1
$\begingroup$

Prove: For any $x,y \in \mathbb{N}, y \neq x+y$.

I'm only suppose to use the Peano axioms as defined here http://aleph0.clarku.edu/~djoyce/numbers/peano.pdf and the properties of addition in $\mathbb{N}$.

I've already proved most of the properties of addition excepting this one and I don't know how to approach to this.

$\endgroup$
  • $\begingroup$ Prove by induction. $\endgroup$ – Berci Aug 24 '14 at 22:39
  • $\begingroup$ @user143201 Do you need further help with this? $\endgroup$ – Git Gud Aug 26 '14 at 10:06
  • $\begingroup$ I did solved it with your hint. But I don't know if I have to click the answer button, because what you gave me was only a hint $\endgroup$ – Keith Aug 27 '14 at 13:55
  • $\begingroup$ @user143201 It is acceptable to not do anything, to accept my answer even though is was only a hint and also to just type your own answer and accept it. $\endgroup$ – Git Gud Sep 30 '14 at 21:22
3
$\begingroup$

Hints: Take an arbitrary $x\in \mathbb N$ and prove the statement $\forall y\in \mathbb N(x\neq x+y)$ by induction. You'll have to use the definition of $+$, the injectivity of the successor function and the fact that $1$ is not the successor of any number.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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