# $λ$-Calculus & Curry-Howard correspondence ###### Younesse Kaddar

I. $λ$-Calculus

1) Definition 2) $𝛼$, $𝜂$-conversions and $𝛽$-reduction 3) Alligators

II. Fixed-point combinator

1) Turing-completeness of $𝜆$-calculus 2) Curry and Turing combinators 3) Curry-Howard correspondence

I.1 - $λ$-Calculus : definition

Let $𝒱 \; ≝ \; \; \lbrace x, y, z, \ldots \rbrace$ be a fixed set of variables

$𝛬$ expressions :
 $$Λ ::=V Λ Λ λV·Λ$$ the smallest set of functions such that
• $𝒱 \; ⊆ 𝛬$
• Application : $∀u, v∈𝛬, \underbrace{uv}_{\text{interpreted as } u(v)} ∈ 𝛬$
• $𝜆$-abstraction : $∀x∈𝒱, u∈𝛬, \underbrace{𝜆x. u}_{\text{interpreted as } x \mapsto u} ∈ 𝛬$
> **NB** : *Such a set exists* : apply Knaster-Tarski to the following monotonic function in the complete lattice $𝒫(𝛴^\ast)$, where $𝛴 \; ≝ \; \; \lbrace 𝜆, . \rbrace ∪ 𝒱$ $$\begin{cases} 𝒫(𝛴^\ast) ⟶ 𝒫(𝛴^\ast) \\ 𝛬 \mapsto \lbrace uv \rbrace_{u,v∈𝛬} ∪ \lbrace 𝜆x. u \rbrace_{\substack{x∈𝒱 \\ u∈𝛬}} ∪ 𝒱 \end{cases}$$

I.2 - $𝛼$ / $𝜂$-conversions, $𝛽$-reduction

$𝛼$-conversion : : $$\lambda x. u= _\alpha \lambda y. (\underbrace{u[x :=y]}_{\rlap{\text{all instances of x in u are replaced by y}}})$$

> ![/!\\](/assets/Slides/Lambda-Calculus/warning_icon.svg) $x=y$ OR $x, y∉ bv(u)$ and $y∉ fv(u)$, as illustrated by these examples : > > |$(λx·y)[x:=uv] ≠_𝛼 λuv·y$|$(λx·x)[x:=y] ≠_𝛼 λx·y$ > |-|- > $(λy·x)[x:=y] ≠_𝛼 λy·y$ |$λx·y[y:=x] ≠_𝛼 λx·x$

$𝛽$-reduction : : $$( \lambda x . u ) v \to u [ x := v ]$$

$𝜂$-conversion : : $$\underbrace{\lambda x . (u x) =_𝜂 u}_{\rlap{\text{If x does not appear free in u}}}$$

Or… in terms of alligators (an idea by Victor Bret)

__

![Alligator family](/assets/Slides/Lambda-Calculus/families_3.png)

Eating process ≡ $𝛽$-reduction

![](eating_1.png)
⇓ ![](eating_3.png)
![](eating_4.png) ⇓ ![](eating_5.png)
⇓ ![](eating_6.png)
![](eating_7.png) ⇓ ![](eating_9.png)
⇓ ![](eating_10.png)

$𝛽$-reduction

1. ![](eatingrule_2.png)
⇓ 2. ![](eatingrule_3.png)
⇓ 3. ![](eatingrule_4.png)

$𝛼$-conversion

1. ![](eatingrule_4.png)
⇓ 2. ![](colorrule_1.png)
⇓ 3. ![](colorrule_2.png)

old alligators ≡ parentheses

![](old_1.png)
⇓ ![](old_3.png) ⇓ ![](old_4.png)

II.1 - Turing-completeness of $𝜆$-calclus

Church Numerals :

∀n∈ℕ, \; \lceil n \rceil \; ≝ \; \; 𝜆 f x. f^n(x) \; ⟹ \; ℕ ≃ \left\lbrace 𝜆 f x. f^n(x) \right\rbrace
• Constant function : $∀n, k∈ℕ, \; f(x_1,\ldots,x_k) = n \\ ≡ 𝜆 x_1 ,\ldots, x_k. \lceil n \rceil$

• Successor function S : $S(x) \stackrel{\mathrm{def}}{=} x + 1 \\ ≡ 𝜆 n. f(n f x)$

• Projection function : $P_i^k(x_1,\ldots,x_k) \stackrel{\mathrm{def}}{=} x_i \\ ≡ 𝜆 x_1 ,\ldots, x_k. \, x_i$

II.1 - Turing-completeness of $𝜆$-calculus

Operators:

• Composition operator : $𝜆$-abstraction

• the primitive recursion operator $\rho$ :
\begin{align} \rho(g, h) &\stackrel{\mathrm{def}}{=} f \quad\text{where}\\ f(0,x_1,\ldots,x_k) &= g(x_1,\ldots,x_k) \\ f(y+1,x_1,\ldots,x_k) &= h(y,f(y,x_1,\ldots,x_k),x_1,\ldots,x_k)\,.\end{align}
• the minimisation operator $\mu$ :
\begin{align} \mu(f)(x_1, \ldots, x_k) = z \stackrel{\mathrm{def}}{\iff}\ f(z, x_1, \ldots, x_k)&=0\quad \text{and}\\ f(i, x_1, \ldots, x_k)&>0 \quad \text{for}\quad i=0, \ldots, z-1. \end{align}

?

