User blog comment:Bubby3/Letter notation extension to R and higher letters/@comment-39541634-20191216055735/@comment-24725252-20191218225116

Sorry for the late reply, because it said "2 days ago." I don't check the comments of the blog posts every day

The expression becomes ((w*2~w*3)[0.5])|10. Because w*3[0] = w*2, m = 0.

We follow the "If m = 0, then c = 10*n and b{d} is b[d]" rule, so c = 5, and w*3{n} = w*2+n.

Now, it becomes ((w*2+5~w*2+6)[0])|10. It would be more convienent to make a (a~b)[0] = a rule, but the function works even without the rule.

We use the sucessor rule to get the expression as (w*2+6)|2, which is (w*2+5)|(w*2+5)|1, which eventually evalulates to (w*2+5)|10.