* and false
has a different intepretation than false and *
. (The same phenomenon is already present with the material conditional in bivalent logics; but seeing that a nonsymmetric semantics for `and` is available even for functional languages is instructive.)
+Another set of operations we have are:
Another way in which order can matter that's present even in functional languages is that the interpretation of some complex expressions can depend on the order in which subexpressions are evaluated. Evaluated in one order, the computations might never terminate (and so semantically we interpret them as having "the bottom value"we'll discuss this). Evaluated in another order, they might have a perfectly mundane value. Here's an example, though we'll reserve discussion of it until later:
+ and, or, not
 (\x. y) ((\x. x x) (\x. x x))
+The first two of these are infix functions that expect two boolean arguments, and gives a boolean result. The third is a function that expects only one boolean argument. Our earlier function `!=` means "doesn't equal", and:
Again, these facts are all part of the metatheory of purely functional languages. But *there is* a different sense of "order matters" such that it's only in imperatival languages that order so matters.
+ x != y
 x := 2
 x := x + 1
 x == 3
+will be just another way to write:
Here the comparison in the last line will evaluate to true.
+ not (x == y)
 x := x + 1
 x := 2
 x == 3
+You see that you can use parentheses in the standard way. By the way, `<=` means ≤ or "less than or equals to", and `>=` means ≥. Just in case you haven't seen them written this way before.
Here the comparison in the last line will evaluate to false.
+I've started throwing in some **variables**. We'll say variables are any expression that's written with an initial lowercase letter, then is followed by a sequence of zero or more upper or lowercase letters, or numerals, or underscores (`_`). Then at the end you can optionally have a `?` or `!` or a sequence of `'`s, understood as "primes." Hence, all of these are legal variables:
One of our goals for this course is to get you to understand *what is* that new
sense such that only so matters in imperatival languages.
+ x
+ x1
+ x_not_y
+ xUBERANT
+ x'
+ x''
+ x?
+ xs
Finally, you'll see the term **dynamic** used in a variety of ways in the literature for this course:
+We'll follow a *convention* of using variables with short names and a final `s` to represent collections like sequences (to be discussed below). But this is just a convention to help us remember what we're up to, not a strict rule of the language. We'll also follow a convention of only using variables ending in `?` to represent functions that return a boolean value. Thus, for example, `zero?` will be a function that expects a single number argument and returns a boolean corresponding to whether that number is `0`. `odd?` will be a function that expects a single number argument and returns a boolean corresponding to whether than number is odd. Above, I suggested we might use `lessthan?` to represent a function that expects *two* number arguments, and again returns a boolean result.
* dynamic versus static typing
+We also conventionally reserve variables ending in `!` for a different special class of functions, that we will explain later in the course.
* dynamic versus lexical scoping
+In fact you can think of `succ` and `pred` and `not` and the rest as also being variables; it's just that these variables have been predefined in our language to be bound to functions we agreed upon in advance. You can even think of `==` and `<` as being variables, too, bound to other functions. But I haven't given you parsing rules yet which would make them legal variables, because they don't start with a lowercase letter. We can make the parsing rules more liberal later.
* dynamic versus static control operators
+Only a few simple expressions in our language aren't variables. These include the literal values, and also **keywords** like `let` and `case` and so on that we'll discuss below. You can't use `let` as a variable, else the syntax of our language would become too hard to mechanically parse. (And probably too hard for our meager brains to parse, too.)
* finally, we're used ourselves to talking about dynamic versus static semantics
+The rule for symbolic atoms is that a single quote `'` followed by any single word that could be a legal variable expresses such an atom, a different atom for each different expression.
+Thus `'false` is a symbolic atom, but so too are `'x` and `'succ`. For the time being, I'll restrict myself to only talking about the symbolic atoms `'true` and `'false`. These constitute a special subclass of symbolic atoms that we call the **booleans** or truthvalues. Nothing deep hangs on them being a subclass of a larger type in this way; it just seems elegant. Some other languages make booleans their own special type, not a subclass of another type. Others make them a subclass of the numbers (yuck). We will think of them this way.
For the most part, these uses are only loosely connected to each other. We'll tend to use "imperatival" to describe the kinds of semantic properties made available in dynamic semantics, languages which have robust notions of sequencing changes, and so on.
+Note that when writing a symbolic atom there is no closing `'`, just a `'` at the beginning. That's enough to make the whole word, up to the next space (or whatever) count as expressing a symbolic atom. We use the initial `'` to make it easy for us to have a rich set of symbolic atoms, as well as a rich set of variables, without getting them mixed up. Variables never begin with `'`; symbolic atoms always do.
Map
===
+We call these things symbolic *atoms* because they aren't collections. Thus numbers are also atoms, but not symbolic ones. And functions are also atoms, but again, not symbolic ones.











