Lambda Calculus: Why is the second equation more general than the first?

Question

I was reading up on lambda calculus from Introduction to Lambda Calculus by Henk Barendregt and Erik Barendsen and came across this:

$(λ~x.f[\textbf{x}])\textbf{x} = f[\textbf{x}]$ more generally one has $(λ~x.f[\textbf{x}])\textbf{N} = f[\textbf{N}]$

Why is the second equation more general than the first?

Perhaps $x$ is a single variable while $N$ represents an arbitrary term? It would be helpful if you mentioned the number of the page where you found this so that we could see the context. — MJD, Oct 25 '20 at 03:43
You're new, so you simply don't know... but I hope you can learn: This is a terrible title, which barely refers to the content of your question. (Is this about topology? Differential equations? Complex analytic sheaf cohomology? Oh... and "latter" what??) Change it. — David G. Stork, Oct 25 '20 at 06:14
I am sorry about the title. I will change it as soon as possible. — KMR, Oct 25 '20 at 06:31

score 2 · Accepted Answer · answered Oct 25 '20 at 06:21

2

My guess in the comment seems to be correct: In

$$(λ\vec x.f[\vec x])\vec x = f[\vec x]$$

the λ-expression is being applied to a list of variables $\vec x$, but in

$$(λ\vec x.f[\vec x])\vec N = f[\vec N]$$

it is being applied to a list of arbitrary terms $\vec N$.

A single variable is the simplest possible special case of a term, as explained in definition 2.1 on page 9.

answered Oct 25 '20 at 06:21

MJD

Thank you. I think I get it now. – KMR Oct 25 '20 at 06:33
So while x is a list of variables, N is a list of expressions? – KMR Oct 25 '20 at 06:43
Yes, if you understand "expression" in the special limited sense of definition 2.1. – MJD Oct 25 '20 at 17:01

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