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

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:

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

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

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

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

Yes, it is.

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

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

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

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

Yep.