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)    
 ∀x. (F x or ∀x (not (F x)))

+which evaluates `f (10, 1)` to `(10, 11, 12, 13)`, which it matches against the complex pattern `(a, b, c, d)`, binding all four of the contained variables, and then evaluates `(b, d)` under those bindings, giving us the result `(11, 13)`.
 When a previouslybound variable is rebound in the way we see here, that's called **shadowing**: the outer binding is shadowed during the scope of the inner binding.
+Notice that in the preceding expression, the variables `a` and `c` were never used. So the values they're bound to are ignored or discarded. We're allowed to do that, but there's also a special syntax to indicate that this is what we're up to. This uses the special pattern `_`:
+`let`
+ `f match` λ`(x, y). (x, x + y, x + 2*y, x + 3*y);`
+ `(_, b, _, d) match f (10, 1)`
+`in (b, d)`
Some more comparisons between Scheme and OCaml

+The role of `_` here is just to occupy a slot in the complex pattern `(_, b, _, d)`, to make it a multivalue of four values, rather than one of only two.
11. Simple predefined values
+One last wrinkle. What if you tried to make a pattern like this: `[x, x]`, where some variable occurs multiple times. This is known as a "nonlinear pattern". Some languages permit these (and require that the values being bound against `x` in the two positions be equal). Many languages don't permit it. Let's agree not to do this.
 Numbers in Scheme: `2`, `3`
 In OCaml: `2`, `3`
 Booleans in Scheme: `#t`, `#f`
 In OCaml: `true`, `false`
+### Case and if ... then ... else ... ###
 The eighth letter in the Latin alphabet, in Scheme: `#\h`
 In OCaml: `'h'`
+In class we introduced this form of complex expression:
12. Compound values
+`if` φ `then` ψ `else` χ
 These are values which are built up out of (zero or more) simple values.
+Here φ should evaluate to a boolean, and ψ and χ should evaluate to the same type. The result of the whole expression will be the same as ψ, if φ evaluates to `'true`, else to the result of χ.
 Ordered pairs in Scheme: `'(2 . 3)`
 In OCaml: `(2, 3)`
+We said that that could be taken as shorthand for the following `case`expression:
 Lists in Scheme: `'(2 3)`
 In OCaml: `[2; 3]`
 We'll be explaining the difference between pairs and lists next week.
+`case` φ `of`
+ `'true then` ψ`;`
+ `'false then` χ
+`end`
 The empty list, in Scheme: `'()`
 In OCaml: `[]`
+The `case`expression has a list of patterns and expressions. Its initial expression φ is evaluated and then attempted to be matched against each of the patterns in turn. When we reach a pattern that can be matchedthat doesn't result in a matchfailurethen we evaluate the expression after the `then`, using any variable bindings in effect from the immediately preceding match. (Any match that fails has no effect on future variable bindings. In this example, there are no variables in our patterns, so it's irrelevant.) What that righthand expression evaluates to becomes the result of the whole `case`expression. We don't attempt to do any further patternmatching after finding a pattern that succeeds.
 The string consisting just of the eighth letter of the Latin alphabet, in Scheme: `"h"`
 In OCaml: `"h"`
+If a `case`expression gets to the end of its list of patterns, and *none* of them have matched its initial expression, the result is a patternmatching failure. So it's good style to always include a final pattern that's guaranteed to match anything. You could use a simple variable for this, or the special pattern `_`:
 A longer string, in Scheme: `"horse"`
 In OCaml: `"horse"`
+ case 4 of
+ 1 then 'true;
+ 2 then 'true;
+ x then 'false
+ end
 A shorter string, in Scheme: `""`
 In OCaml: `""`
+ case 4 of
+ 1 then 'true;
+ 2 then 'true;
+ _ then 'false
+ end
13. Function application
+will both evaluate to `'false`, without any patternmatching failure.
 Binary functions in OCaml: `foo 2 3`

