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

1. bluebrandon

I'm supposed to use the 3 axioms of the Hilbert System and Modus Ponens to prove this

2. LolWolf

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

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

sorry if the way I wrote that made it confusing

5. LolWolf

Oh, all right, sorry, that's not all too difficult, let me retype the above:

6. LolWolf

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

are you assuming p->q at the beginning?

8. LolWolf

Yes, you have to, otherwise nothing follows.

9. LolWolf

Because we would then have no relation between instance $$p$$ and $$q$$.

10. bluebrandon

well the only assumption I think I can make is not q

11. bluebrandon

and I can only use things I can derive from the 3 axioms or modus ponens

12. bluebrandon

for example I could use axiom 1 to write ((~q) ->(p->(~q))

13. bluebrandon

and use that like an assumption

14. bluebrandon

at least this is my understanding of the hilbert proof system

15. bluebrandon

This is my first time using it

16. LolWolf

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

I don't understand how you went from (p→q)→(¬q→¬p) to the end

18. bluebrandon

I can get to this point by manipulating the axioms

19. bluebrandon

how can you get the end to be just (not p)

20. bluebrandon

and I've never seen an ^ in any Hilber proofs before

21. LolWolf

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

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

Well, yeah, I've already proven it using only those.

24. bluebrandon

sorry for not responding, I had to catch a bus

25. bluebrandon

hmm maybe I'm confused about how you wrote it then

26. LolWolf

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

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

I'm a little confused how you begin with (p→q)→(¬q→¬p)

29. bluebrandon

actually that is axiom 3 isn't it with A=¬p and B=¬q

30. LolWolf

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

Yes, it is.

32. LolWolf

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

could you please explain where this part comes from? ¬q→¬p,¬q

34. LolWolf

We know by our previous deduction that: $\neg q\to \neg p$Since we have: $\neg q$Therefore: $\neg p$(Modus Ponens)

35. bluebrandon

can you apply modus ponens to the end of a statement like that?

36. bluebrandon

From what I've seen they only apply modus ponens to the whole thing

37. LolWolf

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

38. bluebrandon

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

or are you getting the assumption that (¬q→¬p) from somewhere else?

40. LolWolf

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

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

so you are assuming (p→q) to infer that (¬q→¬p) then

43. LolWolf

Yep.

44. bluebrandon

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

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?