects (subsets of $\mathbb N$).

In FOL, the "schematic" $∀x(P(x)→P(x))$ can be instantiated with any (unary) predicate expressible in the language of f-o arithmetic.

An unary predicate is an open formula with only the variable $x$ free. A formula of a language with a finite number of symbols is a finite string of synbols; thus, in f-o language for arithmetic we can form only countable many open formulae, i.e. we can express only countable many unary predicates.

Thus, FOL has not the same power as SOL.

Mauro ALLEGRANZA
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  • I generally understand the the difference between FOL and SOL stems from the fact that "An unary predicate is an open formula with only the variable x free". To be frank I did know this fact and I can not explain why it is so. On syntactic side predicate is just a symbol and on semantic it is just relation. Thus unary predicate in FOL is element of power set of domain. This means that for domain of naturals I would have uncountably many one place predicates contrary to what you said that there are only countably many predicates in FOL. – Trismegistos Oct 02 '14 at 11:11
  • @Trismegistos - you are mixing "$predicates_1$" as symbols of the language with "$predicates_2$ as (unary) relations (i.e. subsets of the domain). The first one are symbols and expressions of the language (an open formula) while the second are existsing "out there" in the interpretation, independently of the fact that we have a name in the language denoting them or not. – Mauro ALLEGRANZA Oct 02 '14 at 11:30
  • In my answer, predicates stands for "$predicate_1$", i.e. is always referring to a syntactical object (symbol or expression). – Mauro ALLEGRANZA Oct 02 '14 at 11:32
  • I know that in FOL syntax set of predicate symbols is countable and $P$ must belong to this set. What I mean is that $P$ can be modeled in FO formula ∀x(P(x)→P(x)) as any predicate because my model can assign any of domain susbsets to $P$. So it looks like second order formula ∀P∀x(P(x)→P(x)) is not stronger because I can assign exaclty same subset to $P$. So first order model must assign any element of power-set of Domain to P and standard second order model must assign any element of power-set of Domain to P. So I have exactly same relations at my disposal. Thus proof has same power. – Trismegistos Oct 02 '14 at 11:40
  • @Trismegistos - I would suggest you to review the basic concepts of Second-order Logic : "Second-order logic is an extension of first-order logic where, in addition to quantifiers such as 'for every object (in the universe),' one has quantifiers such as 'for every property of objects (in the universe).' This augmentation of the language increases its expressive strength, without adding new non-logical symbols, such as new predicate symbols." 1/2 – Mauro ALLEGRANZA Oct 02 '14 at 11:49
  • "In first-order languages,[...] a difficulty arises if we want to express the 'well-ordering property' that any non-empty set of natural numbers has a smallest member. [...] in a second-order language for arithmetic, we can say that the natural numbers are well ordered. We know that the well-ordering property is not expressible by any first-order sentence [...]". 2/2 – Mauro ALLEGRANZA Oct 02 '14 at 11:51
  • Article on first order schema which you provided states: "That is, there are sentences of first-order arithmetic that can be deduced from the second-order induction axiom (together with the other axioms of arithmetic, which are common to first-order and second-order arithmetic) but not from the instances of the first-order induction schema (see Shapiro 1991: 110)." So I am going to read this book for an answer. – Trismegistos Oct 02 '14 at 17:04
  • @Trismegistos - See Stewart Shapiro, Foundations without foundationalism (1991), page 110: The instances of these schemes [e.g.axiom scheme of induction] can be derived from the corresponding second-order axiom. The second-order theories of arithmetic and set theory are not conservative extensions of their first-order counterparts. That is, in each case there is a formula $\Phi$ of the first-order language, such that $\Phi$ can be deduced from the second-order axiom, but $\Phi$ is not a deductive onsequence of the first-order axioms (nor,by completeness,is it a semantic consequence). 1/2 – Mauro ALLEGRANZA Oct 03 '14 at 08:00
  • For example, in second-order arithmetic and second-order set theory, one can formulate a 'truth' definition for the corresponding first-order theory and thus one can deduce the consistency of the first-order theory. 2/2 – Mauro ALLEGRANZA Oct 03 '14 at 08:01
  • I understand that cited fragment of Saphiro holds. I do not understand why it holds. I know that my thesis that first-order axiom scheme is equivalent to second-order axiom is wrong it is just that I can convince my self (using deduction I presented) that they are equivalent and I do not see gap in presented deduction.

    Probably if I saw some counter example, some concrete sentence that is not instance of FO schema but is concrete instance of second-order axiom that would help.

     – Trismegistos Oct 03 '14 at 08:35
  • @Trismegistos - you will not find a "simple" formula. See P.Smith, An Introduction to Gödel's Theorems(2007),page 193: Th.22.1 $PA_2$ can prove every $L_A$ wff that $PA$ can prove, and some that $PA$ can’t prove. The first half of that theorem is, of course, trivial. As for the second half, we just report e.g. that $PA$’s canonical Godel sentence $G$ which is unprovable in $PA$ is provable in $PA_2$. Of course, (assuming that $PA_2$ is consistent) while it can prove $PA$’s Godel's sentence $G$, it can’t prove its own canonical Godel sentence $G_2$. – Mauro ALLEGRANZA Oct 03 '14 at 09:00