0
$\begingroup$

Is the following statement correct ?

A is a formula in PA without a quantifier and A is true for the standard model of arithmetic, i.e. the model |N = (N,+,×,0,1,<) This means: |N |= A

==>

A is proofable in PA. This means: PA |- A

Example: |N |= (x+y)^2 =x^2 + 2xy + y^2 ===> PA |- (x+y)^2 =x^2 + 2xy + y^2

If this statement is wrong, please give me a counterexample

cu cx

$\endgroup$
1
$\begingroup$

EDIT: It looks like I misread your question. There is an important difference between quantifier-free sentences, which is what my original answer (now below the fold) addressed, and quantifier-free formulas. Namely, when you speak of PA proving a formula with free variables (or of a structure satisfying such a formula), you're really asking about PA proving (or a structure satisfying) the universal closure of a quantifier-free formula: e.g. $$PA\vdash (x+y)^2=x^2+2x+y^2$$ is really shorthand for $$PA\vdash\forall x,y((x+y)^2=x^2+2x+y^2).$$

Such statements are no longer provable in PA in general. For example, it's essentially a consequence of the MRDP theorem that there is a Diophantine equation $t_1(x_1,..,x_n)=t_2(x_1,...,x_n)$ which has no solutions but which PA can't prove has no solutions; then the quantifier-free formula $$\neg(t_1(x_1,...,x_n)=t_2(x_1,...,x_n))$$ is true in $\mathbb{N}$ but not PA-provable.


What if we don't work with implicit universal quantifiers - that is, we restrict attention to sentences (= formulas with no free variables)? This was the original - incorrect - way I interpreted your question.

Here we get a positive result: PA proves every true $\Sigma_1$ sentence (that is, every sentence of the form $\exists x_1...\exists x_n\varphi(x_1,...,x_n)$, where $\varphi$ uses only bounded quantifiers). This fact ("$\Sigma_1$-completeness") about PA does not depend on PA being consistent - it is provable in PA itself, or indeed much less. The proof is tedious but not hard. Under the further assumption that PA is $\Sigma_1$-sound, this implies that the true $\Sigma_1$-sentences are exactly those which PA proves.

So this exactly delineates PA's power in terms of the coarse arithmetic hierarchy: PA proves exactly the correct $\Sigma_1$ sentences (under a reasonable assumption on PA), but there are correct $\Pi_1$ sentences which PA doesn't prove.

$\endgroup$
  • $\begingroup$ Thank you for your feedback. I tried to proof my statement, but I could not do it. My idee: every term represent as a polynom. Please can you give me a link to the proof of my statement. cu cx $\endgroup$ – user508589 Jan 29 at 21:29
  • $\begingroup$ @user508589 Looking at your question again, I misread it (the issue being "formula" versus "sentence" and the implicit universal quantifier when we talk about proving/satisfying the former). I've edited. $\endgroup$ – Noah Schweber Jan 29 at 22:35
-1
$\begingroup$

$$ \text{You gave a counterexample: true in N but not PA-provable.} \\ \text{But there are so many examples with: true in N and PA-provable: } \\ \text{I define: } \\ \text{I_N} = \{0,1,2, ...\} \text{ is the set of intuitive natural numbers} \\ \bar 0 \; and \; \bar 1 \text{ are the constants from PA} \\ \oplus and \circ \text{ are the functions in PA} \\ \text{when z is element from I_N, I define:} \\ \bar z := \bar 1 \oplus ... \oplus + \bar 1 \quad \text{n-times} \\ \text{I proofed the following statement "S1"} \\ \text{When n is element from I_N then we have:} \\ PA \vdash \forall x \forall y \; (\bar z = x \oplus y \quad .\rightarrow. \quad \bigvee_{i=0}^z (x=\bar i \land y=\overline{z-i}) \\ \text{I assume, that there are "many" of "similar" statements like S1.} \\ \text{Is it possible to formulate a statement "S", so that "S1" follows from "S" ?} \\ $$

cu cx

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.