# Is it possible to give a purely syntactic proof of : $|x| <N$ is equivalent to $x>-N \text{ AND } x<N$?

In the document accessible at the following link [1]: https://i.stack.imgur.com/P2AvU.jpg, I've tried to explain through purely logical means ( mainly DeMorgan's law) why a $$<$$ absolute value inequality must be translated by an $$\text{AND}$$ statement :

• Have you considered a specific example? For instance, if $|x|<3$, what is the possible range of valid $x$ values? – Arthur Mar 17 at 20:40
• The question is simple but a little bit Tricky: IMO it is not possible ti use only distributivity (see your comment below). We have a conditional definition of abs that - when used with the inequality - gives : $( x \ge 0 \land x < N) \lor (x < 0 \land -x < N)$. – Mauro ALLEGRANZA Apr 10 at 15:12
• Consider 1st disjunct : from $x \ge 0$ we derive (using properties of numbers) $x > -N$. Thus we have : $(x < N) \text { and } (x > -N)$. – Mauro ALLEGRANZA Apr 10 at 15:14
• 2nd disjunct : from $-x < N$ we derive (using properties of numbers) : $x > -N$. And from $x < 0$ we have (again by arithmetic) : $x < N$. Thus we have : $(x < N) \text { and } (x > -N)$. – Mauro ALLEGRANZA Apr 10 at 15:16
• Now we may conclude by logic alone, having derived the same result under both disjuncts, using disjunction elimination. – Mauro ALLEGRANZA Apr 10 at 15:18

We have $$|x|. Whenever you have an absolute value you can do a case by case investigation.
Case 1: Assume $$x>0$$ then $$|x|=x \implies x. Combining $$x>0$$ and $$x yields $$0.
Case 2: Assume $$x\leq 0$$ then $$|x|=-x \implies -x. Combining $$x\leq 0$$ and $$-N results in $$-N.
If we combine both cases we get $$|x| is equivalent to $$-N, which is equivalent to $$-N.
• $-N<x \text{ OR } x<N$ would imply that $x = 2N$ (if we look at $-N<x$) is also a feasible solution. Which is not the case. Both conditions need to be satisfied for any feasible solution. – MachineLearner Mar 17 at 21:10