-They permit you to appreviate the first λ-expression as simply `(10 - )`. We know there's an argument missing, because the infix operator `-` demands two arguments, but we've only supplied one. So `(10 - )` expresses a function that takes an argument `x` and evaluates to `10 - x`. In other words, it expresses λ`x. 10 - x`.Similarly, `( & ys)` expresses a function that takes an argument `x` and evaluates to `x & ys`. And --- can you guess what the last one will be? --- `( + )` expresses a function that takes two arguments `(x, y)` and evaluates to `x + y`.
+They permit you to appreviate the first λ-expression as simply `(10 - )`. We know there's an argument missing, because the infix operator `-` demands two arguments, but we've only supplied one. So `(10 - )` expresses a function that takes an argument `x` and evaluates to `10 - x`. In other words, it expresses λ`x. 10 - x`. Similarly, `( & ys)` expresses a function that takes an argument `x` and evaluates to `x & ys`.
+
+All of this only works with infix operators like `-`, `&` and `+`. You can't write `(1 swap)` or `(swap 1)` to mean λ`x. swap (1, x)`.
+
+Can you guess what our shortcut for the last function will be? It's `( + )`. That
+expresses a function that takes two arguments `(x, y)` and evaluates to `x + y`.