I don't have numbers for the lines yet, but here's the bones of my current strategy.

Exists-x (~P(x) || Q(x)) ......... premise. (I'm pretty sure that's right.

)

Open up an "Arb-Box"... x

_{0} ~P(x

_{0}) || Q(x

_{0})

.... deal with the cases here? Or take it to the next level?

Open up a box in which to get a contradiction of P(x

_{0}) && Q(x

_{0})

...

...

It ends in "bottom".

So after that inner box we have ~ P(x

_{0}) && Q(x

_{0}) ....... ~i (the previous box)

....

This is a "Chi" to end the exists-+-"arb-that-exists" box, so we're allowed to take it out? ... This is NOT such a "Chi", is it? A little bell rings saying you can't use the arbitrary x in your Chi....

OK. Enough doubt! Carry on.

The idea Was, come out of the box with ~ P(x) && Q(x), which I'm almost certain is wrong.

Obviously were it not wrong, one would easily move from there to "Exists" to reach the conclusion sought.

There you go. I won't edit this. Mistakes are often far more interesting than correct answers.