-
Evaluation Strategies and Normalization
=======================================
-In the assignment we asked you to reduce various expressions until it wasn't possible to reduce them any further. For two of those expressions, this was impossible to do. One of them was this:
+In last week's homework, we asked you to reduce various expressions until it wasn't possible to reduce them any further. For two of those expressions, this was impossible to do. One of them was this:
(\x. x x) (\x. x x)
-As we saw above, each of the halves of this formula are the combinator <code>ω</code>; so this can also be written:
+As we discuss in other notes for this week, each of the halves of this formula are the combinator `ω`; so this can also be written:
-<pre><code>ω ω</code></pre>
+ ω ω
-This compound expression---the self-application of <code>ω</code>---is named Ω. It has the form of an application of an abstract (<code>ω</code>) to an argument (which also happens to be <code>ω</code>), so it's a redex and can be reduced. But when we reduce it, we get <code>ω ω</code> again. So there's no stage at which this expression has been reduced to a point where it can't be reduced any further. In other words, evaluation of this expression "never terminates." (This is the standard language, however it has the unfortunate connotation that evaluation is a process or operation that is performed in time. You shouldn't think of it like that. Evaluation of this expression "never terminates" in the way that the decimal expansion of π never terminates. These are static, atemporal facts about their mathematical properties.)
+This compound expression---the self-application of `ω`---is named `Ω`. It has the form of an application of an abstract (`ω`) to an argument (which also happens to be `ω`), so it's a redex and can be reduced. But when we reduce it, we get `ω ω` again. So there's no stage at which this expression has been reduced to a point where it can't be reduced any further. In other words, evaluation of this expression "never terminates." (This is the standard language, however it has the unfortunate connotation that evaluation is a process or operation that must be performed over time. You shouldn't think of it like that. Evaluation of this expression "never terminates" in the way that the decimal expansion of π never terminates. These are static, atemporal facts about their mathematical properties.)
-There are infinitely many formulas in the lambda calculus that have this same property. Ω is the syntactically simplest of them. In our meta-theory, it's common to assign such formulas a special value, <big><code>⊥</code></big>, pronounced "bottom." When we get to discussing types, you'll see that this value is counted as belonging to every type. To say that a formula has the bottom value means that the computation that formula represents never terminates and so doesn't evaluate to any orthodox, computed value.
+There are infinitely many formulas in the Lambda Calculus that have this same property. `Ω` is the syntactically simplest of them. In our meta-theory, it's common to assign such formulas a special meaning, <big><code>⊥</code></big>, pronounced "bottom." When we get to discussing types, you'll see that expressions with this meaning are counted as belonging to every type. To say that a term "is bottom" means that the computation that term represents never terminates, and so the term doesn't evaluate to any orthodox, computed value.
From a "Fregean" or "weak Kleene" perspective, if any component of an expression fails to be evaluable (to an orthodox, computed value), then the whole expression should be unevaluable as well.
However, in some such cases it seems *we could* sensibly carry on evaluation. For instance, consider:
-<pre><code>
-(\x. y) (ω ω)
-</code></pre>
+ (\x. y) (ω ω)
-Should we count this as unevaluable, because the reduction of <code>(ω ω)</code> never terminates? Or should we count it as evaluating to `y`?
+Should we count this as unevaluable, because the reduction of `(ω ω)` never terminates? Or should we count it as evaluating to `y`?
This question highlights that there are different choices to make about how evaluation or computation proceeds. It's helpful to think of three questions in this neighborhood:
-> Q1. When arguments are complex, as <code>(ω ω)</code> is, do we reduce them before substituting them into the abstracts to which they are arguments, or later?
+> Q1. When arguments are complex, as `(ω ω)` is, do we reduce them before substituting them into the abstracts to which they are arguments (`(\x. y)`, in the expression displayed above), or later?
> Q2. Are we allowed to reduce inside abstracts? That is, can we reduce:
> where x does not occur free in `M`, to `M`?
-With regard to Q3, it should be intuitively clear that `\x. M x` and `M` will behave the same with respect to any arguments they are given. It can also be proven that no other functions can behave differently with respect to them. However, the logical system you get when eta-reduction is added to the proof theory is importantly different from the one where only beta-reduction is permitted.
