## anonymous one year ago Provide a proof using standard predicate logic (without using the deduction method, proof by contradiction, or proof by contraposition) for the validity of the following argument: (∀x)A(x) ∧ (∃x)B(x) → (∃x)[A(x) ∧ B(x)]

1. anonymous

I'm trying to solve this now. If anyone wants to work with me then that would be great; otherwise, I have no way to check my proof.

2. anonymous

If you care to show your solution I am happy to read it for you. What I remember from my course (which was very long ago) was to in order to show such statements we had to use axioms, in fact we had 13. It is indeed hard to judge if you're working with the same axiom system as I did. If you work with an axiom system you need to share your axioms in order to get help. Aside from that, I can only tell that the implication is indeed true. Because if a Formula A(x) is true for all its arguments and there exists an x such that B(x) is true then of course there exists an x such that A(x) and B(x) are true. This is however far from a formal proof. If you need a formal one, we need your axioms.

3. anonymous

@Spacelimbus

4. anonymous

@Spacelimbus Thank you for replying. I like your informal proof :) I am still working on trying to figure this out.

5. anonymous

@zepdrix how are you with predicate logic?

6. zepdrix

Ok sorry sorry was finishing up a game :d What's going on now?

7. zepdrix

If For all x, A(x) and there exists an x such that B(x), then there exists an x such that A(x) and B(x). This is really weird stuff 0_o I don't remember ever doing this nonsense with these quantifiers.. hmm

8. anonymous

The good news is-- I'm fine with induction and propositional logic now... this is the last piece of the puzzle.

9. anonymous

what game were you playing?

10. zepdrix

Rocket League XD lol

11. zepdrix

Had to give up the addicting games so I could really really focus on schooling this semester :P Rocket League is mindless fun though, nothing that I can't let go of.

12. anonymous

that's cray. i've never seen a vehicle based soccer game before, lol

13. anonymous

i only play chess and starcraft :X

14. zepdrix

Hmm I gotta think about this problem a bit :O

15. zepdrix

Ummm ok let's see if this works or not...

16. zepdrix

The first piece: $$\large\rm \forall x, A(x)$$ For all x, A(x). So any x's that we have, all of them, satify this statement A. So all together we have the union of a bunch of statements A involving different x's. $\large\rm \forall x, A(x)\quad\equiv\quad A(x_1)\cup A(x_2)\cup...\cup A(x_n)$

17. zepdrix

This next piece: $$\large\rm \exists x: B(x)$$ Which translates to There exists an x such that B(x), tells us that there is a single x in existence that satisfies the statement B. Let's call it x_i, just one of the many x's we could list. $\large\rm \exists x: B(x)\quad\equiv\quad B(x_i)$

18. zepdrix

And of course, the caret shape is to denote "and" or "intersection.

19. zepdrix

So on the left side of our thing we have,$\large\rm \left[A(x_1)\cup A(x_2)\cup...\cup A(x_n)\right]\cap B(x_i)$

20. zepdrix

They only intersect at the same x,$\large\rm A(x_i)\cap B(x_i)$And we need to see if that implies the right part. I don't think I'm doing this right -_- it's been too long...

21. zepdrix

On the right side, $$\large\rm \exists x:\left[A(x)\wedge B(x)\right]$$ Translates to: There exists an x such that, both A(x) and B(x). Again, let's call this x, x_i. So the right side says that some x exists, so that the intersection of A(x) and B(x) is satisfied.$\large\rm \exists x:\left[A(x)\wedge B(x)\right]\quad\equiv\quad A(x_i)\cap B(x_i)$

22. zepdrix

But again, I don't think I'm doing this correctly -_- There are probably some familiar easy steps that I'm forgetting about. I just don't remember how to deal with these XD lol