Define a set $\Gamma$ to be consistent if $\nvdash \bigwedge\Sigma \rightarrow \bot$ for each finite $\Sigma \subseteq \Gamma$.
How does this definition relate to the usual one, i.e. that you can't derive both $\varphi$ and $\neg \varphi$ from $\Gamma$ ? I know that they shoud be equal but I can't see how to get from one to the other.
On
There is an example in the Stanford Encyclopedia of Philosophy’s entry on Provability Logic that will summarize here.
Take $\Gamma$={$\Diamond P_0, \Box(P_0 \implies \Diamond P_1), \Box(P_1 \implies \Diamond P_2),…$}.
This set entails a contradiction in the provability logic G.L., but each finite subset is satisfiable. This set entails a contradiction in G.L. ultimately because of the fact that formal systems of arithmetic cannot prove their own consistency. The intuitive reason is that the ability for formal systems to make self referential statements about their own provability imposes a restriction on what kinds of sets of formulas may be considered consistent. Our $\Gamma$ is inconsistent precisely because of this limitation; in essence, $\Gamma$ as a whole entails an infinite chain of worlds between which the binary accessibility relation between states holds. The accessibility relation corresponds to frames that are finite, irreflexive, conversely well-founded, and transitive; it is ultimately converse well-foundedneas that is defied in this case.
$$\nvdash \bigwedge\Sigma \rightarrow \bot\\\iff \Gamma \nvdash \bot \text{ (deduction theorem)}\\\iff \Gamma \nvdash \phi \land \neg \phi \text{ for some } \phi \text{ (logical equivalence $\phi \land \neg \phi \equiv \bot$)}\\\iff \text{not: } \Gamma \vdash \phi \land \neg \phi \text{ (def. $\nvdash$)}\\\iff \text{not: } (\Gamma \vdash \phi \text{ and } \Gamma \vdash \neg \phi) \text{ (follows from rule def. for $\land$)}$$