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anonymous
 4 years ago
prove using the Hilbert System
{not q} ⊢ (p→q)→(not p)
anonymous
 4 years ago
prove using the Hilbert System {not q} ⊢ (p→q)→(not p)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm supposed to use the 3 axioms of the Hilbert System and Modus Ponens to prove this

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Using 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think I need to assume (not q) and use that with the Hilbert System to come to the conclusions that (p→q)→(not p)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry if the way I wrote that made it confusing

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, all right, sorry, that's not all too difficult, let me retype the above:

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It'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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0are you assuming p>q at the beginning?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, you have to, otherwise nothing follows.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Because we would then have no relation between instance \(p\) and \(q\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well the only assumption I think I can make is not q

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and I can only use things I can derive from the 3 axioms or modus ponens

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for example I could use axiom 1 to write ((~q) >(p>(~q))

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and use that like an assumption

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0at least this is my understanding of the hilbert proof system

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is my first time using it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't understand how you went from (p→q)→(¬q→¬p) to the end

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I can get to this point by manipulating the axioms

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how can you get the end to be just (not p)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and I've never seen an ^ in any Hilber proofs before

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You 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: \[ \frac{\neg q\to\neg p,\neg q}{\therefore \neg p} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, yeah, I've already proven it using only those.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry for not responding, I had to catch a bus

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm maybe I'm confused about how you wrote it then

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's all right... but I've already shown the proof with only those 3.. I can rewrite it wholly, if you wish...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm a little confused how you begin with (p→q)→(¬q→¬p)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0actually that is axiom 3 isn't it with A=¬p and B=¬q

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The Hilbert System allows one to replace any single instance \(\phi\) with whatever other instance we wish, so long as it holds true.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0could you please explain where this part comes from? ¬q→¬p,¬q

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We know by our previous deduction that: \[ \neg q\to \neg p \]Since we have: \[ \neg q \]Therefore: \[ \neg p \](Modus Ponens)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you apply modus ponens to the end of a statement like that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0From what I've seen they only apply modus ponens to the whole thing

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, the statement I gave IS Modus Ponens (MP). We're simply taking our final logical deduction given by our previous.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but 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)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or are you getting the assumption that (¬q→¬p) from somewhere else?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Treat \(p \to q\) 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: \[ \frac{\Gamma\vdash a}{\Gamma \vdash b} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so you are assuming (p→q) to infer that (¬q→¬p) then

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OK 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?
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