[Math] Natural deduction proofs without a premise

logic

How do I prove a formula without a premise?
The question looks like this

⊢ (P→Q)

I have started by making the assumption NOT(P→Q) to get a contradiction, and have no idea where to go from here.

If anyone could offer me some guidance, or even direct me towards some similar questions(preferably with answers) that would also be much appreciated.

Any help is appreciated and thanks in advance.

Best Answer

If you want to prove $P \rightarrow Q$ then you should assume $P$ and try to deduce $Q$ in some way.

In the comment you way that you want to prove $⊢ ∃x(Px → ∀y Py)$, however this is only true if we have a constantsymbol already inthe language we are studying, since else we could have the empty model as a counter example.

Here is one strategy to prove that $⊢ ∃x(Px → ∀y Py)$:

  1. Show that $\forall x P(x) \vee \neg \forall x P(x)$ hold.
  2. Do $\vee-$elimination on this formula. The case $\forall x(P(x))$ is then quite straight forward (use the fact that we have a constant for some element $a$ and for this element $P(a)$ has to hold).
  3. In the Case $\neg\forall x P(x)$ derive $\exists x \neg P(x)$.
  4. Do $\exists-$elimination to derive an element $c$ such that $\neg P(c)$ hold, now assuming $P(c)$ you can derive $\forall x P(x)$ and thus finnish the proof.

This may possibly be done in a simpler way. I hope that these "strategy hints" are enough for you to figure it out.

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