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.
(\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 formula 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.