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Is it true that a conservative extension of a theory is equiconsistent with it, and, if so, why?

WP says: "In mathematical logic, a logical theory T2 is a (proof theoretic) conservative extension of a theory T1 if the language of T2 extends the language of T1; every theorem of T1 is a theorem of T2; and any theorem of T2 which is in the language of T1 is already a theorem of T1. [...] Note that a conservative extension of a consistent theory is consistent."

I have a copy of Hodges, A shorter model theory, which gives a definition on p. 58 that seems to be exactly equivalent, although the claim about equiconsistency isn't made.

So what's wrong with the following counterexample?

T1 is a theory in which the only sentence is A, and the only axiom is that A is true.

T2 is a theory in which the only sentences are A and B. It has three axioms: (1) A is true. (2) B is true. (3) B is false.

T2 is a conservative extension of T1, but although T1 is consistent, T2 is inconsistent.

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How do you know $T_2$ is a conservative extension? For all is known $A\to B$ might holds. –  azarel Feb 26 '12 at 4:56
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Those are not logical theories. –  Ricky Demer Feb 26 '12 at 4:57
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2 Answers

up vote 6 down vote accepted

The general claim that conservative extensions are equiconsistent depends on the theories extending ordinary propositional calculus.

So in your $T_2$ we can prove $\neg A$ from $B$ and $\neg B$ (because everything follows from a contradiction), and since $\neg A$ is a sentence in the language of $T_1$ that is not a theorem of $T_1$, $T_2$ is actually not a conservative extension.

If we really want, we can weaken the assumption of including propositional calculus to "allows every sentence to be negated, as a matter of syntax" together with "allows everything to be inferred from a sentence together from its negation". The latter condition is implicit in the common definition of consistency as "cannot derive both a sentence and its negation".

Also, without these minimal assumptions, it is hard to imagine how you're going to state "B is false" as an axiom in the first place.

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Should the last sentence go "without the first of those assumptions"? –  Ricky Demer Feb 26 '12 at 5:05
    
Yes, perhaps. However, I thend to think that the second of those assumptions is essential for making sure that the thing you do to the sentence in the first assumption is actually negation, as opposed to something completely different that just happens to be written with a bent-stick symbol. –  Henning Makholm Feb 26 '12 at 5:08
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You’ve not actually specified languages or theories, so you really have no example at all. However, the idea behind your example could be turned into something that makes sense, so I’ll address the fundamental misunderstanding that remains even after that has been done: your $T_2$ is not a conservative extension of $T_1$. Because $T_2$ is inconsistent, every formula in its language is provable in $T_2$, including the formula $\lnot A$. However, $\lnot A$ is not provable in $T_1$.

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I get your point, but I'm still a little confused. I thought I'd specified the languages of the theories. T1's language has only one sentence in it: A. T2's has two sentences in it: A and B. Is the problem that you need a certain minimum structure for expressing truth, falsehood, and logical inference, or it's not considered a theory? You say that in an inconsistent theory, every sentence is provable. Doesn't this require that you define some rule of inference such as $A \vee \neg A$? I simply intended not to have $\vee$ or $\neg$ in my language. –  Ben Crowell Feb 26 '12 at 5:11
    
How do you define "B is false" without having $\lnot$ in your language? –  Ricky Demer Feb 26 '12 at 5:20
    
@Ben: If you don’t have some notion of inference and negation, the usual notion of consistency isn’t even meaningful, and I’d hesitate to say that you have anything that could meaningfully be called a logical theory. Also, see Ricky Demer’s comment (if you haven’t already). –  Brian M. Scott Feb 26 '12 at 6:31
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