# The Eightfold Path to the Calculus of Constructions

Has a dog Buddha-nature?

This is the most serious question of all.

If you say yes or no,

— from Chapter 1 of the Gateless Gate

In the beginning, there was nothing.

Then Alonzo Church came and said, “Let there be:”

variable [x]
a symbol or string of symbols that represents a particular value
λ-abstraction [(λx.E)]
a function on the λ-term E, with the variable x bound to E
application [(E F)]
applying the λ-term E to the argument F

And thus all three and only they attained λ-nature (i.e., are valid λ-terms).

Whence cometh these λ-terms?

Consider the noble function $f(x) = x^3 + 1$. We note that the name $f$ is illusory and thus we can rewrite the same function as $x \mapsto x^3 + 1$. A λ-abstraction then is simple another way of writing this expression:

$(\lambda x\;.\;x^3 + 1)$

This is not a trivial transformation, however, as this also binds the variable $x$ to the λ-term $x^3 + 1$. Their destinies are forever intertwined within their confining parentheses and hence the expression $(\lambda y\;.\;((\lambda x \;.\; x^2)\; y+1))$ is equivalent to $(\lambda x\;.\;((\lambda x \;.\; x^2)\; x+1))$ even though the variables have the same names.

To keep the expressions as elegant as possible, we shall adopt the following rules:

1. Application is left-associative: $(E \; F \; G)$ is equivalent to $((E \; F) \; G)$
2. The outermost parentheses are omitted: write $\lambda x . x$ over $(\lambda x . x)$
3. The term in a λ-abstraction extends up until a same-level closing parenthesis is found: $(\lambda x . (\lambda y.E) \; F)$ is equivalent to $\lambda x.((\lambda y.E) \; F)$, not $(\lambda x . (\lambda y.E)) \; F$
4. Successive λ-abstractions can be combined under a single λ: $\lambda x.\lambda y.\lambda z.E$ is equivalent to $\lambda xyz.E$.

#### Meditation: Which of the following do not have λ-nature?1

1. $\lambda x.x$
2. $\lambda x^2.x$
3. $\lambda x . 1$
4. $\lambda xy . (\lambda zx . z^2 + x^2) \; y$

Any λ-term may be further transformed according to the following reduction rules:

α-conversion
from $\lambda x . E$ to $\lambda y.F$, where $F$ is the same as $E$ but with all instances of $x$ replaced by $y$; only bound variables may be replaced in this manner
β-reduction
application to arguments leads to substitution by those arguments, e.g., $(\lambda x.x^2) \; 3$ is equivalent to the λ-term $3^2$
η-conversion
functions are completely determined by their arguments, i.e., $\lambda x . (f \; x)$ is equivalent to $f$ iff all instances of $x$ in $f$ are bound

We are now ready to contemplate the truth2. Consider the following definitions:

1. $\text{TRUE} := \lambda x . \lambda y . x$
2. $\text{FALSE} := \lambda x . \lambda y. y$
3. $\text{AND} := \lambda p . \lambda q . p \; q \; p$
4. $\text{OR} := \lambda p . \lambda q . p \; p \; q$
5. $\text{NOT} := \lambda p . p \; \text{FALSE} \; \text{TRUE}$

#### Meditation: Show that OR FALSE TRUE evaluates to TRUE3.

Do not be fooled: the simplicity of our system hides a profound strength. For instance, contemplate the following:

1. $\lambda fx.x$
2. $\lambda fx.f \; x$
3. $\lambda fx.f \; (f \; x)$
4. (etc.)

What do they do with $f$ and $x$? What happens if I apply the two-argument λ-abstraction $\lambda xyf.x \; (y \; f)$ to (2) and (3)? To (3) and (3)? To (1) and any of the succeeding λ-abstractions in the list?

Indeed, we have defined an operation that behaves like multiplication on a list that behaves like numbers. We have:

$0 := \lambda fx.x$
$1:= \lambda fx.f \; x$
$2 := \lambda fx.f \; (f \; x)$
(etc.)

And in general these numerals (called Church numerals) repeat the functions to which they are arguments. So for instance, if we rename our multiplication λ-abstraction above to $\text{MULT} := \lambda xyf.x \; (y \; f)$, then:

\begin{aligned} \text{MULT} \; 2 \; 1 &= (\lambda x y z . x \; (y \; z)) 2 1 \\ &= (\lambda z . 2 \; (1 \; z)) \\ &= (\lambda z . 2 \; (\; (\lambda f x . f \; x) \; z\;)) \\ &= (\lambda z . 2 \; (\lambda x . z \; x)) \\ &= (\lambda z . 2 \; (\lambda \alpha . z \; \alpha)) \\ &= (\lambda z . (\lambda f x . f \; (f \; x)) (\lambda \alpha . z \; \alpha)) \\ &= (\lambda z . (\lambda x . (\lambda \alpha . z \; \alpha) \; ((\lambda \alpha . z \; \alpha) \; x))) \\ &= (\lambda z . (\lambda x . (\lambda \alpha . z \; \alpha) \; (z \; x))) \\ &= (\lambda z . (\lambda x . z \; (z \; x))) \\ &= (\lambda zx . z \; (z \; x)) \\ &= (\lambda fx . f \; (f \; x)) \\ &= 2 \end{aligned}

Note: it may be best if you do this calculation by yourself using only the reduction rules above.

We may now describe the machinations of any sort of calculating apparatus by defining what “computability” is:

Computability
a function $f$ on the natural numbers is computable if, for all its outputs, there is a corresponding λ-term that β-reduces to the output’s Church numeral

Or in the Platonic realm of mathematics, if I can find a λ-expression $F \; x' =_\beta y'$ that is equivalent to $f(x)=y$ where $x'$ and $y'$ are the Church numeral representations of $x$ and $y$ respectively, and by $E =_\beta F$ we mean you can get $F$ from $E$ by a series of β-reductions and vice versa.

Or in plain English, if anything I can do using $f$ in the world of math, I can also do using a corresponding $F$ in the world of λ-calculus.

However, our symbolic excursion hides an uncomfortable truth. Can you guess what it is? (Hint: it involves the operator $NOT$.)

Consider the following derivation:

Let $k := \lambda x.(\text{NOT} \; (x \; x))$.

\begin{aligned} k \; k &= \lambda x.(\text{NOT} \; (x \; x)) \; k \\ &= \text{NOT} \; (k \; k)\\ \end{aligned}

We have just demonstrated a paradox, an inescapable death knell to our naïve λ-calculus.

Wither our quest to bring forth computation? Is computer science a vacuous discipline after all?

Stay tuned for the next in this series.

1. Answer: (2), which is only valid if you consider “$x^2$” a string; note that (4) evaluates to a function and can still be applied
2. I’m just gonna follow the exposition on Wikipedia here.
3. Just use β-reduction until you arrive at $\text{FALSE}$