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bluebrandon
prove using the Hilbert System {not q} ⊢ (p→q)→(not p)
I'm supposed to use the 3 axioms of the Hilbert System and Modus Ponens to prove this
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.
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)
sorry if the way I wrote that made it confusing
Oh, all right, sorry, that's not all too difficult, let me retype the above:
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.
are you assuming p->q at the beginning?
Yes, you have to, otherwise nothing follows.
Because we would then have no relation between instance \(p\) and \(q\).
well the only assumption I think I can make is not q
and I can only use things I can derive from the 3 axioms or modus ponens
for example I could use axiom 1 to write ((~q) ->(p->(~q))
and use that like an assumption
at least this is my understanding of the hilbert proof system
This is my first time using it
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 singly-defined tautology.
I don't understand how you went from (p→q)→(¬q→¬p) to the end
I can get to this point by manipulating the axioms
how can you get the end to be just (not p)
and I've never seen an ^ in any Hilber proofs before
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 explicit-axiom proof would look like this, for \[ (p\to q)\to (\neg q \to \neg p) \]Thus, MP: \[ \frac{\neg q\to\neg p,\neg q}{\therefore \neg p} \]
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
Well, yeah, I've already proven it using only those.
sorry for not responding, I had to catch a bus
hmm maybe I'm confused about how you wrote it then
It's all right... but I've already shown the proof with only those 3.. I can re-write it wholly, if you wish...
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
I'm a little confused how you begin with (p→q)→(¬q→¬p)
actually that is axiom 3 isn't it with A=¬p and B=¬q
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.
The Hilbert System allows one to replace any single instance \(\phi\) with whatever other instance we wish, so long as it holds true.
could you please explain where this part comes from? ¬q→¬p,¬q
We know by our previous deduction that: \[ \neg q\to \neg p \]Since we have: \[ \neg q \]Therefore: \[ \neg p \](Modus Ponens)
can you apply modus ponens to the end of a statement like that?
From what I've seen they only apply modus ponens to the whole thing
Well, the statement I gave IS Modus Ponens (MP). We're simply taking our final logical deduction given by our previous.
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)?
or are you getting the assumption that (¬q→¬p) from somewhere else?
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.
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: \[ \frac{\Gamma\vdash a}{\Gamma \vdash b} \]
so you are assuming (p→q) to infer that (¬q→¬p) then
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
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?