II.1 - Turing-completeness of $𝜆$-calculus

If

\mathbf{TRUE} \; ≝ \; \; λx.λy.x \\ \mathbf{FALSE} \; ≝ \; \; λx.λy.y

then :

\mathbf{AND} \; ≝ \; \; λp.λq.p q p \\ \mathbf{OR} ≝ λp.λq.p\; p \; q \\ \mathbf{NOT} \; ≝ \; \; λp.p \; \mathbf{FALSE} \; \mathbf{TRUE} \\ \mathbf{IFTHENELSE} \; ≝ \; \; λp.λa.λb.p \; a \; b \\ \mathbf{ISZERO} \; ≝ \; \; λn. \; n \; (λx.\, \mathbf{FALSE}) \; \mathbf{TRUE}

![](descartes.png)
![](escher-hands.gif)

II.2 - Fixed-point combinators

Astounding fact : each $𝜆$-term actually has a fixed point !

and there even exist fixed-point combinators $y$ :

\displaystyle y\ f = f\ (y\ f) \ \ \text{ for all } f
The Curry combinator :
\displaystyle Y \; ≝ \; \; \lambda f.(\lambda x.f\ (x\ x))\ (\lambda x.f\ (x\ x))
The Turing combinator :
\displaystyle 𝛳 \; ≝ \; \; \big(λf. λg. \, g \, (f \, f \, g) \big) \big(λf. λg. \, g \, (f \, f \, g)\big)

\begin{align*} Y\ f & = (\lambda f'.(\lambda x.f'\ (x\ x))\ (\lambda x.f'\ (x\ x)))\ f \\ & = (\lambda x.f\ (x\ x))\ (\lambda x.f\ (x\ x)) \\ & = f (\underbrace{(\lambda x.f\ (x\ x))\ (\lambda x.f\ (x\ x))}_{= Y\ f}) \end{align*}
fact \; n = \begin{cases} 1 \text{ if } n=0 \\ n × (fact \, (n-1)) \text{ else} \end{cases}

F f \; ≝ \; \; 𝜆f. n \, (𝜆x. \, n × f (Pred \, n)) \, 1 \\ Succ \, n \; ≝ \; \; 𝜆n f x. f(n f x) \\ Pred \, n \; ≝ \; \; 𝜆n. 𝜋_1 \Big( n \, \big(𝜆c. ⟨𝜋_2 c, Succ \, 𝜋_2 c⟩\big) \, ⟨0,0⟩ \Big) \\ n × m \; ≝ \; \; 𝜆n, m. n (m f)
\begin{align*} fact & = YF \\ & = 𝜆f. (𝜆x. f (xx)) (𝜆x. f (xx)) \Bigg( 𝜆f. n \, \Big(𝜆x. \, n × f \Big(𝜋_1 \\ &\bigg( n \, \big(𝜆c. ⟨𝜋_2 c, Succ \, 𝜋_2 c⟩\big) \, ⟨0,0⟩ \bigg) \Big)\Big) \, 1 \Bigg) \\ \end{align*}

II.3 - Curry-Howard isomorphism

It’s great, but… you can’t do that in Haskell (for example) :

y :: (a -> a) -> a
y = \f -> (\x -> f (x x)) (\x -> f (x x))


Occurs check: cannot construct the infinite type: r0 ~ r0 -> t
Expected type: r0 -> t
Actual type: (r0 -> t) -> t

![WHY?](/assets/Slides/Lambda-Calculus/rage-why.png)

Because Haskell (and other functional programming languages such as OCaml, Lisp, Clojure, etc…) is strongly typed.

What is a type ? :

A type is a property of a program.

Ex :

• the type of the $𝜆$-term $\lceil 1 \rceil$ can be be thought as “Integer” : int.
• the program $fact$ takes a natural number and produces another natural number : its type is $int ⟶ int$

The CURRY-HOWARD CORRESPONDENCE :

Proofs can be represented as programs ($𝜆$-terms), and the formulas it prove are the types for the program

Intuitionistic implicational natural deduction Lambda calculus type assignment rules
$\displaystyle \frac{}{\Gamma_1, \alpha, \Gamma_2 \vdash \alpha} \text{Ax}$ $\displaystyle \frac{}{\Gamma_1, x:\alpha, \Gamma_2 \vdash x:\alpha}$
$\displaystyle\frac{\Gamma, \alpha \vdash \beta}{\Gamma \vdash \alpha \rightarrow \beta} \rightarrow I$ $\displaystyle\frac{\Gamma, x:\alpha \vdash t:\beta}{\Gamma \vdash \lambda x.t : \alpha \rightarrow \beta}$
$\displaystyle\frac{\Gamma \vdash \alpha \rightarrow \beta \qquad \Gamma \vdash \alpha}{\Gamma \vdash \beta} \rightarrow E$ $\displaystyle\frac{\Gamma \vdash t:\alpha \rightarrow \beta \qquad \Gamma \vdash u:\alpha}{\Gamma \vdash t\;u:\beta}$

From a logical standpoint, the $Y$ combinator corresponds to the Curry paradox.

“If this sentence is true then I am an alligator” is true : why ?

Let’s suppose that the sentence is true : it must be shown that I am an alligator.

But since the sentence is true, modus ponens can be applied, which yields the conclusion.