1. ## inconsistent set

Hey everyone,
I am stuck on this proof and don't know what direction to take. I know what inconsistent means, but I don't know how to show $G_{\sigma} \vdash \lnot \sigma$.

Here is the problem...

Suppose that $G$ is an inconsistent set of sentences. For each $\sigma \in G$, let $G_{\sigma}$ be the set obtained by removing $\sigma$ from $G$. Prove
for any $\sigma \in G, G_{\sigma} \vdash \lnot \sigma$

Thanks!

2. Hi

Does inconsistent mean "no model" for you? If so, what can you say of a model of $G_{\sigma}$, can it be model of $\sigma$? Conclude. ( $\sigma$ is a sentence: in any structure, either it is true or false)

3. Inconsistent to me means there is a $\sigma$ such that $G \vdash \sigma$ and $G \vdash \lnot \sigma$. How do you know that by removing $\sigma$, you can still derive $\lnot \sigma$? Why couldn't you derive $\sigma$?

4. Originally Posted by Kiwili49
Hey everyone,
I am stuck on this proof and don't know what direction to take. I know what inconsistent means, but I don't know how to show $G_{\sigma} \vdash \lnot \sigma$.

Here is the problem...

Suppose that $G$ is an inconsistent set of sentences. For each $\sigma \in G$, let $G_{\sigma}$ be the set obtained by removing $\sigma$ from $G$. Prove
for any $\sigma \in G, G_{\sigma} \vdash \lnot \sigma$

Thanks!
You're given that G is an inconsistent set of formulas.
This means that from G a contradiction can be derived.

Suppose you pull any one of the formulas, say P, out of G to form G-{P}. Call it G'.

Now I think there are two cases to consider:

Case 1: G' is still an inconsistent set. But then G' |- "any formula you like". So surely, G' |- ~P.

Case 2: G' is now a consistent set. But clearly, G'U{P} is simply G (our inconsistent set). All we've done is to put P back into G'.
Now all we need to show is that if G'U{P} is inconsistent (which we know it is), then G' |- ~P. We can work RAA on this one.
Suppose G'U{P} is inconsistent. Then for some formula say Q we can derive both G',P |- Q and G',P |- ~Q
But by RAA, we can then conclude that G' |- ~P.

5. Thank you so much. I really appreciate all of your help!

I am still confused though on case two. Does it work because you can assume anything from the contradiction?

6. Originally Posted by Kiwili49
Thank you so much. I really appreciate all of your help!

I am still confused though on case two. Does it work because you can assume anything from the contradiction?
The wording of your question is a bit of a problem. We don't assume anything from a contradition. Replace the word assume with conclude.

Case 1 should be immediate.

I thought case 2 had enough detail to make it clear what was done. So maybe it wasn't enough.

Here's case 2 in more detail. (Suppose we're in some NDS.)
We know by pulling P out of G that we've made the result, G', consistent (this is case 2).
Now take G' as our premiss-set and see if we can reach ~P from G'.
So our derivation initially is nothing more than a listing of the formulas in G'.
At that point we can introduce an assumption, namely, P.
Essentially, we've now opened a subderivation within the our main derivation.
Why choose P? It's because we know that G' plus P is an inconsistent set.
Let's get all our premisses (i.e., formulas in G') into the subderivation by reiteration.
So now, in this subderivation we have the means for deriving a contradiction (say Q and ~Q).
Once we obtain the contradiction, what can we say about our assumption, P? Well, I'd say it doesn't look to good.
So RAA (or typically ~I in the NDS) allows us to conclude, ~P.
At that point, because ~I is a discharge rule, we have in effect abandoned the assumption P and closed (or sealed) the subderivation.
What's left? We're left with G' taking us to ~P, i.e., G' |- ~P.