Definition of (in)consistency in propositional logic
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
$\endgroup$2 Answers
$\begingroup$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.)
$\endgroup$ $\begingroup$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.
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