3
$\begingroup$

I've came across an example and I'm not quite sure on how the solution was met after performing beta-reduction on the following expression. It doesn't show any of the steps. Any help is appreciated!

(𝜆x. (𝜆y. y)) (𝜆a. (𝜆b.a))

(𝜆y. y)

$\endgroup$
2
$\begingroup$

The $(\beta)$ rule tells us that an expression of the form $(\lambda x . s)t$ reduces to $s[t/x]$.

In your case we have $t \equiv (\lambda a .(\lambda b. a))$, and $s \equiv \lambda y . y$.

Then one $\beta$-reduction gives us $s[t/x]$, which is just equivalent to $s$ because $x$ doesn't appear free in $s$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.