 Or: `( + ) 2 3`
+There's a superficial similarity between the `let`constructions and the `case`constructions. Each has a list whose lefthand sides are patterns and righthand sides are expressions. Each also has an additional expression that stands out in a special position: in `let`expressions at the end, in `case`expressions at the beginning. But the relations of these different elements to each other is different. In `let`expressions, the righthand sides of the list supply the values that get bound to the variables in the patterns on the lefthand sides. Also, each pattern in the list will get matched, unless there's a patternmatch failure before we get to it. In `case`expressions, on the other hand, it's the initial expression that supplies the value (or multivalues) that we attempt to match against the pattern, and we stop as soon as we reach a pattern that we can successfully match against. Then the variables in that pattern are thereby bound when evaluating the corresponding righthand side expression.
 These are the same as: `((foo 2) 3)`. In other words, functions in OCaml are "curried". `foo 2` returns a `2`fooer, which waits for an argument like `3` and then foos `2` to it. `( + ) 2` returns a `2`adder, which waits for an argument like `3` and then adds `2` to it.
 In Scheme, on the other hand, there's a difference between `((foo 2) 3)` and `(foo 2 3)`. Scheme distinguishes between unary functions that return unary functions and binary functions. For our seminar purposes, it will be easiest if you confine yourself to unary functions in Scheme as much as possible.
+### Recursive let ###
 Additionally, as said above, Scheme is very sensitive to parentheses and whenever you want a function applied to any number of arguments, you need to wrap the function and its arguments in a parentheses.
+Given all these tools, we're (almost) in a position to define functions like the `factorial` and `length` functions we defined in class.
+Here's an attempt to define the `factorial` function:
What "sequencing" is and isn't

+`let`
+ `factorial match` λ `n. if n == 0 then 1 else n * factorial (n1)`
+`in factorial`
We mentioned before the idea that computation is a sequencing of some changes. I said we'd be discussing (fragments of, and in some cases, entire) languages that have no native notion of change.
+or, using `case`:
Neither do they have any useful notion of sequencing. But what this would be takes some care to identify.
+`let`
+ `factorial match` λ `n. case n of 0 then 1; _ then n * factorial (n  1) end`
+`in factorial`
First off, the mere concatenation of expressions isn't what we mean by sequencing. Concatenation of expressions is how you build syntactically complex expressions out of simpler ones. The complex expressions often express a computation where a function is applied to one (or more) arguments,
+But there's a problem here. What value does `factorial` have when evaluating the expression `factorial (n  1)`?
Second, the kind of rebinding we called "shadowing" doesn't involve any changes or sequencing. All the precedence facts about that kind of rebinding are just consequences of the compound syntactic structures in which it occurs.
+As we said in class, the natural precedent for this with nonfunction variables would go something like this:
Third, the kinds of bindings we see in:
+ let
+ x match 0;
+ y match x + 1;
+ x match x + 1;
+ z match 2 * x
+ in (y, z)
 (define foo A)
 (foo 2)
+We'd expect this to evaluate to `(1, 2)`, and indeed it does. That's because the `x` in the `x + 1` on the righthand side of the third binding (`x match x + 1`) is evaluated under the scope of the first binding, of `x` to `0`.
Or even:
+We should expect the `factorial` variable in the righthand side of our attempted definition to behave the same way. It will evaluate to whatever value it has before reaching this `let`expression. We actually haven't said what is the result of trying to evaluate unbound variables, as in:
 (define foo A)
 (define foo B)
 (foo 2)
+ let
+ x match y + 0
+ in x
don't involve any changes or sequencing in the sense we're trying to identify. As we said, these programs are just syntactic variants of (single) compound syntactic structures involving "let"s and "lambda"s.
+Let's agree not to do that. We can consider such expressions only under the implied understanding that they are parts of larger expressions that assign a value to `y`, as for example in:
Since Scheme and OCaml also do permit imperatival constructions, they do have syntax for genuine sequencing. In Scheme it looks like this:
+ let
+ y match 1
+ in let
+ x match y + 0
+ in x
 (begin A B C)
+Hence, let's understand our attempted definition of `factorial` to be part of such a larger expression:
In OCaml it looks like this:
+`let`
+ `factorial match` λ `n. n`
+`in let`
+ `factorial match` λ `n. case n of 0 then 1; _ then n * factorial (n  1) end`
+`in factorial 4`
 begin A; B; C end
+This would evaluate to what `4 * factorial 3` does, but with the `factorial` in the expression bound to the identity function λ `n. n`. In other words, we'd get the result `12`, not the correct answer `24`.
Or this:
+For the time being, we will fix this solution by just introducing a special new construction `letrec` that works the way we want. Now in:
 (A; B; C)
+`let`
+ `factorial match` λ `n. n`
+`in letrec`
+ `factorial match` λ `n. case n of 0 then 1; _ then n * factorial (n  1) end`
+`in factorial 4`
In the presence of imperatival elements, sequencing order is very relevant. For example, these will behave differently:
+the initial binding of `factorial` to the identity function gets ignored, and the `factorial` in the righthand side of our definition is interpreted to mean the very same function that we are hereby binding to `factorial`. Exactly how this works is a deep and exciting topic, that we will be looking at very closely in a few weeks. For the time being, let's just accept that `letrec` does what we intuitively want when defining functions recursively.
 (begin (print "under") (print "water"))

