*	In def 2.5, they say the denotation of an e-type constant <code>&alpha;</code> wrt a discourse possibility `(r, g, w)` is whatever object the world `w` associates with <code>&alpha;</code>. Since we don't have worlds, this will just be an object.

*	They say the denotation of a predicate is whatever extension the world `w` associates with the predicate. Since we don't have worlds, this will just be an extension.

*	They say the denotation of a variable is the object which the store `g` assigns to the index that the assignment function `r` assigns to the variable. In other words, if the variable is `'x'`, its denotation wrt `(r, g, w)` is `g[r['x']]`.

We're going to keep all of that, except dropping the worlds. And instead of talking about

>	\[[expression]] in possibility `(r, g, w)`

we'll just talk about \[[expression]] and let that be a monadic object, implemented in part by a function that takes `(r, g)` as an argument.

More specifically, \[[expression]] will be a set of `'a discourse possibility` monads, where `'a` is the appropriate type for *expression*, and the discourse possibility monads are themselves state monads where `(r, g)` is the state that gets updated. Those are implemented as functions from `(r, g)` to `(a, r', g')`, where `a` is a value of type `'a`, and `r', g'` are possibly altered assignment functions and stores.

*	In def 2.7, GS&V talk about an operation that takes an existing set of discourse possibilities, and extends each member in the set by allocating a new location in the store, and assigning a variable `'x'` to that location, which holds some object `d` from the domain. It will be useful to have a shorthand way of referring to this operation:

		let newpeg_and_bind (variable : char) (d : entity) =
			fun ((r, g) : assignment * store) ->
				let newindex = List.length g
				(* first we store d at index newindex in g, which is at the very end *)
				(* the following line achieves that in a simple but very inefficient way *)
				in let g' = List.append g [d]
				(* next we assign 'x' to location newindex *)
				in let r' = fun variable' ->
					if variable' = variable then newindex else r variable'
				(* the reason for returning a triple with () in first position will emerge *)
				in ((), r',g')

*	At the top of p. 13 (this is in between defs 2.8 and 2.9), GS&V give two examples, one for \[[&exist;xPx]] and the other for \[[Qx]]. In fact it will be easiest for us to break \[[&exist;xPx]] into two pieces, \[[&exist;x]] and \[[Px]]. Let's consider expressions like \[[Px]] (or \[[Qx]]) first.