@francopalombo No, this is no theorem of this system of modal logic. But I don't think he is saying that. He says that if A is possible, then it is necessarily possible (axiom 5) and that PA and PPA are equivalent, which is a theorem of this system.
Hi Dr. Campbell, I know you made these videos seven years ago, but if you're reading this, would you mind making videos on how to derive rules of inference in modal logic through proofs (ex: how to get tilde in and tilde out?) I had a really hard time understanding my professor the other day in class, and you explain things very well. Thank you, Isabela
You should make it clear that your S4 is not equivalent to the conjunction of your (m) and your 4: it is strictly weaker. Your S4, NA iff NNA, does not imply your axiom (m): if NA, then A; because 'if NNA, then NA' does not imply 'if NA, then A'. The point is that while A stands for any formula whatsoever, that is not the case for NA. In addition, the claim that if NA, then NA is a tautology. You don't need any modal axiom to derive this.
@francopalombo
No, this is no theorem of this system of modal logic. But I don't think he is saying that. He says that if A is possible, then it is necessarily possible (axiom 5) and that PA and PPA are equivalent, which is a theorem of this system.
Hi Dr. Campbell,
I know you made these videos seven years ago, but if you're reading this, would you mind making videos on how to derive rules of inference in modal logic through proofs (ex: how to get tilde in and tilde out?) I had a really hard time understanding my professor the other day in class, and you explain things very well.
Thank you,
Isabela
Question: Is modal logic derived from an abstraction of the subjunctive and optative moods of Koine Greek?
¿ if A is posible, then it is posible that A is necesari?
just asking
¿ if A is posible, then it is posible that A is necesari? ´{ min 7:35}
just asking
You should make it clear that your S4 is not equivalent to the conjunction of your (m) and your 4: it is strictly weaker.
Your S4, NA iff NNA, does not imply your axiom (m): if NA, then A; because 'if NNA, then NA' does not imply 'if NA, then A'. The point is that while A stands for any formula whatsoever, that is not the case for NA.
In addition, the claim that if NA, then NA is a tautology. You don't need any modal axiom to derive this.
S4 is actually more along the lines of metalogic, isn't it? Using logic to describe the un-necessarity of redundant logic?
@piplepipo LOL...I sense the NERD is strong in this one! :-)