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A function type ${\tau }_1\to {\tau }_2$ classifies functions that accept arguments of type ${\tau }_1$ and produce results of type ${\tau }_2$ if they return. For example, $\lambda x.x+1$ has function type $\mathrm{int}\to \mathrm{int}$ in our teaching programming language, while $\lambda b.\mathrm{if}b\mathrm{then}1\mathrm{else}2$ has function type $\mathrm{bool}\to \mathrm{int}$. A function type specifies how a function may be used; it does not guarantee that a call returns.

The two central typing judgment rules areThe lambda abstraction rule checks the body under an extended typing context; the lambda application rule checks that the argument has exactly the required type. These rules describe our simple type examples. ^[ref]

For $(\lambda x.x+1)3$, assume $x:\mathrm{int}$; both $x$ and $1$ have type int, so $x+1$ has type int. The abstraction therefore has function type $\mathrm{int}\to \mathrm{int}$. Since $3$ has type int, the call has type int and evaluates to $4$. Replacing $3$ by true fails the argument requirement. This local derivation explains each step without running the program.

Arrows associate to the right: ${\tau }_1\to {\tau }_2\to {\tau }_3$ means ${\tau }_1\to ({\tau }_2\to {\tau }_3)$, a function returning another function. For instance, $\lambda x.\lambda y.x+y$ has function type $\mathrm{int}\to (\mathrm{int}\to \mathrm{int})$. Supplying one integer yields a function awaiting the second. Functions can also be arguments: $(\mathrm{int}\to \mathrm{int})\to \mathrm{int}$ accepts an integer-to-integer function and has an integer result type.

// Target: computer-science.bigb

The University of Oxford computer science Finals paper dated Thursday 4 June 2015, 2:30 pm–4:00 pm, Trinity Term 2015. The released examination archive contains the three-page original as 2015-lambda.pdf. Questions 1–2 occupy printed p. 2 and Question 3 printed p. 3. The local sources in the Oxford examinations repository are pdf/computer-science/finals/lambda-calculus-and-types/2015.pdf and tex/computer-science/finals/lambda-calculus-and-types/2015.tex. Question 2(d)'s literal claim is false: its proposed typing judgment holds precisely when $B=A\Rightarrow C$ for some simple type $C$.

The occurs check tests whether a type variable occurs inside the expression proposed as its replacement during type unification. After discarding a trivial equation such as $\alpha =\alpha$, solving $\alpha =\tau$ by a type substitution requires that $\alpha$ not occur in $\tau$. Thus $\alpha =\mathrm{int}\to \beta$ is permitted when $\alpha$ and $\beta$ are distinct, but $\alpha =\alpha \to \beta$ is rejected in our finite simple type grammar.

There is a short reason for the rejection. If a finite replacement $T$ for $\alpha$ existed, with replacement $B$ for $\beta$, the equation would become $T=T\to B$. The right side's abstract syntax tree has one new arrow node plus the nodes of $T$ and $B$, so it has strictly more nodes than $T$. Equal finite type expressions cannot have different sizes. Recursive types are a different design: they explicitly describe self-reference and specify how it interacts with typing judgments.

Begin with the variable typing judgment $\{x:A\}\mapsto x:A$. Because $z$ is absent from this typing context, the lambda abstraction rule permits $z$ to have type $\star$:Assign the entire lambda application $yx$ type $\star$ using the universal typing axiom, without attempting to assign a function type to $y$. ThenAlthough $y$ is a free variable of the body, it is absent from its recorded typing context. Consequently the lambda abstraction rule allows any simple type $B$ for $y$. Abstract over $y$ and then $x$:The side condition for $y$ holds because $y$ is absent; the side condition for $x$ holds because its recorded simple type is $A$. Thus

The exact restriction in the ordinary simply typed lambda calculus isThus the unrestricted choice of $A,B$ available in part (c) fails. A reading of the printed sentence as asserting impossibility for every choice of $A,B$ is false. This issue occurs in the original PDF as well as its transcription.^[ref]

To prove necessity, first observe why the ordinary rules cannot hide assumptions. Structural induction on a derivation provesFor a variable, both sides are the singleton containing that variable. For lambda application, both sides are the union of the two premise supports. For lambda abstraction, both sides delete the bound variable. There is no universal typing axiom to interrupt this equality.

Now invert the proposed derivation. Each outer lambda abstraction must have been introduced by the abstraction rule, since the other ordinary rules conclude a variable or a lambda application. The body must therefore have the typing judgmentThe support equality just proved ensures that both displayed assumptions occur; consistency fixes their simple types to those specified by the outer function type. Inverting the lambda application gives a simple type $C$ for its argument and the following premises:There is no flexibility in the supports: $\lambda z.x$ has the single free variable $x$, whereas $yx$ has both free variables $y,x$. Invert the second lambda application. Its two premises must befor some simple type $D$. The variable rule is the only possible final rule for each premise. It forces $D=A$ and $B=D\Rightarrow C$, yielding $B=A\Rightarrow C$.

Conversely, suppose $B=A\Rightarrow C$. The ordinary rules giveThe typing contexts are consistent, so another lambda application gives the body type $A$. Two lambda abstractions discharge $y$ and $x$, establishingThis proves sufficiency and supplies a family of counterexamples to the literal impossibility claim. For a concrete failure of the unrestricted claim, choose $A=\alpha$ and $B=\beta$ as atomic type variables. The type variable $\beta$ is not a function type $\alpha \Rightarrow C$, so no derivation exists for that fixed pair. Here the atomic type variables are fixed type expressions, rather than unknowns that one may change during type inference.

The reason for the restriction is the otherwise discarded lambda application $yx$: the ordinary rules still type-check it. Its subsequent disappearance under beta reduction does not license reversing that beta reduction while preserving every simple type of the result. The beta-normal form $\lambda xy.x$ admits arbitrary $A\Rightarrow (B\Rightarrow A)$, whereas the unreduced lambda term has the additional constraint just derived.

A recursive type explicitly names a type in terms of itself. Notation $\mu \alpha .\tau$ binds $\alpha$ within $\tau$. A programming language must specify whether unfolding gives type equality or requires explicit operations; this is outside the finite simple type grammar used here. See Benjamin C. Pierce's treatment of recursive types in chapters 20–21. ^[ref]

In the examples here, a simple type is a finite expression built from the base types int and bool using function types:For example, $\mathrm{int}\to \mathrm{bool}$ describes a function accepting an integer and producing a Boolean value. During type inference, the grammar is extended with type variables such as $\alpha$ for types not yet determined. A type scheme additionally binds such type variables; it is distinct from the individual simple type used at one call.

Type unification finds a type substitution making specified type expressions equal, or reports incompatible constraints. In our simple type grammar, matching two function types reduces to matching their argument types and their result types. Distinct base types int and bool fail to match; solving a type variable against a nontrivial type expression uses the occurs check.

// Target: computer-science.bigb

An extension of simple type assignment that assigns a distinguished atomic type variable $\star$ to every lambda term in the empty typing context. The recorded assumptions therefore need not contain all free variables of the subject. This is a rule admitting a typing judgment, rather than an implicit conversion to every simple type or a general subtyping rule.