+With regard to Q3, it should be intuitively clear that `\x. M x` and
+`M` will behave the same with respect to any arguments they are
+given. It can also be proven that no other functions can behave
+differently with respect to them (that is, no function can interact differently with further arguments depending on whether it has been applied to `M` versus `\x. M x`). However, the logical system you get when eta-reduction is added to the proof theory is importantly different from the one where only beta-reduction is permitted.
-If we answer Q2 by permitting reduction inside abstracts, and we also permit eta-reduction, then where none of <code>y<sub>1</sub>, ..., y<sub>n</sub></code> occur free in M, this:
+If we answer Q2 by permitting reduction inside abstracts, and we also permit eta-reduction, then where none of <code>y<sub>1</sub>, ..., y<sub>n</sub></code> occur free in `M`, this:
-<pre><code>\x y<sub>1</sub>... y<sub>n</sub>. M y<sub>1</sub>... y<sub>n</sub></code></pre>
+<pre><code>\x y<sub>1</sub>...y<sub>n</sub>. M y<sub>1</sub>...y<sub>n</sub></code></pre>
will eta-reduce by n steps to:
\x. M
-When we add eta-reduction to our proof system, we end up reconstruing the meaning of `~~>` and `<~~>` and "normal form", all in terms that permit eta-reduction as well. Sometimes these expressions will be annotated to indicate whether only beta-reduction is allowed (<code>~~><sub>β</sub></code>) or whether both beta- and eta-reduction is allowed (<code>~~><sub>βη</sub></code>).
+When we add eta-reduction to our proof system, we end up reconstruing the meaning of `~~>`, and `<~~>`, and "normal form" (we will discuss this below), all of these in terms that permit eta-reduction as well. Sometimes these expressions will be annotated to indicate whether only beta-reduction is allowed (<code>~~><sub>β</sub></code>) or whether both beta- and eta-reduction is allowed (<code>~~><sub>βη</sub></code>).
The logical system you get when eta-reduction is added to the proof system has the following property:
-> if `M`, `N` are normal forms with no free variables, then <code>M ≡ N</code> iff `M` and `N` behave the same with respect to every possible sequence of arguments.
+> if `M`, `N` are normal forms with no free variables, then `M ≡ N` iff `M` and `N` behave the same with respect to every possible sequence of arguments.
This implies that, when `M` and `N` are (closed normal forms that are) syntactically distinct, there will always be some sequences of arguments <code>L<sub>1</sub>, ..., L<sub>n</sub></code> such that:
See Hindley and Seldin, Chapters 7-8 and 14, for discussion of what should count as capturing the "extensionality" of these systems, and some outstanding issues.
-The evaluation strategy which answers Q1 by saying "reduce arguments first" is known as **call-by-value**. The evaluation strategy which answers Q1 by saying "substitute arguments in unreduced" is known as **call-by-name** or **call-by-need** (the difference between these has to do with efficiency, not semantics).
+The evaluation strategy which answers Q1 by saying "reduce arguments first" is known as **call-by-value**. The evaluation strategy which answers Q1 by saying "substitute arguments in unreduced" is known as **call-by-name** or **call-by-need** (in the absence of side-effects, the difference between these has only to do with efficiency, not semantics).
-When one has a call-by-value strategy that also permits reduction to continue inside unapplied abstracts, that's known as "applicative order" reduction. When one has a call-by-name strategy that permits reduction inside abstracts, that's known as "normal order" reduction. Consider an expression of the form:
+When one has a call-by-value strategy that *also* permits reduction to continue inside unapplied abstracts, that's known as "applicative order" reduction. When one has a call-by-name strategy that permits reduction inside abstracts, that's known as "normal order" reduction. Consider an expression of the form:
((A B) (C D)) (E F)
Applicative order evaluation does what's called a "post-order traversal" of the tree: that is, we always go down when we can, first to the left, and we process a node only after processing all its children. So `(C D)` gets processed before `((A B) (C D))` does, and `(E F)` gets processed before `((A B) (C D)) (E F)` does.
-Normal order evaluation, on the other hand, will substitute the expresion `(C D)` into the abstract that `(A B)` evaluates to, without first trying to compute what `(C D)` evaluates to. That computation may be done later.
+Normal order evaluation, on the other hand, will substitute the expression `(C D)` into the abstract that `(A B)` evaluates to, without first trying to compute what `(C D)` evaluates to. That computation may be done later.
