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Let "COF-" be the result of conjoining statements that together assert that we have a complete ordered field, except that we omit the assertion that we have associativity for the addition operation. In other words, the usual symbols (for a set of numbers, the constant number values zero and one, and operations and relations) would be a complete ordered field if we also were to assume that the addition operation satisfies associativity. Specifically, we would assert that the real numbers under addition are an abelian group, but we omit the associativity requirement that is part of the definition of what it means to be an algebraic group.

Is it possible to formulate a statement (call is "S") such that the conjunction of S and COF- implies that we have a complete ordered field, and such that if S implies something that isn't true by logic alone, then the assertion implied by S isn't a logical consequence of COF-?

Observe that the associative property for addition doesn't fulfill our requirements for S.

For example, the associative property for addition implies:

$$\sqrt{2} + (2\sqrt{2} + 3\sqrt{2}) = 6\sqrt{2} = (\sqrt{2} + 2\sqrt{2}) + 3\sqrt{2}$$

... and we can get that by applying the distributive property for real numbers, so we get it from COF-.

So if S asserts x + (y + z) = (x + y) + z for (x,y,z) an element of G, then we cannot have $(\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}) \in G$.

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Derp, I'm silly - I read an additional condition into the question for no good reason! The answer is no if we add that condition, and yes otherwise, each time for a rather silly general reason.


The answer to your question is yes.

Simply take $S$ to be "$COF-\rightarrow Assoc.$" Clearly $COF-+S$ characterizes exactly the complete ordered fields, so the nontrivial part is to show:

There is no non-tautology which each of $COF-$ and $S$ entail individually.

(This is just a rephrasing of your condition " if S implies something that isn't true by logic alone, then the assertion implied by S isn't a logical consequence of COF-.")

To see this, suppose $S\vdash N$. Then any structure in which $COF-$ fails must satisfy $N$, since $\neg COF-\vdash COF-\rightarrow Assoc.$. But if we also were to have $COF-\vdash N$ then $N$ would additionally hold in any structure in which $COF-$ holds. Combining these, we would have that $N$ holds in every structure whatsoever.

More generally, suppose $X,Y$ are sentences such that $X\not\vdash Y$ (here my $X$ is $COF-$ and my $Y$ is $Assoc.$). Then setting $S = X\rightarrow Y$ we have:

  • $X+S\vdash X+Y$, but

  • if $X\vdash N$ and $S\vdash N$ then $\vdash N$.


If we add the condition that $\neg COF-\not\vdash S$ (that is, that $COF-\vee S$ not be a tautology) then the answer is no.

This is because taking $N$ to be $COF-\vee S$ we have $COF-\vdash N$, $S\vdash N$, but - by our added hypothesis - $\not\vdash N$. And as above, this lifts to more general contexts.

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  • $\begingroup$ It was not my intention to either assert or imply the following: "whenever $C\vdash X$ and $B\vdash X$ we have $\vdash X$". On the contrary, given that C alone doesn't assert as much as B, the conjunction $(C\vdash X and B\vdash X)$ should simply be equivalent to $C\vdash X$. Maybe we should replace your second bulletpoint with the following: "whenever $C\vdash X$ and $A\vdash X$ we have $\vdash X$", meaning that the overlap of C and A is as close as we can get to nothing. $\endgroup$ Mar 3, 2020 at 17:51
  • $\begingroup$ @RenEhDaycart I think you've mixed up what my letters are referring to; I've rewritten my answer to be clearer. $\endgroup$ Mar 3, 2020 at 18:03
  • $\begingroup$ "whenever $C\vdash X$ and $A\vdash X$ we have $\vdash X$" should be equivalent to "whenever $C\vdash X$ and $not(\vdash X)$ we have $not(A\vdash X)$", and switching back from the Schweber letters to my letters, that becomes "whenever $S\vdash X$ and $not(\vdash X)$ we have $not(COF-\vdash X)$", where X is something not true by logic alone that is implied by S. $\endgroup$ Mar 3, 2020 at 18:06
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    $\begingroup$ @RenEhDaycart I did make a mistake - I assumed that we also want $COF-\vee S$ to not be a tautology. If we add that then what I wrote is correct, but that isn't what you asked. $\endgroup$ Mar 3, 2020 at 18:15
  • $\begingroup$ @RenEhDaycart I've fixed my answer. $\endgroup$ Mar 3, 2020 at 18:28

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