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Modal logic 1.4 - truth trees for invalid arguments in K 

Kane B
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21 окт 2024

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Комментарии : 14   
@JonSebastianF
@JonSebastianF 3 года назад
3:50 explained: A 'trick' is: Moving the ~ through the operators, □ or ◇, flips their respective values around, so that ~□p flips to ◇~p, and ~◇p flips to □~p. Using this trick here on “~□◇p” leaves you two options: Moving the ~ once or twice: · Moving it once, gives ◇~◇p, from which we derive “~◇p” in w2. · Moving it twice, gives ◇□~p, from which we derive “□~p” in w2.
@Aglaophamus1
@Aglaophamus1 8 лет назад
1:15 I think you mean the fallacy of affirming the consequent. Thanks for making these videos!
@bertrandrussell6470
@bertrandrussell6470 10 лет назад
Forgive me if you already explained this in another video, but why is it that you write ~p in the counterexample when neither p nor ~p occurs in the corresponding world?
@orangereplyer
@orangereplyer 4 года назад
He could've chosen to put p instead; either choice would've worked.
@JonSebastianF
@JonSebastianF 3 года назад
5:54 _“(...) If neither the variable, nor its negation, occurs, then we put the negation in the corresponding world.”_ ...Why is this so? Why does that imply the negation?
@KaneB
@KaneB 11 лет назад
If ~[]p is true, then, obviously, []p is false. And *any* material conditional with a false antecedent is true. So since []p is false, []p -> []q is true.
@Alkis05
@Alkis05 4 года назад
9:50 The other counter example would be the same, but with Valuation(w3,~p & q)
@wenaolong
@wenaolong 7 лет назад
I think it should be said that in w2, it is possibly not possibly p, or possibly necessarily not p. And so that means there is at least one possible world where it is necessarily true that not-p. Therefore it is necessarily true in all possible worlds that it is possibly true in some world, in this case w2, that necessarily not-p, and so it is necessarily not-p in all possible worlds. But in w1 p is true. So by "revival", we can import not-p back to w1, and have a contradiction. This appeals to the intuition that if it is possibly true, then it is true in all possible worlds that it is possibly true, and so it is necessarily true that it is possibly true. Ah, but I suppose that K is very different from S5, and I am thinking in S5... K is quite weird. Accessibility becomes a key issue in order to make it "reasonable" for our use in any practical way it seems.
@physics_philosophy_faith
@physics_philosophy_faith 4 года назад
"possibly not possibly p" would only belong in w0, not w2. The "possibly" is removed because you are opening up a new world, w2, and then putting "not possibly p" there. That is how we have been defining and using "possibly(p)" by opening up a new world and putting p in it, not putting possibly(p) in the new world
@ktube2482
@ktube2482 7 лет назад
Why can you infer ~diamond(p) from ~box,diamond(p)? Please if someone sees this, help me out.
@physics_philosophy_faith
@physics_philosophy_faith 4 года назад
This took me a minute too. ~necessary(x) is identical to possibly(~x). This means you open up a new world with ~x in it. Here, x = diamond(p). So the new world should have ~diamond(p).
@JonSebastianF
@JonSebastianF 3 года назад
A 'trick' is: Moving the ~ through the operators, □ or ◇, flips their respective values around, so that ~□p flips to ◇~p, and ~◇p flips to □~p. Using this trick here on “~□◇p” leaves you two options: Moving the ~ once or twice: · Moving it once, gives ◇~◇p, from which we derive “~◇p” in w2. · Moving it twice, gives ◇□~p, from which we derive “□~p” in w2.
@turnz153
@turnz153 Год назад
5:25
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