With normal-order evaluation (or call-by-name more generally), if we have an expression like:
(\x. y) (C D)
-the computation of `(C D)` won't ever have to be performed. Instead, `(\x. y) (C D)` reduces directly to `y`. This is so even if `(C D)` is the non-evaluable <code>(ω ω)</code>!
+the computation of `(C D)` won't ever have to be performed. Instead, `(\x. y) (C D)` reduces directly to `y`. This is so even if `(C D)` is the non-evaluable `(ω ω)`!
-Call-by-name evaluation is often called "lazy." Call-by-value evaluation is also often called "eager" or "strict". Some authors say these terms all have subtly different technical meanings, but I haven't been able to figure out what it is. Perhaps the technical meaning of "strict" is what I above called the "Fregean" or "weak Kleene" perspective: if any argument of a function is non-evaluable or non-normalizing, so too is the application of the function to that argument.
+Call-by-name evaluation is often called "lazy." Call-by-value evaluation is also often called "eager" or "strict". In some contexts, these terms all have subtly different technical meanings. But those differences are hard to identify in the first place, plus mostly the terms are used in a way that doesn't adhere carefully to such subtleties. One association it may be helpful for you to remember is between the term "strict" and what we above called the "Fregean" or "weak Kleene" perspective: if any argument of a function is non-evaluable or non-normalizing, so too should be the application of the function to that argument.
+Most programming languages, including Scheme and OCaml, default to using the strict/call-by-value evaluation strategy. (But they *don't* permit evaluation to continue inside unappplied functions.) There are techniques for making these languages perform lazy/call-by-name evaluation, when necessary. But by default, arguments will always be evaluated *before* being bound to the parameters (the `\x`s) of a function.
-Most programming languages, including Scheme and OCaml, use the call-by-value evaluation strategy. (But they don't permit evaluation to continue inside an unappplied function.) There are techniques for making them model call-by-name evaluation, when necessary. But by default, arguments will always be evaluated before being bound to the parameters (the `\x`s) of a function.
+When functions take more than one argument at a time, as is the norm with Scheme, a further question arises: whether the multiple arguments should be evaluated left-to-right, or right-to-left, or is nothing guaranteed about what order they are evaluated in? Different languages make different choices about this.
-For languages like Scheme that permit functions to take more than one argument at a time, a further question arises: whether the multiple arguments are evaluated left-to-right, or right-to-left, or nothing is guaranteed about what order they are evaluated in. Different languages make different choices about this.
+Some functional programming languages, such as Haskell, default to using the lazy/call-by-name evaluation strategy. There are techniques for making Haskell perform strict/call-by-value evaluation, when necessary.
-Some functional programming languages, such as Haskell, use the call-by-name evaluation strategy.
+The Lambda Calculus can be evaluated either way. You have to decide what the rules shall be.
-The lambda calculus can be evaluated either way. You have to decide what the rules shall be.
+As we'll see later in the seminar, there are techniques by which we can specify --- in an order-independent way --- that a computation should be evaluated in call-by-value order; and also techniques for specifying call-by-name order. If you liked, you could even have a nested hierarchy, where blocks at each level were forced to be evaluated in alternating ways.
-As we'll see in several weeks, there are techniques for *forcing* call-by-value evaluation of a computation, and also techniques for forcing call-by-name evaluation. If you liked, you could even have a nested hierarchy, where blocks at each level were forced to be evaluated in alternating ways.
+Call-by-value and call-by-name evaluation have different pros and cons.
-Call-by-value and call-by-name have different pros and cons.
+One important advantage of normal-order evaluation in particular is
+that it can compute orthodox values for `(\x. y) (ω ω)`.
-One important advantage of normal-order evaluation in particular is that it can compute orthodox values for:
+To see how normal-order evaluation delivers a normal
+form, consider all conceivable evaluation paths of the following term:
-<pre><code>
-(\x. y) (ω ω)<p>
-\z. (\x. y) (ω ω)
-</code></pre>
+<pre>
+(\x.I) ((\x.xxx)(\x.xxx))
+ | \
+ | (\x.I) ((\x.xxx)(\x.xxx)(\x.xxx))
+ | / \
+ |-------+ (\x.I) ((\x.xxx)(\x.xxx)(\x.xxx)(\x.xxx))
+ | / \
+ I ------------+ etc.
+</pre>
-Indeed, it's provable that if there's *any* reduction path that delivers a value for a given expression, the normal-order evalutation strategy will terminate with that value.
