let x be 5 in x + 1
-which evaluates to `6`. Okay, fair enough, so I am moving beyond the ∀ `x==5.` φ idea when I do this. But the rule for how to interpret this are just a straightforward generalization of our existing understanding for how to interpret bound variables. So there's nothing fundamentally novel here.
+which evaluates to `6`. Okay, fair enough, so I am moving beyond the ∀ `x==5.` φ idea when I do this. But the rules for how to interpret this are just a straightforward generalization of our existing understanding for how to interpret bound variables. So there's nothing fundamentally novel here.
We can have multiple `let`-expressions embedded, as in:
The `x + 1` that is evaluated to give the value that `y` gets bound to uses the (more local) binding of `x` to `5`, not the (previous, less local) binding of `x` to `0`. By the way, the parentheses in that displayed expression were just to focus your attention. It would have parsed and meant the same without them.
+Now we can allow ourselves to introduce λ-expressions in the following way. If a λ-expression is applied to an argument, as in: `(`λ `x.` φ`) M`, for any (simple or complex) expressions φ and `M`, this means the same as: `let x be M in` φ. That is, the argument to the λ-expression provides (when evaluated) a value for the variable `x` to be bound to, and then the result of the whole thing is whatever φ evaluates to, under that binding to `x`.
+
+If we restricted ourselves to only that usage of λ-expressions, that is when they were applied to all the arguments they're expecting, then we wouldn't have moved very far from the decidable fragment of arithmetic we began with.
+
+However, it's tempting to help ourselves to the notion (at least partly) *unapplied* λ-expressions, too. If I can make sense of what:
+
+ (λ x. x + 1) 5
+
+means, then I can make sense of what:
+
+ (λ x. x + 1)
+
+means, too. It's just *the function* that waits for an argument and then returns the result of `x + 1` with `x` bound to that argument.
+
*More to come.*