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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.

seriallchiller
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  • There seems to be some context missing? They way you put it, would mean that EVERY proposition P implies EVERY proposition Q, which is nonsensical. In general, to prove something without a premise, using natural deduction, you can make any assumptions you want, so long as you eventually eliminate all of them. – elWanderero Apr 02 '15 at 10:51
  • Yes that is true, the actual problem i am trying to solve is ⊢ ∃x(Px → ∀y Py). to which i have gotten stuck at the point ~(Pa → Pb) (through ∃ and ∀ elimination from an assumption intended to find a contradiction). I just thought I would ask the question as (P→Q) because i thought it was my underlying problem? – seriallchiller Apr 02 '15 at 10:52
  • An outline of a proof then (assuming it is not natural deduction itself you have a problem with, but just the winding path to prove this particular statement): Either $\forall y P y$ or $\neg\forall y P y$. In the first case, the statement follows. In the second case, $\exists x \neg P x$. For this $x$, $Px$ is false, so that it can imply anything your heart desires... – elWanderero Apr 02 '15 at 11:27
  • Cool thanks for your help, i'll give it a go :) – seriallchiller Apr 02 '15 at 11:33
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    You can see this post. – Mauro ALLEGRANZA Apr 02 '15 at 11:57
  • This is now just a a duplicate as @m pointed out. Got me interested and I posted my answer there - https://math.stackexchange.com/a/2422151/432081 – CopyPasteIt Sep 09 '17 at 14:04

1 Answers1

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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.

Ove Ahlman
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  • The standard rules for first-order logic assume that the universe is not empty, no matter whether there is a constant symbol in the language or not. In particular, $\exists$-introduction allow you to replace any term, not only a closed one, by a new existential variable. So when you're assuming $\forall x.Px$ you can derive $Px\to\forall x.Px$ (by propositional tautology) and apply $\exists$-introduction to get $\exists x.(Px\to \forall x.Px)$. – hmakholm left over Monica Sep 08 '17 at 13:24
  • @HenningMakholm I agree with your calculations given that we do not allow for an empty universe. However "Standard rules" just depend on which book you look in or from which area of mathematics you come from. I come from model theory, where the empty structure is "standard" to exist. – Ove Ahlman Oct 02 '17 at 16:22