+If we start by evaluating the leftmost redex, we're done in one step.
+If we unwisely start by evaluating the rightmost redex, we arrive at a
+term that still has two redexes. Once again, we have a choice: either
+reduce the leftmost redex, in which case we're done. Or reduce the
+rightmost redex, creating an even longer term. Clearly, in this
+situation, we want to prioritize reducing the leftmost redex in order
+to arrive at a stable result more quickly.
+
+Indeed, it's provable for the untyped Lambda Calculus that if there's *any* reduction path that delivers a value for a given expression, the normal-order evaluation strategy will terminate with that value.
An expression is said to be in **normal form** when it's not possible to perform any more reductions (not even inside abstracts).
-There's a sense in which you *can't get anything more out of* <code>ω ω</code>, but it's not in normal form because it still has the form of a redex.
+There's a sense in which you *can't get anything more out of* `ω ω`, but it's not in normal form because it still has the form of a redex.
A computational system is said to be **confluent**, or to have the **Church-Rosser** or **diamond** property, if, whenever there are multiple possible evaluation paths, those that terminate always terminate in the same value. In such a system, the choice of which sub-expressions to evaluate first will only matter if some of them but not others might lead down a non-terminating path.
-The untyped lambda calculus is confluent. So long as a computation terminates, it always terminates in the same way. It doesn't matter which order the sub-expressions are evaluated in.
+The diamond property is named after the shape of diagrams like the following:
-A computational system is said to be **strongly normalizing** if every permitted evaluation path is guaranteed to terminate. The untyped lambda calculus is not strongly normalizing: <code>ω ω</code> doesn't terminate by any evaluation path; and <code>(\x. y) (ω ω)</code> terminates only by some evaluation paths but not by others.
+<pre>
+ ((\x.x) ((\y.z) y))
+ / \
+ / \
+ / \
+ / \
+ ((\y.z) y) ((\x.x) z)
+ \ /
+ \ /
+ \ /
+ \ /
+ \ /
+ \ /
+ z
+</pre>
+
+Because the starting term contains more than one redex, we can imagine
+reducing the leftmost redex first (the left branch of the diagram) or
+else the rightmost redex (the right branch of the diagram). But
+because the Lambda Calculus is confluent, no matter which lambda you
+choose to target for reduction first, you end up at the same place.
+It's like traveling in (much of) Manhattan: if you walk uptown first and then
+head east, you end up in the same place as if you walk east and then
+head uptown.
+
+As we mentioned, the untyped Lambda Calculus is confluent. So long as a computation terminates, it always terminates in the same way. It doesn't matter which order the sub-expressions are evaluated in.
-But the untyped lambda calculus enjoys some compensation for this weakness. It's Turing complete! It can represent any computation we know how to describe. (That's the cash value of being Turing complete, not the rigorous definition. There is a rigrous definition. However, we don't know how to rigorously define "any computation we know how to describe.") And in fact, it's been proven that you can't have both. If a computational system is Turing complete, it cannot be strongly normalizing.
+A computational system is said to be **strongly normalizing** if *every* permitted evaluation path is guaranteed to terminate. The untyped Lambda Calculus is *not* strongly normalizing: `ω ω` doesn't terminate by any evaluation path; and `(\x. y) (ω ω)` terminates only by some evaluation paths but not by others.
-A computational system is said to be **weakly normalizing** if there's always guaranteed to be *at least one* evaluation path that terminates. The untyped lambda calculus is not weakly normalizing either, as we've seen.
+But the untyped Lambda Calculus enjoys some compensation for this weakness. It's Turing Complete! It can represent any computation we know how to describe. (That's the cash value of being Turing Complete, not the rigorous definition. There is a rigorous definition. However, we don't know how to rigorously define "any computation we know how to describe.") And in fact, it's been proven that you can't have both. If a computational system is Turing Complete, it cannot be strongly normalizing.
-The *typed* lambda calculus that linguists traditionally work with, on the other hand, is strongly normalizing. (And as a result, is not Turing complete.) It has expressive power (concerning types) that the untyped lambda calculus lacks, but it is also unable to represent some (terminating!) computations that the untyped lambda calculus can represent.
+A computational system is said to be **weakly normalizing** if there's always guaranteed to be *at least one* evaluation path that terminates. The untyped Lambda Calculus is not weakly normalizing either, as we've seen.
