# Using the μ (mu) operator

### Problem

I've got this function:

$$f(x,y)=(6-3\cdot x)\cdot(y+2)$$, with $$(x,y)\in\mathbb{N}^2$$

Now I have to find $$g=\mu f$$.

### Proposed solution

My solution was to find the smallest $$n\in\mathbb{N}$$ to find $$f(n,y)=0$$ and show that $$\forall 0\leq m\leq n : f(m,y)$$ is defined:

$$f(0,y)=(6-0)(y+2) = 6y+12$$, defined $$\forall y\in\mathbb{N}$$

$$f(1,y)=(6-3)(y+2) = 3y+6$$, defined $$\forall y\in\mathbb{N}$$

$$f(2,y)=(6-6)(y+2) = 0$$, defined $$\forall y\in\mathbb{N}$$

So I've found $$g=\mu f=2$$.

### Questions

• Is above solution correct?
• Is it always the first parameter that becomes $$n$$?
• If $$f(x,y)=(6-3\cdot x)\cdot(y-2)$$ and $$f(0,y) = (6-0)(y-2) = 6y-12$$, wouldn't be $$f(0,y)=0$$, if $$y=2$$ and therefore my $$n=0$$ for this $$y$$, but for other $$y$$, my $$n$$ would be different?
• Please restrain your self to one question per post; in particular, the last bullet is independent of the others. You may want to check out other questions about mu-calculus. Regarding "real world examples", not that mu-calculus is Turing complete so every computer programis a real world example, even if they are written in a weird way.
– Raphael
Jul 20, 2014 at 13:51
• @Raphael I moved the last bullet to a new question. If you can answer any of these questions (here or in the other thread) please don't hesitate to explain :) Jul 20, 2014 at 14:15
• Could you clarify what you mean by "$\mu$-calculus"? The $\mu$-calculus I'm familiar with is an extension of modal propositional logic with least and greatest fixed-point operators but you seem to be talking about something else. Jul 20, 2014 at 14:22
• @DavidRicherby I am talking about mu recursive functions. Might be that the tag is wrong. Is it? The mu operator is defined here for example. Jul 20, 2014 at 14:31
• @polym: That's how we have been using the tag. Not everyone encounters "classic" recursion theory these days, though, especially outside of Europe (?).
– Raphael
Jul 20, 2014 at 14:39