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bluebrandon
Group Title
prove using the Hilbert System
{not q} ⊢ (p→q)→(not p)
 one year ago
 one year ago
bluebrandon Group Title
prove using the Hilbert System {not q} ⊢ (p→q)→(not p)
 one year ago
 one year ago

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bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
I'm supposed to use the 3 axioms of the Hilbert System and Modus Ponens to prove this
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Using the Hilbert Axioms. We state our given: \[ \neg q \vdash (p\to q)\to\neg p \]Using the fourth (or third primary axiom): \[ (\neg p\to \neg q)\to(q \to p) \]We contraposition the first clause, since they are equivalent statements, by the above logical axiom: \[ \neg q \vdash (\neg q\to \neg p)\to\neg p \]Since\[ \neg p\to \neg p \]By a simple reduction, then: \[ \neg q \to \neg p \]QED.
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
I think I need to assume (not q) and use that with the Hilbert System to come to the conclusions that (p→q)→(not p)
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
sorry if the way I wrote that made it confusing
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Oh, all right, sorry, that's not all too difficult, let me retype the above:
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
It's essentially the same proof, just that we assume only the first given to denote the second. We assume: \[ \neg q, (p\to q) \]To be true, along with the three logical actions in the Hilbert system. We begin: \[ (p\to q)\to (\neg q \to \neg p) \]By (3): \[ \left ( \lnot \phi \to \lnot \psi \right) \to \left( \psi \to \phi \right) \]So: \[ (\neg q \to \neg p)\to \neg p\wedge \neg q\to \neg p \]Consequentially: \[ \neg q \vdash (p\to q) \to \neg p \]QED.
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
are you assuming p>q at the beginning?
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Yes, you have to, otherwise nothing follows.
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Because we would then have no relation between instance \(p\) and \(q\).
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
well the only assumption I think I can make is not q
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
and I can only use things I can derive from the 3 axioms or modus ponens
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
for example I could use axiom 1 to write ((~q) >(p>(~q))
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
and use that like an assumption
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
at least this is my understanding of the hilbert proof system
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
This is my first time using it
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Yeah, I only used axioms, the third axiom states the manipulation above, and there needs to be some relation for p, q, otherwise we can only imply \(q\to q\), or some other singlydefined tautology.
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
I don't understand how you went from (p→q)→(¬q→¬p) to the end
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
I can get to this point by manipulating the axioms
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
how can you get the end to be just (not p)
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
and I've never seen an ^ in any Hilber proofs before
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
You can take the first axiom (or a conjunction/disjunction operator, I don't know if you're allowed to use these, but that's what my proof contains): An explicitaxiom proof would look like this, for \[ (p\to q)\to (\neg q \to \neg p) \]Thus, MP: \[ \frac{\neg q\to\neg p,\neg q}{\therefore \neg p} \]
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
I don't know what a conjunction/disjunction operator is. the only thing I have available to me are these axioms A1 (A > (B>A)) A2 ((A > (B>C)) > ((A>B)>(A>C))) A3 (((~A) > (~B)) > (B>A)) and modus ponens hich states if we have A>B and A we can infer B
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Well, yeah, I've already proven it using only those.
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
sorry for not responding, I had to catch a bus
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
hmm maybe I'm confused about how you wrote it then
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
It's all right... but I've already shown the proof with only those 3.. I can rewrite it wholly, if you wish...
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
well can you explain how you use each axiom at each step? From the examples I've been provided it seems like you need to start with an assumption like (~q) or one of the axioms with values substituted into it
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
I'm a little confused how you begin with (p→q)→(¬q→¬p)
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
actually that is axiom 3 isn't it with A=¬p and B=¬q
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
We assume: \[\neg q, (p\to q) \]To be true, along with the three logical actions in the Hilbert system. We begin: As given by Axiom (3) which states: \[ \left ( \lnot \phi \to \lnot \psi \right) \to \left( \psi \to \phi \right) \]Then: \[ (p\to q)\to (\neg q \to \neg p) \](Since, we can replace it the logical inverses of each operator) So, by the above: \[ \frac{\neg q\to\neg p,\neg q}{\therefore \neg p} \]And we're done.
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Yes, it is.
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
The Hilbert System allows one to replace any single instance \(\phi\) with whatever other instance we wish, so long as it holds true.
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
could you please explain where this part comes from? ¬q→¬p,¬q
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
We know by our previous deduction that: \[ \neg q\to \neg p \]Since we have: \[ \neg q \]Therefore: \[ \neg p \](Modus Ponens)
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
can you apply modus ponens to the end of a statement like that?
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
From what I've seen they only apply modus ponens to the whole thing
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Well, the statement I gave IS Modus Ponens (MP). We're simply taking our final logical deduction given by our previous.
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
but we are only applying it to (¬q→¬p) rather than the whole (p→q)→(¬q→¬p) can we assume (¬q→¬p) based on (p→q)→(¬q→¬p)?
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
or are you getting the assumption that (¬q→¬p) from somewhere else?
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
No, the assumption is reached by the previous statement. I think it's easier to write it out. We have that (p implies q) implies (not q implies not p). Therefore, by our assumption, we can therefore say that, for anything that holds in (p implies q), such will also hold in (not q implies not p). With this, we give the statement that not p is true based upon the fact that not q is true as the statement previously given follows.
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Treat \(p \to q\) as our premise, and \(\neg q \to \neg p\) as our conclusive statement. Because we can say that, if \(\Gamma\vdash a\to b \), then, by MP: \[ \frac{\Gamma\vdash a}{\Gamma \vdash b} \]
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
so you are assuming (p→q) to infer that (¬q→¬p) then
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
I don't think I can use (p→q) as an assumption though. I think the only thing I can use as an assumption is (¬q) unless I can derive (p→q) from one of the axioms
 one year ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
OK so I completely understand your solution now. But I need to write an explicit proof so I can't just convert (¬¬p→¬¬q)→(¬q→¬p) to (p→q)→(¬q→¬p). I need o show I can get the (¬¬p→¬¬q) with the axioms as well. Can you help me out with that please?
 one year ago
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