Scheme (functional part)  OCaml (functional part)  C, Java, Pasval +Scheme (imperative part) +OCaml (imperative part) 
lambda calculus +combinatorial logic  
 Turing complete   
+  more advanced type systems, such as polymorphic types +  + 
+  simplytyped lambda calculus (what linguists mostly use) +  + 
**Variables**: `x`, `y`, `z`, ... +Variables:Each variable is an expression. For any expressions M and N and variable a, the following are also expressions:x
,y
,z
...
**Abstract**:We'll tend to write(λa M)
+Abstract:(λa M)
(λa M)
as just `( \a M )`.
+We'll tend to write (λa M)
as just `(\a M)`, so we don't have to write out the markup code for the λ
. You can yourself write (λa M)
or `(\a M)` or `(lambda a M)`.
**Application**: `(M N)`
+Application: (M N)
Some authors reserve the term "term" for just variables and abstracts. We won't participate in that convention; we'll probably just say "term" and "expression" indiscriminately for expressions of any of these three forms.
@@ 142,7 +166,7 @@ Examples of expressions:
(x (\x x))
((\x (x x)) (\x (x x)))
The lambda calculus has an associated proof theory. For now, we can regard the proof theory as having just one rule, called the rule of "betareduction" or "betacontraction". Suppose you have some expression of the form:
+The lambda calculus has an associated proof theory. For now, we can regard the proof theory as having just one rule, called the rule of **betareduction** or "betacontraction". Suppose you have some expression of the form:
((\a M) N)
@@ 150,7 +174,7 @@ that is, an application of an abstract to some other expression. This compound f
The rule of betareduction permits a transition from that expression to the following:
 M {a:=N}
+ M [a:=N]
What this means is just `M`, with any *free occurrences* inside `M` of the variable `a` replaced with the term `N`.
@@ 167,7 +191,7 @@ For instance:
> T is defined to be `(x (\x (\y (x (y z)))))`
The first occurrence of `x` in `T` is free. The `\x` we won't regard as being an occurrence of `x`. The next occurrence of `x` occurs within a form that begins with `\x`, so it is bound as well. The occurrence of `y` is bound; and the occurrence of `z` is free.
+The first occurrence of `x` in T is free. The `\x` we won't regard as being an occurrence of `x`. The next occurrence of `x` occurs within a form that begins with `\x`, so it is bound as well. The occurrence of `y` is bound; and the occurrence of `z` is free.
Here's an example of betareduction:
@@ 185,33 +209,33 @@ Different authors use different notations. Some authors use the term "contractio
M ~~> N
We'll mean that you can get from M to N by one or more reduction steps. Hankin uses the symbol > for onestep contraction, and the symbol >> for zeroormore step reduction. Hindley and Seldin use (triangle..sub1) and (triangle).
+We'll mean that you can get from M to N by one or more reduction steps. Hankin uses the symbol →
for onestep contraction, and the symbol ↠
for zeroormore step reduction. Hindley and Seldin use ⊳_{1}
and ⊳
.
When M and N are such that there's some P that M reduces to by zero or more steps, and that N also reduces to by zero or more steps, then we say that M and N are **betaconvertible**. We'll write that like this:
M <~~> N
This is what plays the role of equality in the lambda calculus. Hankin uses the symbol `=` for this. So too do Hindley and Seldin.
+This is what plays the role of equality in the lambda calculus. Hankin uses the symbol `=` for this. So too do Hindley and Seldin. Personally, I keep confusing that with the relation to be described next, so let's use this notation instead. Note that `M <~~> N` doesn't mean that each of `M` and `N` are reducible to each other; that only holds when `M` and `N` are the same expression. (Or, with our convention of only saying "reducible" for one or more reduction steps, it never holds.)
In the metatheory, it's also sometimes useful to talk about formulas that are syntactically equivalent *before any reductions take place*. Hankin uses the symbol (three bars) for this. So too do Hindley and Seldin. We'll use that too, and will avoid using `=` when discussing metatheory for the lambda calculus. Instead we'll use `<~~>` as we said above. When we want to introduce a stipulative definition, we'll write it out longhand, as in:
+In the metatheory, it's also sometimes useful to talk about formulas that are syntactically equivalent *before any reductions take place*. Hankin uses the symbol ≡
for this. So too do Hindley and Seldin. We'll use that too, and will avoid using `=` when discussing metatheory for the lambda calculus. Instead we'll use `<~~>` as we said above. When we want to introduce a stipulative definition, we'll write it out longhand, as in:
> T is defined to be `(M N)`.
We'll regard the following two expressions:
 (\x x y)
+ (\x (x y))
 (\z z y)
+ (\z (z y))
as syntactically equivalent, since they only involve a typographic change of a bound variable. Read Hankin section 2.3 for discussion of different attitudes one can take about this.
Note that neither of those expressions are identical to:
 (\x x w)
+ (\x (x w))
because here it's a free variable that's been changed. Nor are they identical to:
 (\y y y)
+ (\y (y y))
because here the second occurrence of `y` is no longer free.
