f (f (f (z, 10), 20), 30)
+<a id=flipped-cons></a>
With a commutative operator like `(+)`, it makes no difference whether you say `fold_right ((+), z) xs` or `fold_left ((+), z) xs`. But with other operators it will make a difference. We can't say `fold_left ((&), []) [10, 20, 30]`, since that would start by trying to evaluate `[] & 10`, which would crash. But we could do this:
let
<code>3 ≡ \f z. f (f (f z)) ; or \f z. f<sup>3</sup> z</code>
<code>...</code>
-The encoding for `0` is equivalent to `\f z. z`, which we've also proposed as the encoding for `[]` and for `false`. Don't read too much into this.
+The encoding for `0` is what we also proposed as the encoding for `[]` and for `false`. Don't read too much into this.
Given the above, can you figure out how to define the `succ` function? We already worked through the definition of `cons`, and this is just a simplification of that, so you should be able to do it. We'll make it a homework.