Thread: Reversing quantifiers

Any hints on the strategy to derive ∃y∀x(Ax & By) from ∀x∃y(Ax & By) in fitch-format? I've been stuck on this one for weeks.

"There exists x (property of x)" can be changed to "not For all x, not (property of x)." Same thing in converse. Is that enough of a hint?

Originally Posted by Annatala

"There exists x (property of x)" can be changed to "not For all x, not (property of x)." Same thing in converse.

Yes, but I am not sure how this helps here.

Starting from $\displaystyle \forall x\exists y\,(Ax\land By)$, we can instantiate x with an arbitrary term t (e.g., a fresh variable) and get a y such that $\displaystyle At\land By$, in particular, $\displaystyle By$. So, we are left to show $\displaystyle \forall x\,(Ax\land By)$. Fix an arbitrary x, instantiate the universal variable in $\displaystyle \forall x\exists y\,(Ax\land By)$ with x and thus get $\displaystyle Ax$.

Emakarov, I'm not sure it's that easy to do in a fitch-format derivation. A violation of rules involving the eigen-variable occur. So, for example, I can't do this:
∀x∃y(Ax & By)
∃y(Az & By)
-Az&Bw
-Az
-∀xAx

('-' indicates a sub-derivation). The last line in the sub-derivation is a violation. I can't universally quantify anything that's free in an assumption.

Here is a derivation in tree format.

Originally Posted by emakarov

Yes, but I am not sure how this helps here.

My bad. I don't understand this question at all, although I thought I did.