[fexl]

Oh I love questions like these, thanks!

You went off the rails because you replaced (C z) with z. That is not a valid equivalence. (C z) does not equal z. But (C z) applied to anything else does equal z. So in particular, (C z) (C z) equals z.

Here's how the reduction sequence goes:

  : S C C z
  = (C z) (C z)
  = z

If you think about it, that second C can be replace with anything and it still works:

  : S C anything z
  = (C z) (anything z)
  = z

That C there simply "holds on" to its first argument, and then when it encounters any second argument, it ignores it completely and yields up the thing it was holding onto.