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As was established in my question here, one reason that $Q$ + induction on formulas with bounded quantifiers is stronger than $Q$ + induction on quantifier-free formulas is that the variable that we're doing induction on could be the bound of one of the bounded quantifiers. ($Q$ denotes Robinson arithmetic.) My question, is that the only reason for the difference in strength? Would the have equal strength if this possibility were disallowed? In other words, is $Q$ + induction on quantifier-free formulas just as strong as $Q$ + induction on formulas with bounded quantifiers, with the bounds being either numerals or bound variables?

To me, the answer is pretty obviously yes, since a formula with bounded quantifiers with the bounds being either numerals or bound variables can very easily be written as a formula without quantifiers, and we can apply induction to the latter formula, and then derive a formula of the former type.

Any help would be greatly appreciated.

Thank You in Advance.

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Keshav Srinivasan
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Your thinking is correct. A formula in the language of natural number arithmetic in which all quantifiers are bound by a constant or a bound variable is equivalent to a quantifier-free formula. To see this work down from the top of formula, transforming existential quantifiers into disjunctions and universal quantifiers into conjunctions. E.g., transform $\exists x < 2\cdot\phi$ into $\phi[0/x] \lor \phi[1/x] \lor \phi[2/x]$, where $\phi[i/x]$ means the result of substituting $i$ for $x$ in $\phi$. These transformations preserve the property that all quantifiers are bounded by either constants or bound variables and reduce the number of quantifiers. When no more transformations are possible, you have a quantifier-free formula equivalent to the original. This equivalence holds in any theory that can prove all instances of: $$x < S^n(0) \implies x = 0 \lor x = S(0) \lor x = S^2(0) \lor \ldots \lor x = S^{n-1}(0)$$ (when $n = 0$, the empty disjunction should be read as a false statement, e.g. $0 \not= 0$).

In particular, I claim the equivalence holds in Robinson's system $Q$. In $Q$ one defines $x < y$ to mean $\exists b\cdot x + S(b) = y$. It is easy to prove (by induction in the metalanguage) that the axioms of $Q$ imply all instances of: $$ x + S(b) = S^n(0) \implies x = 0 \lor x = S(0) \lor x = S^2(0) \lor \ldots \lor x = S^{n-1}(0)$$ Hence the claimed equivalence does indeed hold in $Q$.

Rob Arthan
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  • Thanks, that was exactly my reasoning. – Keshav Srinivasan Sep 14 '13 at 22:48
  • The question is to show that this can be formalized into Robinson arithmetic. – Carl Mummert Sep 14 '13 at 23:19
  • @Carl Mummert: I thought that in a model of Robinson arithmetic, for each numeric constant $n = S^n(0)$, the set ${i: i < n}$ only had the "standard" elements $0$, $S(0)$, ..., $S^{n-1}(0)$. Is that not correct? – Rob Arthan Sep 15 '13 at 00:21
  • I believe that the claim in my comment above is correct and have amended the answer to outline a (syntactic, rather than model-theoretic) proof. Comments are welcome. – Rob Arthan Sep 15 '13 at 15:23