-Other more-powerful type systems we'll look at in the course will also fail to be Turing complete, though they will turn out to be pretty powerful.
+The *simply-typed* lambda calculus that linguists traditionally work with, on the other hand, is strongly normalizing. (And as a result, is not Turing Complete.) It has expressive power (concerning types) that the untyped Lambda Calculus does not natively possess, but it is also unable to represent some (terminating!) computations that the untyped Lambda Calculus can represent.
+
+(Interestingly, since the untyped Lambda Calculus *is* Turing Complete, any algorithmic manipulations we can do the simply-typed (or any other) lambda calculus can in principle be *encoded in* operations in the untyped Lambda Calculus. It then becomes an interesting question what we could mean, when we want to say things like "This typed lambda calculus has an expressive power not (natively) present in the untyped Lambda Calculus.")
+
+Other more-powerful type systems we'll look at in the course will also fail to be Turing Complete, though they will turn out to be pretty powerful.
Further reading:
* [[!wikipedia Strict programming language]]<p>
* [[!wikipedia Church-Rosser theorem]]
* [[!wikipedia Normalization property]]
-* [[!wikipedia Turing completeness]]
+* [[!wikipedia Turing Completeness]]
Decidability
============
-The question whether two formulas are syntactically equal is "decidable": we can construct a computation that's guaranteed to always give us the answer.
+The question whether two formulas are syntactically equal is "decidable": we can construct a computation that's guaranteed to always give us the correct answer.
What about the question whether two formulas are convertible? Well, to answer that, we just need to reduce them to normal form, if possible, and check whether the results are syntactically equal. The crux is that "if possible." Some computations can't be reduced to normal form. Their evaluation paths never terminate. And if we just kept trying blindly to reduce them, our computation of what they're convertible to would also never terminate.
So it'd be handy to have some way to check in advance whether a formula has a normal form: whether there's any evaluation path for it that terminates.
-Is it possible to do that? Sure, sometimes. For instance, check whether the formula is syntactically equal to Ω. If it is, it never terminates.
+Is it possible to do that? Sure, sometimes. For instance, check whether the formula is syntactically equal to `Ω`. If it is, it never terminates.
-But is there any method for doing this in general---for telling, of any given computation, whether that computation would terminate? Unfortunately, there is not. Church proved this in 1936; Turing also essentially proved it at the same time. Geoff Pullum gives a very reader-friendly outline of the proofs here:
+But is there any method for doing this in general---for telling, of any given computation, whether that computation would terminate? Unfortunately, there is not. Church proved this in 1936; Turing also essentially proved it at the same time. Geoff Pullum gives a very reader-friendly outline of Turing's proof here:
* [Scooping the Loop Snooper](http://www.cl.cam.ac.uk/teaching/0910/CompTheory/scooping.pdf), a proof of the undecidability of the halting problem in the style of Dr Seuss by Geoffrey K. Pullum
-Interestingly, Church also set up an association between the lambda calculus and first-order predicate logic, such that, for arbitrary lambda formulas `M` and `N`, some formula would be provable in predicate logic iff `M` and `N` were convertible. So since the right-hand side is not decidable, questions of provability in first-order predicate logic must not be decidable either. This was the first proof of the undecidability of first-order predicate logic.
-
-
-##More on evaluation strategies##
-
-Here are notes on [[evaluation order]] that make the choice of which
-lambda to reduce next the selection of a route through a network of
-links.
+Interestingly, Church also set up an association between the Lambda Calculus and first-order predicate logic, such that, for arbitrary lambda formulas `M` and `N`, some formula would be provable in predicate logic iff `M` and `N` were convertible. So since the right-hand side is not decidable, questions of provability in first-order predicate logic must not be decidable either. This was the first proof of the undecidability of first-order predicate logic.
+## Efficiency
-##More on evaluation strategies##
-
-Here (below) are notes on [[evaluation order]] that make the choice of which
-lambda to reduce next the selection of a route through a network of
-links.
-
-
-More on Evaluation Order
-========================
-
-This discussion elaborates on the discussion of evaluation order in the
-class notes from week 2. It makes use of the reduction diagrams drawn
-in class, which makes choice of evaluation strategy a choice of which
-direction to move through a network of reduction links.
