# I Heloise and Abelard

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1. Sep 21, 2016

The question is about a logic game between two competing players "Abelard and Heloise", Abelard is the name given to the "For all" quantifier and "Heloise" is the name given for the "There exists" quantifier. The statement is:

Is $(\forall x)(\forall y)(\forall z)(\exists u)(\exists v)(\forall t) xu - yt > v^z$ for x, y, z, u, v $\in N$

So where do I start for this? reading from left to right, does Abelard get 3 turns to move x, y, z, then Heloise can move u and v...?

I am completely lost, any reference to a website or book that would help would be greatly appreciated.

2. Sep 21, 2016

### andrewkirk

What are you trying to do?

3. Sep 21, 2016

Sorry, determine whether or not the statement is true given those existential quantifiers.

4. Sep 22, 2016

### Zafa Pi

What is N? What is the domain of t?

5. Sep 22, 2016

### Heinera

I guess N is the natural numbers. But is t supposed to be in N as well?

6. Sep 22, 2016

### andrewkirk

The crunch step is the $\forall t$, as you have to, given values for $x,y,z$, select values for $u$ and $v$ such that the inequality holds for all $t$. That sounds unlikely to be possible, but it's easy to test.

Since $t$ is the crunch, rearrange the inequality with $t$ on its own on one side. To reduce the number of moving parts, select values of $x,y,z$ that make the calcs easy, such as their all being 1. What is the inequality now?

Having done all that, it should be immediately obvious whether it's possible to choose $u,v$ such that the inequality holds for all possible values of $t$. I get the same answer regardless of whether $t$'s range of possible values is $\mathbb R,\mathbb Z$ or $\mathbb N$.

7. Sep 22, 2016

### Zafa Pi

If you look around you will see that sometimes the naturals contain 0 and sometimes not. That is what I was trying to find out.

Last edited: Sep 22, 2016
8. Sep 22, 2016

### Zafa Pi

If t ∊ {0} and N = {1,2,3, ...} then the expression is valid by letting u = 2, and v = 1. If instead N = {0,1,2, ...} then the expression is invalid.
Of course by changing the domain of t we may get different answers. If t ∊ N then the expression is invalid. The OP is naively vague.

9. Sep 22, 2016

Yeah my professor sent out an email today saying he forgot to define t in the set of natural numbers =/. @Zafa Pi, I need a safe space right now.

10. Sep 22, 2016

### Zafa Pi

Oh dear, you should have given us a trigger warning.

11. Sep 22, 2016

So I kind of get it now, Eloise wants the statement to be true and Abelard wants it to be false. So since Eloise chooses v, she can choose 1 which makes the RHS 1 for any value of Z. Here is my solution:

A: x =1. A y = 10. A: z = 10. E: u = 2*y*t. E: v = 1. A: t = 10. Giving 10(2*10-10) > 1^10, making the statement true so Eloise wins. The only part I'm uncertain on is whether or not Eloise is allowed to choose that value for u since t is not chosen yet. Thanks for the help everyone.

12. Sep 22, 2016

### Zafa Pi

E can can only make u depend on x, y, and z. A always wins. I said this in post 8.

I would like to express my regrets if this hurts your feelings, but that's what you get for leaving your safe space. Perhaps you could sue PF under Title IX and they might suspend me.

Last edited: Sep 22, 2016
13. Sep 22, 2016

### Erland

If x=y=0, then the left side is 0 and the inequality is false no matter how we choose the natural numbers z, t, u, and v.

14. Sep 22, 2016

### Zafa Pi

Indeed if 0 is in N (see my post #7), however you are not free to make x and y be 0. But from post #11 it seems the OP has N = {1,2,3, ...}. Yet from my post #12 the result is the same. My guess is that you didn't read the previous posts before you posted.

15. Sep 22, 2016

### Zafa Pi

It might help if we translate the OP into English.
All variables are in N, the natural numbers.
For each each choice of x, y, and z we can choose a u and a v such that xu - yt > v² for all t.
The above is false whether N contains 0 or not.

16. Sep 22, 2016

Yes, Abelard is the actual winner. I think my error was letting Eloise play out of turn. Since Abelard gets the last turn, he can choose a high value for t to make it false.

17. Sep 22, 2016

### Zafa Pi

That's correct assuming N = {1, 2, 3, ...}. If N = {0, 1, 2, ...} then Erland's post #13 has a short proof.

18. Sep 23, 2016

### PeroK

I would have experimented with some numbers. The obvious starting point was $x = y = z = 1$. Now, I need to find $u, v$ such that $\forall t$:

$u - t > v$

Which means that $\forall t$

$t < u-v$

Which is absurd. Whatever $u, v$ are choosen, we can always find a large enough $t$. So, the statement is false. It failed for the most simple choice of $x, y, z$ (assuming the Naturals do not include 0).

If I had been able to find $u, v$ for my first choice of $x, y, z$ that might have given me some clues on choosing a different set of $x, y, z$ to show the statement is false or on how to prove it is true.

The important point is that to show that something like this is false, you only have to find a single set of $x, y, z$. To show it is true, you have to prove it for all $x, y, z$. If you're stuck, try some numbers and see what happens. Especially if you have it described as a game - you are allowed to try playing the game, you don't just have to sit and read the rules to work it out!