@@ 224,11 +248,11 @@ Shorthand
The grammar we gave for the lambda calculus leads to some verbosity. There are several informal conventions in widespread use, which enable the language to be written more compactly. (If you like, you could instead articulate a formal grammar which incorporates these additional conventions. Instead of showing it to you, we'll leave it as an exercise for those so inclined.)
Dot notation: dot means "put a left paren here, and put the right
+**Dot notation** Dot means "put a left paren here, and put the right
paren as far the right as possible without creating unbalanced
parentheses". So:
 (\x (\y (xy)))
+ (\x (\y (x y)))
can be abbreviated as:
@@ 236,23 +260,23 @@ can be abbreviated as:
and:
 (\x \y. (z y) z)
+ (\x (\y. (z y) z))
would abbreviate:
 (\x \y ((z y) z))
+ (\x (\y ((z y) z)))
This on the other hand:
 ((\x \y. (z y) z)
+ (\x (\y. z y) z)
would abbreviate:
 ((\x (\y (z y))) z)
+ (\x (\y (z y)) z)
Parentheses: outermost parentheses around applications can be dropped. Moreover, applications will associate to the left, so `M N P` will be understood as `((M N) P)`. Finally, you can drop parentheses around abstracts, but not when they're part of an application. So you can abbreviate:
+**Parentheses** Outermost parentheses around applications can be dropped. Moreover, applications will associate to the left, so `M N P` will be understood as `((M N) P)`. Finally, you can drop parentheses around abstracts, but not when they're part of an application. So you can abbreviate:
 (\x x y)
+ (\x. x y)
as:
@@ 266,7 +290,7 @@ and:
z (\x. x y)
Merging lambdas: an expression of the form `(\x (\y M))`, or equivalently, `(\x. \y. M)`, can be abbreviated as:
+**Merging lambdas** An expression of the form `(\x (\y M))`, or equivalently, `(\x. \y. M)`, can be abbreviated as:
(\x y. M)
@@ 281,14 +305,14 @@ Lambda terms represent functions
All (recursively computable) functions can be represented by lambda
terms (the untyped lambda calculus is Turing complete). For some lambda terms, it is easy to see what function they represent:
(\x x) represents the identity function: given any argument M, this function
simply returns M: ((\x x) M) ~~> M.
+> `(\x x)` represents the identity function: given any argument `M`, this function
+simply returns `M`: `((\x x) M) ~~> M`.
(\x (x x)) duplicates its argument:
((\x (x x)) M) ~~> (M M)
+> `(\x (x x))` duplicates its argument:
+`((\x (x x)) M) ~~> (M M)`
(\x (\y x)) throws away its second argument:
(((\x (\y x)) M) N) ~~> M
+> `(\x (\y x))` throws away its second argument:
+`(((\x (\y x)) M) N) ~~> M`
and so on.
@@ 309,13 +333,11 @@ both represent the same function, the identity function. However, we said above
(\z z)
yet when applied to any argument M, all of these will always return M. So they have the same extension. It's also true, though you may not yet be in a position to see, that no other argument can differentiate between them when they're supplied as an argument to it. However, these expressions are all syntactically distinct.
+yet when applied to any argument M, all of these will always return M. So they have the same extension. It's also true, though you may not yet be in a position to see, that no other function can differentiate between them when they're supplied as an argument to it. However, these expressions are all syntactically distinct.
The first two expressions are *convertible*: in particular the first reduces to the second. So they can be regarded as prooftheoretically equivalent even though they're not syntactically identical. However, the proof theory we've given so far doesn't permit you to reduce the second expression to the third. So these lambda expressions are nonequivalent.
There's an extension of the prooftheory we've presented so far which does permit this further move. And in that extended proof theory, all computable functions with the same extension do turn out to be equivalent (convertible). However, at that point, we still won't be working with the traditional mathematical notion of a function as a set of ordered pairs. One reason is that the latter but not the former permits uncomputable functions. A second reason is that the latter but not the former prohibits functions from applying to themselves. We discussed this some at the end of seminar (and further discussion is best pursued in person).


+There's an extension of the prooftheory we've presented so far which does permit this further move. And in that extended proof theory, all computable functions with the same extension do turn out to be equivalent (convertible). However, at that point, we still won't be working with the traditional mathematical notion of a function as a set of ordered pairs. One reason is that the latter but not the former permits uncomputable functions. A second reason is that the latter but not the former prohibits functions from applying to themselves. We discussed this some at the end of Monday's meeting (and further discussion is best pursued in person).
@@ 325,7 +347,11 @@ Booleans and pairs
Our definition of these is reviewed in [[Assignment1]].

+It's possible to do the assignment without using a Scheme interpreter, however
+you should take this opportunity to [get Scheme installed on your
+computer](/how_to_get_the_programming_languages_running_on_your_computer), and
+[get started learning Scheme](/learning_scheme). It will help you test out
+proposed answers to the assignment.