I'm thinking (dreaming) that the post-boil gravity would be greater than .5Z because the extra water would lead to a more efficient or thorough extraction.

There is only a finite amount of extract in X lbs or kg of grain. While it's true that a more dilute mash will have more of the available extract in solution, the dilution, by it's very nature lowers the gravity of the wort.

I see what you are trying to do but you are getting stuck on

~~available~~ extract in solution versus gravity of the solution.

I ran some numbers:

Let's negate losses except for grain absorption. Let's also assume 100% conversion efficiency and assume no-sparge so that we only see first wort extract. I assumed 80% Fine Grind Extract and 4% moisture.

If I mash 5.67 kg (12.50 lbs) of grain into 32 l (8.45 gal) of water, I could reasonably expect ~ 12P (~ 1.048 S.G.) of first wort extract.

If I mash 5.67 kg (12.50 lbs) of grain into 64 l (16.91 gal) of water, I could reasonably expect ~ 6.4P (~ 1.025 S.G.) of first wort extract.

So you can see that the second case is marginally more than the 0.5Z gravity i roughed up before. Either way you shake it, more of the available extract in solution does not translate into an appreciable benefit in gravity, i.e. it is swamped about by the massive drop in gravity of the more dilute mash.

Now granted, that's an extreme case. There is a whole range of more practical increases in volume between Y and 2Y. The real point is that you wouldn't increase volume without a calculated increase in grain amount to offset the change in gravity.