Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In my textbook I have this (syntactic) definition of inconsistency:

We say that $\Sigma$ ($\subseteq\textrm{Prop}(A)$) is inconsistent if $\Sigma\vdash\bot$.

I think the intuitive definition of inconsistency is that some set $\Sigma$ of propositional formulas is inconsistent if it contains contradiction (to be more precise, if the contradiction is provable), that is if there exists a formula $p$, such that $\{p,\neg p\}\subseteq\Sigma$.

I would like to check whether the intuitive definition of consistency is consistent with the one from my text book. (:

I would do it like this:

Lets suppose that there exists a proposition $p$, such that $\{p,\neg p\}\subseteq\Sigma$. Now, among my (the ones defined in my textbook) propositional axioms I have this one $$p\rightarrow(\neg p \rightarrow \bot).$$ So by applying modus ponens to $p$ and $p\rightarrow(\neg p \rightarrow \bot)$ I infer that $$\Sigma\vdash (\neg p \rightarrow \bot).$$ Now I again apply modus ponens to $\neg p$ and $\neg p\rightarrow \bot$ and infer that $$\Sigma\vdash\bot.$$ This yields a contradiction, because one of my propositional axioms is also $\top$.

Is this a correct way to prove that those definitions are equivalent?

Thank you for your answers!

--pizet

share|improve this question
add comment

2 Answers 2

up vote 5 down vote accepted

Note that $\Sigma$ being inconsistent does not mean that $\Sigma$ contains $\{p,\lnot p\}$ for some $p$.

For example $\Sigma=\{p\land\lnot p\}$ is clearly inconsistent, but does not contain two propositions at all.

You are correct that inconsistency is the same as contradiction being provable. But as I remarked above, it is not the same thing as containing $p$ and $\lnot p$ of some formula.

share|improve this answer
    
Of course, you are right. How would you express the fact that a contradiction is provable in $\Sigma$? –  pizet Apr 14 '13 at 0:38
    
pizet, well... what exactly do you mean by "express"? I find $\Sigma\vdash\bot$ fairly good. –  Asaf Karagila Apr 14 '13 at 0:44
add comment

As a footnote to Asaf's answer, I suspect that what the OP was after by way of an "intuitive definition" was in fact the idea that

(A) A set of wffs $\Sigma$ is inconsistent just when there is a proposition $A$ such that $\Sigma \vdash A$ and $\Sigma \vdash \neg A$.

And the question is how this relates to the textbook definition

(B) A set of wffs $\Sigma$ is inconsistent iff $\Sigma \vdash \bot$.

There's a third familiar definition

(C) A set of wffs $\Sigma$ is inconsistent iff, for all wffs $A$ (of the relevant language) $\Sigma \vdash A$.

Well, these come to just the same in a classical (and intuitionist) setting. The OP's sketched proof-idea, together with the rule that $\bot$ entails anything, can be re-used to show that (A) is equivalent to (B) and to (C). (In fancier logics, or more restricted logics these definitions can of course peel apart.)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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