Activity for kouty
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #293122 |
Post edited: |
— | 2 months ago |
| Comment | Post #293102 |
@#53398 Thanks for all the topic is clear for me now.
It would be good if you write an definitive answer.
(more) |
— | 3 months ago |
| Edit | Post #293102 |
Post edited: |
— | 3 months ago |
| Edit | Post #293122 |
Post edited: |
— | 4 months ago |
| Edit | Post #293122 |
Post edited: |
— | 4 months ago |
| Edit | Post #293122 | Initial revision | — | 4 months ago |
| Answer | — |
A: Compactness of the Propositional Calculus I think that now I understood a couple of points: The following statement is now clearer. > Let $i(P1) = 0$ and suppose $\phi(1)$ does not hold. That is, there is a finite subset $\Gamma'$ of $\Gamma$ which has no model in which $P1$ takes the value $i(P1) = 0$. Then we define $i(P1) = 1$ and show... (more) |
— | 4 months ago |
| Edit | Post #293102 |
Post edited: |
— | 4 months ago |
| Comment | Post #293102 |
@#53398 You wrote `We're unioning sets of axioms`. I was aware of this. From the enunciation of the Theorem I was understanding that formulas are not necessarily axioms and theorems but formulas that are simply not Self Contradictory, I mean, that are satisfied in at least one combination of atomic f... (more) |
— | 4 months ago |
| Comment | Post #293102 |
@#53398 Thanks so much. But for me it's still hard to get. From the multiple models, at least one need to be common for all formulas, right? (more) |
— | 4 months ago |
| Comment | Post #293102 |
OK I got a part of the proof. [This]( https://www.cs.ox.ac.uk/people/james.worrell/lecture08.pdf) paper is written clearer. When there is a subset Gamma ' for which Phi does not give a line in the Truth Table such that every formulas of Gamma' are true, there is nothing to do, simply. The proof of de... (more) |
— | 4 months ago |
| Comment | Post #293102 |
OK , it's now clear that because of the union of two subsets wouldn't have a model If they have contradictory models, we can aggregate only such subsets they have non contradictory models. However if $ \Gamma $ have two 'contradictory subsets', each of them having a model, the theorem fails to apply.... (more) |
— | 4 months ago |
| Edit | Post #293102 |
Post edited: |
— | 4 months ago |
| Edit | Post #293102 |
Post edited: |
— | 4 months ago |
| Comment | Post #293102 |
I was thinking that with an alternation maybe something representing line 1 or line 2, but I how/if it's possible. (more) |
— | 4 months ago |
| Comment | Post #293102 |
@#53398 . Thirst, I am very grateful to you for receiving this answer. I'll try to translate your answer in 'Truth Table speaking ' : If there are two finite subsets $\Gamma'$ and $\Gamma''$ such that everyone has as model, respectively the line 1 and the line 2 of the Table, how can we find a line... (more) |
— | 4 months ago |
| Edit | Post #293102 |
Post edited: |
— | 4 months ago |
| Edit | Post #293102 |
Post edited: |
— | 4 months ago |
| Edit | Post #293102 |
Post edited: typo |
— | 4 months ago |
| Edit | Post #293102 |
Post edited: typo |
— | 4 months ago |
| Comment | Post #293102 |
Right. It's a typo (more) |
— | 4 months ago |
| Edit | Post #293102 |
Post edited: |
— | 4 months ago |
| Edit | Post #293102 | Initial revision | — | 4 months ago |
| Question | — |
Compactness of the Propositional Calculus Here is the theorem and its proof from Mathematical and Philosophical Logic from Harrie de Swart (p.35). > Let $\Gamma$ be a (possibly infinite) set of formulas such that every finite subset of $\Gamma$ has a model[^1]. Then $\Gamma$ has a model. > > Proof: > > Let $\Gamma$ be a (possibly inf... (more) |
— | 4 months ago |