Define T to be `(\x. x y) z`. Then T and `(\x. x y) z` are syntactically equal, and we're counting them as syntactically equal to `(\z. z y) z` as well, which we will write as:
-<pre>
-T ≡ `(\x. x y) z` ≡ `(\z. z y) z`
-</pre>
+<pre><code>T ≡ (\x. x y) z ≡ (\z. z y) z
+</code></pre>
This:
-<pre>
-T ~~> `z y`
-</pre>
+ T ~~> z y
means that T beta-reduces to `z y`. This:
-<pre>
-M <~~> T
-</pre>
+ M <~~> T
means that M and T are beta-convertible, that is, that there's something they both reduce to in zero or more steps.
Lambda expressions that have no free variables are known as **combinators**. Here are some common ones:
-<pre>
-**I** is defined to be `\x x`<p>
-**K** is defined to be `\x y. x`, That is, it throws away its second argument. So `K x` is a constant function from any (further) argument to `x`. ("K" for "constant".) Compare K to our definition of **true**.<p>
-**get-first** was our function for extracting the first element of an ordered pair: `\fst snd. fst`. Compare this to K and true as well.<p>
-**get-second** was our function for extracting the second element of an ordered pair: `\fst snd. snd`. Compare this to our definition of false.<p>
-**ω** is defined to be: `\x. x x`<p>
-</pre>
+> **I** is defined to be `\x x`
+
+> **K** is defined to be `\x y. x`, That is, it throws away its second argument. So `K x` is a constant function from any (further) argument to `x`. ("K" for "constant".) Compare K to our definition of **true**.
+
+> **get-first** was our function for extracting the first element of an ordered pair: `\fst snd. fst`. Compare this to **K** and **true** as well.
+
+> **get-second** was our function for extracting the second element of an ordered pair: `\fst snd. snd`. Compare this to our definition of **false**.
+
+> **ω** is defined to be: `\x. x x`
It's possible to build a logical system equally powerful as the lambda calculus (and readily intertranslatable with it) using just combinators, considered as atomic operations. Such a language doesn't have any variables in it: not just no free variables, but no variables at all.
-One can do that with a very spare set of basic combinators. These days the standard base is just three combinators: K and I from above, and also one more, S, which behaves the same as the lambda expression `\f g x. f x (g x)`. behaves. But it's possible to be even more minimalistic, and get by with only a single combinator. (And there are different single-combinator bases you can choose.)
+One can do that with a very spare set of basic combinators. These days the standard base is just three combinators: K and I from above, and also one more, **S**, which behaves the same as the lambda expression `\f g x. f x (g x)`. behaves. But it's possible to be even more minimalistic, and get by with only a single combinator. (And there are different single-combinator bases you can choose.)
These systems are Turing complete. In other words: every computation we know how to describe can be represented in a logical system consisting of only a single primitive operation!
Here's more to read about combinatorial logic:
-* [[!wikipedia Combinatory logic]]
+* [[!wikipedia Combinatory logic]] at Wikipedia
* [Combinatory logic](http://plato.stanford.edu/entries/logic-combinatory/) at the Stanford Encyclopedia of Philosophy
* [[!wikipedia SKI combinatory calculus]]
* [[!wikipedia B,C,K,W system]]
(\x. x x) (\x. x x)
-As we saw above, each of the halves of this formula are the combinator ω; so this can also be written:
+As we saw above, each of the halves of this formula are the combinator <code>ω</code>; so this can also be written:
<pre><code>ω ω</code></pre>
-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 <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 <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.)
-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 formula a special value, <code>⊥</code>, 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 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.
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.
(\x. y) (ω ω)
</code></pre>
-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 <code>(ω ω)</code> 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:
> \x. M x
-> where x does not occur free in `M`, to `M`? 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.
+> 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.
MORE on extensionality
/ \ / \
A B C D
-Applicative order evaluation does what's called a "post-order traversal" of the tree: that is, we always go left and down whenever we can, 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.
+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.
(\x. y) (C D)
-the computation of `(C D)` won't ever have to be performed. Instead, it 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 <code>(ω ω)</code>!
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.
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.
-An expression is said to be in **normal form** when it's not possible to perform any more reductions. (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.
+An expression is said to be in **normal form** when it's not possible to perform any more reductions. (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.
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.
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:
-##[[Lists and Numbers]]##
+* [[!wikipedia Evaluation strategy]]
+* [[!wikipedia Eager evaluation]]
+* [[!wikipedia Lazy evaluation]]
+* [[!wikipedia Strict programming language]]<p>
+* [[!wikipedia Church-Rosser theorem]]
+* [[!wikipedia Normalization property]]
+* [[!wikipedia Turing completeness]]
+
+
+Decidability
+============
-How to do with recursion with omega.
+The question whether two formulas are syntactically equal is "decidable": we can construct a computation that's guaranteed to always give us the 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.
+
+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:
+
+* [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
+
+
+
+##[[Lists and Numbers]]##
-Next week: fixed point combinators