 (begin (print "water") (print "under"))
+**It's important to make sure you say letrec when that's what you want.** You may not *always* want `letrec`, though, if you're ever reusing variables (or doing other things) that rely on the bindings occurring in a specified order. With `letrec`, all the bindings in the construction happen simultaneously. This is why you can say, as Jim did in class:
And so too these:
+`letrec`
+ `even? match` λ `n. case n of 0 then 'true; _ then odd? (n1) end`
+ `odd? match` λ `n. case n of 0 then 'false; _ then even? (n1) end`
+`in (even?, odd?)`
 begin x := 3; x := 2; x end
+Here neither the `even?` nor the `odd?` pattern is matched before the other. They, and also the `odd?` and the `even?` variables in their righthand side expressions, are all bound at once.
 begin x := 2; x := 3; x end
+As we said, this is deep and exciting, and it will make your head spin before we're done examining it. But let's trust `letrec` to do its job, for now.
However, if A and B are purely functional, nonimperatival expressions, then:
 begin A; B; C end
+### Comparing recursivestyle and iterativestyle definitions ###
just evaluates to C (so long as A and B evaluate to something at all). So:
+Finally, we're in a position to revisit the two definitions of `length` that Jim presented in class. Here is the first:
 begin A; B; C end
+`letrec`
+ `length match` λ `xs. case xs of [] then 0; _:ys then 1 + length ys end`
+`in length`
contributes no more to a larger context in which it's embedded than C does. This is the sense in which functional languages have no serious notion of sequencing.
+This function accept a sequence `xs`, and if its empty returns `0`, else it says that its length is `1` plus whatever is the length of its remainder when you take away the first element. In programming circles, this remainder is commonly called the sequence's "tail" (and the first element is its "head").
We'll discuss this more as the seminar proceeds.
+Thus if we evaluated `length [10, 20, 30]`, that would give the same result as `1 + length [20, 30]`, which would give the same result as `1 + (1 + length [30])`, which would give the same result as `1 + (1 + (1 + length []))`. But `length []` is `0`, so our original expression evaluates to `1 + (1 + (1 + 0))`, or `3`.
+Here's another way to define the `length` function:
+`letrec`
+ `aux match` λ `(n, xs). case xs of [] then n; _:ys then aux (n + 1, ys) end`
+`in` λ `xs. aux (0, xs)`
Basics of Lambda Calculus
=========================
+This may be a bit confusing. What we have here is a helper function `aux` (for "auxiliary") that accepts *two* arguments, the first being a counter of how long we've counted in the sequence so far, and the second argument being how much more of the sequence we have to inspect. If the sequence we have to inspect is empty, then we're finished and we can just return our counter. (Note that we don't return `0`.) If not, then we add `1` to the counter, and proceed to inspect the tail of the sequence, ignoring the sequence's first element. After the `in`, we can't just return the `aux` function, because it expects two arguments, whereas `length` should just be a function of a single argument, the sequence whose length we're inquiring about. What we do instead is return a λgenerated function, that expects a single sequence argument `xs`, and then returns the result of calling `aux` with that sequence together with an initial counter of `0`.
The lambda calculus we'll be focusing on for the first part of the course has no types. (Some prefer to say it instead has a single typebut if you say that, you have to say that functions from this type to this type also belong to this type. Which is weird.)
+So for example, if we evaluated `length [10, 20, 30]`, that would give the same result as `aux (0, [10, 20, 30])`, which would give the same result as `aux (1, [20, 30])`, which would give the same result as `aux (2, [30])`, which would give the same result as `aux(3, [])`, which would give `3`. (This should make it clear why when `aux` is called with the empty sequence, it returns the result `n` rather than `0`.)
Here is its syntax:
+Programmers will sometimes define functions in the second style because it can be evaluated more efficiently than the first style. You don't need to worry about things like efficiency in this seminar. But you should become acquainted with, and comfortable with, both styles of recursive definition.
Variables:+It may be helpful to contrast these recursivestyle definitons to the way one would more naturally define the `length` function in an imperatival language. This uses some constructs we haven't explained yet, but I trust their meaning will be intuitively clear enough. Each variable is an expression. For any expressions M and N and variable a, the following are also expressions: +`let` + `empty? match` λ `xs.` *this definition left as an exercise*; + `tail match` λ `xs.` *this definition left as an exercise*; + `length match` λ `xs. let` + `n := 0;` + `while not (empty? xs) do` + `n := n + 1;` + `xs := tail xs` + `end` + `in n` +`in length` x
,y
,z
... 
Abstract: (λa M)

