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bluebrandon

  • 2 years ago

prove using the Hilbert System {not q} ⊢ (p→q)→(not p)

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  1. bluebrandon
    • 2 years ago
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    I'm supposed to use the 3 axioms of the Hilbert System and Modus Ponens to prove this

  2. LolWolf
    • 2 years ago
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    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.

  3. bluebrandon
    • 2 years ago
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    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)

  4. bluebrandon
    • 2 years ago
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    sorry if the way I wrote that made it confusing

  5. LolWolf
    • 2 years ago
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    Oh, all right, sorry, that's not all too difficult, let me retype the above:

  6. LolWolf
    • 2 years ago
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    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.

  7. bluebrandon
    • 2 years ago
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    are you assuming p->q at the beginning?

  8. LolWolf
    • 2 years ago
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    Yes, you have to, otherwise nothing follows.

  9. LolWolf
    • 2 years ago
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    Because we would then have no relation between instance \(p\) and \(q\).

  10. bluebrandon
    • 2 years ago
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    well the only assumption I think I can make is not q

  11. bluebrandon
    • 2 years ago
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    and I can only use things I can derive from the 3 axioms or modus ponens

  12. bluebrandon
    • 2 years ago
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    for example I could use axiom 1 to write ((~q) ->(p->(~q))

  13. bluebrandon
    • 2 years ago
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    and use that like an assumption

  14. bluebrandon
    • 2 years ago
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    at least this is my understanding of the hilbert proof system

  15. bluebrandon
    • 2 years ago
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    This is my first time using it

  16. LolWolf
    • 2 years ago
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    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.

  17. bluebrandon
    • 2 years ago
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    I don't understand how you went from (p→q)→(¬q→¬p) to the end

  18. bluebrandon
    • 2 years ago
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    I can get to this point by manipulating the axioms

  19. bluebrandon
    • 2 years ago
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    how can you get the end to be just (not p)

  20. bluebrandon
    • 2 years ago
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    and I've never seen an ^ in any Hilber proofs before

  21. LolWolf
    • 2 years ago
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    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: \[ \frac{\neg q\to\neg p,\neg q}{\therefore \neg p} \]

  22. bluebrandon
    • 2 years ago
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    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

  23. LolWolf
    • 2 years ago
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    Well, yeah, I've already proven it using only those.

  24. bluebrandon
    • 2 years ago
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    sorry for not responding, I had to catch a bus

  25. bluebrandon
    • 2 years ago
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    hmm maybe I'm confused about how you wrote it then

  26. LolWolf
    • 2 years ago
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    It's all right... but I've already shown the proof with only those 3.. I can re-write it wholly, if you wish...

  27. bluebrandon
    • 2 years ago
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    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

  28. bluebrandon
    • 2 years ago
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    I'm a little confused how you begin with (p→q)→(¬q→¬p)

  29. bluebrandon
    • 2 years ago
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    actually that is axiom 3 isn't it with A=¬p and B=¬q

  30. LolWolf
    • 2 years ago
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    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.

  31. LolWolf
    • 2 years ago
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    Yes, it is.

  32. LolWolf
    • 2 years ago
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    The Hilbert System allows one to replace any single instance \(\phi\) with whatever other instance we wish, so long as it holds true.

  33. bluebrandon
    • 2 years ago
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    could you please explain where this part comes from? ¬q→¬p,¬q

  34. LolWolf
    • 2 years ago
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    We know by our previous deduction that: \[ \neg q\to \neg p \]Since we have: \[ \neg q \]Therefore: \[ \neg p \](Modus Ponens)

  35. bluebrandon
    • 2 years ago
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    can you apply modus ponens to the end of a statement like that?

  36. bluebrandon
    • 2 years ago
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    From what I've seen they only apply modus ponens to the whole thing

  37. LolWolf
    • 2 years ago
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    Well, the statement I gave IS Modus Ponens (MP). We're simply taking our final logical deduction given by our previous.

  38. bluebrandon
    • 2 years ago
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    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)?

  39. bluebrandon
    • 2 years ago
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    or are you getting the assumption that (¬q→¬p) from somewhere else?

  40. LolWolf
    • 2 years ago
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    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.

  41. LolWolf
    • 2 years ago
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    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: \[ \frac{\Gamma\vdash a}{\Gamma \vdash b} \]

  42. bluebrandon
    • 2 years ago
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    so you are assuming (p→q) to infer that (¬q→¬p) then

  43. LolWolf
    • 2 years ago
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    Yep.

  44. bluebrandon
    • 2 years ago
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    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

  45. bluebrandon
    • 2 years ago
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    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?

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