Clarify behavior of B~ A~ B A

[shicks]

There are a number of permutations of this that we need to think about. Specifically:

"run, enter, end-run, dispose" - probably should not restore the 2.

const x = v.withValue(1);
v.run(2, () => x[Symbol.enter]());
x[Symbol.dispose]();

---

"enter, run, dispose, end-run" - probably should not restore the 1.

const x = v.withValue(1);
x[Symbol.enter]();
v.run(2, () => x[Symbol.dispose]());

---

"enter A1, enter A2, dispose A1, dispose A2" - probably should not restore the 1.

const x1 = v.withValue(1);
const x2 = v.withValue(2);
x1[Symbol.enter]();
x2[Symbol.enter]();
x1[Symbol.dispose]();
x2[Symbol.dispose]();

---

Two separate variables - very similar to the previous one, but maybe a non-issue if different vars are completely independent here?

const x1 = v1.withValue(1);
const x2 = v2.withValue(2);
x1[Symbol.enter]();
x2[Symbol.enter]();
x1[Symbol.dispose]();
x2[Symbol.dispose]();