+Here there is no recursion. Rather what happens is that we *initialize* the variable `n` with the value `0`, and then so long as our sequence variable `xs` is nonempty, we *increment* that variable `n`, and *overwrite* the variable `xs` with the tail of the sequence that it is then bound to, and repeat in a loop (the `while ... do ... end` construction). This is similar to what happens in our second definition of `length`, using `aux`, but here it happens using *mutation* or *overwriting* the values of variables, and a special looping construction, whereas in the preceding definitions we achieved the same effect instead with recursion.
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)`.
+We will be looking closely at mutation later in the term. For the time being, our focus will instead be on the recursive and *immutable* style of doing thingsmeaning no variables get overwritten.

Application: (M N)

+It's helpful to observe that in expressions like:
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.
+ let
+ x match 0;
+ y match x + 1;
+ x match x + 1;
+ z match 2 * x
+ in (y, z)
Examples of expressions:
+the variable `x` has not been *overwritten* (mutated). Rather, we have *two* variables `x` and its just that the second one is *hiding* the first so long as its scope is in effect. Once its scope expires, the original variable `x` is still in place, with its orginal binding. A different example should help clarify this. What do you think this:
 x
 (y x)
 (x x)
 (\x y)
 (\x x)
 (\x (\y x))
 (x (\x x))
 ((\x (x x)) (\x (x x)))
+ let
+ x match 0;
+ (y, z) match let
+ x match x + 1
+ in (x, 2*x)
+ in ([y, z], 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:
+evaluates to? Well, consider the righthand side of the second binding:
 ((\a M) N)
+ let
+ x match x + 1
+ in (x, 2*x)
that is, an application of an abstract to some other expression. This compound form is called a **redex**, meaning it's a "betareducible expression." `(\a M)` is called the **head** of the redex; `N` is called the **argument**, and `M` is called the **body**.
+This expression evaluates to `(1, 2)`, because it uses the outer binding of `x` to `0` for the righthand side of its own binding `x match x + 1`. That gives us a new binding of `x` to `1`, which is in place when we evaluate `(x, 2*x)`. That's why the whole thing evaluates to `(1, 2)`. So now returning to the outer expression, `y` gets bound to `1` and `z` to `2`. But now what is `x` bound to in the final line,`([y, z], x)`? The binding of `x` to `1` was in place only until we got to `(x, 2*x)`. After that its scope expired, and the original binding of `x` to `0` reappears. So the final line evaluates to `([1, 2], 0)`.
The rule of betareduction permits a transition from that expression to the following:
+This is very like what happens in ordinary predicate logic if you say:
 M [a:=N]
+∃ `x. F x and (` ∀ `x. G x ) and H x`
What this means is just `M`, with any *free occurrences* inside `M` of the variable `a` replaced with the term `N`.

