Partial Application

What Is Partial Application?

Partial application means supplying fewer arguments than a function expects. Since curried functions consume one argument at a time, applying one argument to a two-parameter function produces a new function:

(λx.λy.x y) a →β λy.a y

The outer λx consumed a, but the inner λy remains. The result is a function — not a final answer.

Result Is a Function

When you partially apply a curried function, the result always starts with λ. That leading lambda is the signal: this term is still a function, still expecting input.

(λx.λy.x y) a →β λy.a y

The λy at the front of the result means "I am a function that takes one more argument." If the result had no λ at the front, it would be a fully reduced value, not a waiting function.

Partially Apply

When you apply one argument to a curried function, you perform one beta reduction. The outer lambda is consumed and its parameter is substituted into the body:

(λx.λy.x y) a →β λy.a y

The inner lambda is untouched — it is part of the body, not the parameter being reduced. Write the result as a lambda expression.

How Many Arguments Left?

Each λ at the front of a curried function represents one expected argument. Applying one argument removes one λ. The number of remaining leading lambdas tells you how many more arguments the function still needs:

• λx.λy.λz.M — expects 3 arguments
• After applying one: λy.λz.M' — expects 2
• After applying two: λz.M'' — expects 1

Apply Then Apply

Partial application gives you a function. You can then apply that function to another argument to continue reducing:

Step 1 (partial): (λx.λy.x y) a →β λy.a y Step 2 (complete): (λy.a y) b →β a b

The first application fills one blank. The second fills the last one. After both, no lambdas remain — you have a final result.

Partial vs. Full Application

The difference between partial and full application is what you get back:

• Partial: The result is a function (starts with $\lambda$)
• Full: The result has no leading $\lambda$ — all parameters consumed

For λx.λy.x y:

• Applying one arg: (λx.λy.x y) a →β λy.a y — partial ($\lambda$ remains)
• Applying both: (λx.λy.x y) a b →β a b — full (no $\lambda$ left)

Why Partial Application Matters

Consider a general function λx.λy.x y that applies x to y. Partially applying it to some specific f gives λy.f y — a specialized function that always applies f to whatever argument it receives.

This means every curried function is also a factory for more specific functions. You do not need separate definitions for each specialized case — you build them by supplying arguments one at a time.

Recognize Partial Application

Practice recognizing and performing partial application. Apply one argument to a curried function: reduce the outer lambda by substituting the argument for its parameter in the body. The inner lambda remains.

(λx.λy.M) a →β (λy.M)[x := a]

Write the resulting lambda expression.