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 $\varphi (1)$ does not hold. That is, there is a finite subset ${\mathrm{\Gamma }}^{\prime }$ of $\mathrm{\Gamma }$ 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 $\mathrm{\Gamma }$ 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 ${\mathrm{\Gamma }}^{\prime }$ and ${\mathrm{\Gamma }}^{″}$ 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 $\mathrm{\Gamma }$ be a (possibly infinite) set of formulas such that every finite subset of $\mathrm{\Gamma }$ has a model[^1]. Then $\mathrm{\Gamma }$ has a model. > > Proof: > > Let $\mathrm{\Gamma }$ be a (possibly inf... (more)	—	4 months ago