What is a free occurrence?

> An occurrence of a variable `a` is **bound** in T if T has the form `(\a N)`.

> If T has the form `(M N)`, any occurrences of `a` that are bound in `M` are also bound in T, and so too any occurrences of `a` that are bound in `N`.

> An occurrence of a variable is **free** if it's not bound.

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.

Here's an example of betareduction:

 ((\x (y x)) z)

betareduces to:

 (y z)

We'll write that like this:

 ((\x (y x)) z) ~~> (y z)

Different authors use different notations. Some authors use the term "contraction" for a single reduction step, and reserve the term "reduction" for the reflexive transitive closure of that, that is, for zero or more reduction steps. Informally, it seems easiest to us to say "reduction" for one or more reduction steps. So when we write:

 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 ⊳_{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. 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 ≡
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))

 (\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))

because here it's a free variable that's been changed. Nor are they identical to:

 (\y (y y))

because here the second occurrence of `y` is no longer free.

There is plenty of discussion of this, and the fine points of how substitution works, in Hankin and in various of the tutorials we've linked to about the lambda calculus. We expect you have a good intuitive understanding of what to do already, though, even if you're not able to articulate it rigorously.


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
paren as far the right as possible without creating unbalanced
parentheses". So:

 (\x (\y (x y)))

can be abbreviated as:

 (\x (\y. x y))

and:

 (\x (\y. (z y) z))

would abbreviate:

 (\x (\y ((z y) z)))

This on the other hand:

 (\x (\y. z y) z)

would abbreviate:

 (\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:

 (\x. x y)

as:

 \x. x y

but you should include the parentheses in:

 (\x. x y) z

and:

 z (\x. x y)

**Merging lambdas** An expression of the form `(\x (\y M))`, or equivalently, `(\x. \y. M)`, can be abbreviated as:

 (\x y. M)

Similarly, `(\x (\y (\z M)))` can be abbreviated as:

 (\x y z. M)


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 x))` duplicates its argument:
`((\x (x x)) M) ~~> (M M)`

> `(\x (\y x))` throws away its second argument:
`(((\x (\y x)) M) N) ~~> M`

and so on.

It is easy to see that distinct lambda expressions can represent the same
function, considered as a mapping from input to outputs. Obviously:

 (\x x)

and:

 (\z z)

both represent the same function, the identity function. However, we said above that we would be regarding these expressions as synactically equivalent, so they aren't yet really examples of *distinct* lambda expressions representing a single function. However, all three of these are distinct lambda expressions:

 (\y x. y x) (\z z)

 (\x. (\z z) x)

 (\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 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 Monday's meeting (and further discussion is best pursued in person).



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.





1. Declarative vs imperatival models of computation.
2. Variety of ways in which "order can matter."
3. Variety of meanings for "dynamic."
4. Schoenfinkel, Curry, Church: a brief history
5. Functions as "firstclass values"
6. "Curried" functions

1. Beta reduction
1. Encoding pairs (and triples and ...)
1. Encoding booleans
+The `x` in `F x` and in `H x` are governed by the outermost quantifier, and only the `x` in `G x` is governed by the inner quantifier.
+### That's enough ###
+This was a lot of material, and you may need to read it carefully and think about it, but none of it should seem profoundly different from things you're already accustomed to doing. What we worked our way up to was just the kind of recursive definitions of `factorial` and `length` that you volunteered in class, before learning any programming.
+You have all the materials you need now to do this week's [[assignmentassignment1]]. Some of you may find it easy. Many of you will not. But if you understand what we've done here, and give it your time and attention, we believe you can do it.
+There are also some [[advanced notesweek1 advanced notes]] extending this week's material.