-
-
-Some lambda terms can be reduced in different ways:
+Which evaluation strategy you adopt has implications not only for
+decidability and termination, but for efficiency. (Later in the
+course, it will have implications for the order in which side effects
+occur.)
<pre>
- ((\x.x)((\y.y) z))
- / \
- / \
- / \
- / \
- ((\y.y) z) ((\x.x) z)
- \ /
- \ /
- \ /
- \ /
- \ /
- \ /
- z
-</pre>
-
-But because the lambda calculus is confluent (has the diamond
-property, named after the shape of the diagram above), no matter which
-lambda you choose to target for reduction first, you end up at the
-same place. It's like travelling in Manhattan: if you walk uptown
-first and then head east, you end up in the same place as if you walk
-east and then head uptown.
-
-But which lambda you target has implications for efficiency and for
-termination. (Later in the course, it will have implications for
-the order in which side effects occur.)
-
-First, efficiency:
-
-<pre>
- ((\x.w)((\y.y) z))
- \ \
- \ ((\x.w) z)
+ ((\x.w) ((\y.z) y))
+ \ \
+ \ ((\x.w) z)
\ /
\ /
\ /
\ /
- w
-</pre>
+ w
+</pre>
-If a function discards its argument (as `\x.w` does), it makes sense
-to target that function for reduction, rather than wasting effort
-reducing the argument only to have the result of all that work thrown
-away. So in this situation, the strategy of "always reduce the
-leftmost reducible lambda" wins.
+If a function discards its argument, as `\x.w` does, it makes sense to
+prioritize redexes in which that function serves as the head, rather
+than wasting effort reducing the argument only to have the result of
+all that work thrown away. So in this situation, the strategy of
+"always reduce the leftmost reducible lambda" is more efficient.
But:
(((\y.y) z)((\y.y) z) ((\x.xx) z)
/ | /
/ (((\y.y)z)z) /
- / | /
+ / | /
/ | /
/ | /
- (z ((\y.y)z)) | /
+ (z ((\y.y)z)) | /
\ | /
-----------.---------
|
If we reduce the rightmost lambda first (rightmost branch of the
diagram), the argument is already simplified before we do the
copying. We arrive at the normal form (i.e., the form that cannot be
-further reduced) in two steps.
+further reduced) in two steps.
But if we reduce the rightmost lambda first (the two leftmost branches
of the diagram), we copy the argument before it has been evaluated.
of work we need to do to deal with that argument.
So when the function copies its argument, the "always reduce the
-rightmost reducible lambda" wins.
+rightmost reducible lambda" is more efficient.
So evaluation strategies have a strong effect on how many reduction
steps it takes to arrive at a stopping point (e.g., normal form).
-Now for termination:
+## Side-effects
-<pre>
-(\x.w)((\x.xxx)(\x.xxx))
- | \
- | (\x.w)((\x.xxx)(\x.xxx)(\x.xxx))
- | / \
- | / (\x.w)((\x.xxx)(\x.xxx)(\x.xxx)(\x.xxx))
- | / / \
- .----------------- etc.
- |
- w
-</pre>
+In the systems we're currently working in, there are no side-effects. (Though the possibility of non-terminating computations like `ω ω` has some similarities to side-effects.)
-Here we have a function that discards its argument on the left, and a
-non-terminating term on the right. It's even more evil than Omega,
-because it not only never reduces, it generates more and more work
-with each so-called reduction. If we unluckily adopt the "always
-reduce the rightmost reducible lambda" strategy, we'll spend our days
-relentlessly copying out new copies of \x.xxx. But if we even once
-reduce the leftmost reducible lambda, we'll immediately jump to the
-normal form, `w`.
-
-We can roughly associate the "leftmost always" strategy with call by
-name (normal order), and the "rightmost always" strategy with call by
-value (applicative order). There are fine-grained distinctions among
-these terms of art that we will not pause to untangle here.
-
-If a term has a normal form (a reduction such that no further
-reduction is possible), then a leftmost-always reduction will find it.
-(That's why it's called "normal order": it's the evaluation order that
-guarantees finding a normal form if one exists.)
-
-Preview: if the evaluation of a function has side effects, then the
-choice of an evaluation strategy will make a big difference in which
-side effect occur and in which order.
+When we later consider systems where the evaluation of some expressions *does* have side-effects, then the choice of evaluation strategy will make a big difference to which side-effects occur, and in what order.