# How do I prove that $\forall a \forall b(\neg a<b\iff b\leq a)$ [duplicate]

The universe is the set of natural numbers including 0, which are defined in accordance with the Peano Axioms.

We define the inequalities as:

$$a\leq b \iff \exists x(a+x=b)$$

$$a

Assuming we defined the natural numbers with PA

We proceed via induction on a:

Let $$a = 0$$:

Case 1: $$b=0$$:

Prove that the $$0<0$$ is false (fill in the details) then prove that $$0 \leq 0$$ is true

Case 2: $$b \neq 0$$:

Prove that $$a \lt b$$ is true then prove that $$b \leq a$$ is false

This is your base case, then assume that the theorem holds for some n such that $$a=n$$ And proceed again, using the axioms (specifically about successors) in two cases, where this time $$b=a$$ and $$b \neq a$$ (in the second case you might have to do another induction this time on b, where you start with $$b = S(a)$$ where $$S(n) = n+1$$

This is not a complete proof of course, just a sketch that might help you

• I was assuming that natural numbers are defined as per the Peano Axioms. I will edit my post. May 25, 2022 at 14:29