Church numerals without functions

Question

This is really a second part to my first question, but I felt that this was different enough from the first part that it merited its own question.

So, using Church numerals, we define

$3 = {\lambda} f. {\lambda}x.f(f(f(x)))$,

and

$4 = {\lambda} f. {\lambda}x.f(f(f(f(x))))$.

We can then add with an expression like

$3\ g\ (4\ g\ z)$

And this reduces to:

$(g (g (g (g (g (g (g\ z)))))))$.

But, of course, this is not how we would define $7$ in the scheme above. $7$ would be

${\lambda}g.{\lambda}z.(g (g (g (g (g (g (g\ z)))))))$.

Why is it still legitimate to call the application $3\ g\ (4\ g\ z)$ "7" when we can no longer perform functions with it?

Answer 1

This is just a shorthand, leaving off some things that aren't really needed to understand the concepts. If you want your $7$ to be written as a function again, all you need are a couple more implicit lambdas. Here's how I'd write the Church addition with those lambdas in place:

$$3 + 4 = \lambda g . \lambda z . 3 g (4 g z) = \lambda g . \lambda z . 7 g z$$

But we also know that:

$$\lambda k . f k = f$$

Therefore:

$$\lambda g . \lambda z . 7 g z = 7$$