Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It's known that all recursive functions are representable in Peano arithmetic. I am trying to find representation of subtraction function

$f(x,y)= \left\{\begin{matrix} x-y &if& x>y\\ 0 & if& x \leq y \end{matrix}\right.$

I think its representation formula is $R(x,y,z) \leftrightharpoons (x>y\Rightarrow x+z=y) \wedge (x\leqslant y \Rightarrow z=0)$. I have to show that

1) If $f(a,b)=c$ then $\vdash R(a,b,c)$

2) $\vdash\exists _{1}z R(x,y,z) $

How to show it?

share|cite|improve this question
up vote 2 down vote accepted

Proof sketch

(1) is straightforward. Argue by cases.

If $f(a, b) = 0$, then not-($a > b$). So $PA \vdash \neg(\overline{a} > \overline{b})$ (why?) and hence $PA \vdash \overline{a} > \overline{b} \to \overline{a} + \overline{c} = \overline{b}$. Also $PA \vdash \overline{0} = \overline{0}$ and hence $PA \vdash \overline{a} \leq \overline{b} \to \overline{0} = \overline{0}$. Putting things together, $PA \vdash R(\overline{a}, \overline{b}, \overline{c})$.

Similarly if $f(a, b) = c$, $c > 0$, then $a + c = b$ and so $PA \vdash \overline{a} + \overline{c} = \overline{b}$ hence $PA \vdash \overline{a} > \overline{b} \to \overline{a} + \overline{c} = \overline{b}$. Also not-$a \leq b$, so $PA \vdash \neg(\overline{a} \leq \overline{b})$ hence $PA \vdash \overline{a} \leq \overline{b} \to \overline{c} = \overline{0}$. Putting things together again, here too $PA \vdash R(\overline{a}, \overline{b}, \overline{c})$.

(2) Query: Are you sure you want to show $PA \vdash\exists _{1}z R(x,y,z)$ rather than the more usual $PA \vdash\exists _{1}z R(\overline{a},\overline{b},z)$?

Anyway, to show (2), prove existence and uniqueness separately. I'll leave existence as an exercise.

For uniqueness, assume $R(x, y, z)$ and $R(x, y, z')$. Argue in $PA$ by cases from $x > y \lor x \leq y$. The first disjunct gives $x + z = y$ and $x + z' = y$ and you need to show (or assume) that those entail $z = z'$. The second disjunct gives $z = 0$ and $z' = 0$ and so $z = z'$. So either way $z = z'$, which completes the proof.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.