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By "the literal sentences of basic arithmetic" let us mean sentences like

$$3+4=7,\;\; 2\cdot 3 = 3\cdot 2, \;\;S(2)=3$$

where for example $3$ is shorthand for $S(S(S(0)).$

Note that some literal sentences of basic arithmetic are true (e.g. $3+4=7$) while others are false (e.g. $2\cdot 3 = 1$).

Now let $\mathrm{PA}^-$ denote Peano arithmetic without induction. Does $\mathrm{PA}^-$ prove all true literal sentences of basic arithmetic?

goblin GONE
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  • What do you "exactly" mean about literal sentences? Only equalities of sums and products? Anything different? – MyUserIsThis Aug 16 '13 at 06:43
  • @MyUserIsThis, precisely sentences of the form $x * y = z$ where the variables are stand-ins for numerals and $*$ is a stand-in for either $+$ or $\cdot$, and sentences of the form $S(x)=y.$ – goblin GONE Aug 16 '13 at 06:46
  • @MyUserIsThis, you have my apologies, but actually the comment I wrote above was silly and should be disregarded. It would be better to include sentences like $4+6=2\cdot(3+2).$ – goblin GONE Aug 16 '13 at 07:19
  • "Literal" is conventionally used in logic for atomic wffs and negated atomic wffs. – Peter Smith Aug 16 '13 at 08:22

1 Answers1

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This theory is called Robinson arithmetic and the answer is yes assuming you mean sentences without quantifiers or variables. There are some very simple quantified sentences that cannot be proved such as commutativity of addition ($\forall x, y: x + y = y + x$), although any instance of that such as $3 + 5 = 5 + 3$ can be.

Dan Brumleve
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  • Quick question: if we drop the axiom $x \neq 0 \rightarrow \exists y(x=S(y))$ from Robinson arithmetic, can all the literal sentences of basic arithmetic still be proved? – goblin GONE Aug 16 '13 at 06:54
  • I think so but I'm not sure. – Dan Brumleve Aug 16 '13 at 07:08
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    Usually, $\mathsf{PA}^-$ is taken to be a set of axioms for a discrete ordered semiring, rather than Robinson arithmetic. – Carl Mummert Aug 16 '13 at 13:01
  • @CarlMummert, thank you I was unaware of this convention. – goblin GONE Aug 16 '13 at 14:01
  • @CarlMummert: Is Robinson's Q strictly weaker than PA$^-$ (discrete ordered semiring)? Rautenberg just says the former is interpretable in the latter but nothing else, and I could not find anything that answers my question. – user21820 Dec 13 '17 at 16:00
  • @user21820: yes, it is much weaker; see a list of examples at the end of this answer: https://math.stackexchange.com/a/998914/630 . This is the main motivation for working with $PA^{-}$ instead, in some contexts. Robinson arithmetic is useful for some results in incompleteness, but not so much for studying models of fragments of Peano Arithmetic. – Carl Mummert Dec 13 '17 at 18:23
  • @CarlMummert: Oh right; I'm not sure why I missed the fact that Q does not prove the basic ring properties. Thanks! – user21820 Dec 14 '17 at 01:56