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Given two sets $A$ and $B$ and a general relation $R$, which one is the stronger "assertion"?

1: $∃x ∈ A , ∀y ∈ B, ~~R(x, y)$

2: $∀y ∈ B , ∃x ∈ A, ~~R(x, y)$

I am not sure what "stronger" means.

Thanks in advance.

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    Stronger means it implies the other. Here the first one is the stronger, because in the second one, the $x$ **depends** on the $y$, but in the first one, there is one $x$ that works for **all** the $y$'s :-) – 2017-01-14
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    $(1) \implies (2)$ but not necessarily $(2) \implies (1)$ – 2017-01-14

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Just informally, the more things I claim, the stronger the total of my claims become. For example, if I claim that 'it is going to rain tomorrow', then I not claiming as much as when I say 'It is going to rain for the next two days'. So, the latter is the 'stronger' claim. Similarly, 'I can lift a refrigerator!' is a much stronger claim than 'I can lift a pencil!'.

Notice also that the stronger the claim is, the less likely it is for it to be true: it is likely that I can lift a pencil, but less likely I can lift a refrigerator.

More formally, in logic we say that claim P is stronger than claim Q when P implies Q. This actually makes sense in the examples before: If it rains for the next two days, then it will rain tomorrow, and if I can lift a refigerator, then I can certainly lift a pencil.

To see that in your problem, claim 1 implies claim 2, notice that claim 1 is saying that 'there is some object from set A that stands in relation R to all objects in set B$, while claim 2 says that 'for every object in set B there is some objects from set A that stands in relation R to it'. In other words, in claim 1 we know it is the same one object from set A that stands in relation R to all objects in set B, while in claim 2, all objects in set B once again stand in relation R to some object in set A, but this time, it doesn't have to be the same object ... making claim 2 more more likely to be true .. and thus the weaker claim. Indeed, claim 1 logically implies claim 2, because if there is an object (call it 'a') that stand in relation R to all objects in set B, then for each object b in set B, I can of course find an object in set A ('a'!) that stands in relation R to b. But the other way around does not hold, so claim 1 is strictly stronger than claim 2.

Finally, to provide a concrete example: suppose both sets are natural numbers, and suppose the $R(x,y)$ means $x > y$

Then claim 1 means: There is a natural number that is greater than all natural numbers ... which is clearly false

Claim 2 would mean: For every number there is a greater natural number ... which is true

Claim 1 is false because there is no one number greater than all others ... but claim 2 is true, because for each number we can find a greater number. Claim 1 is therefore such a strong claim that it is false.