You can either have a direct definition or a recursive definition. If you have a recursive definition you need a base case from which all cases arrive.
Your function appears to be recursive but it has no base case.
a < b if a + 1< b + 1 which raises the question what is the definition of a + 1 < b + 1 to which a + 1 < b+a if a + 2 < b +2, and final verification is pushed further and further away.
So if you are going to do recursion, you need a base case involving 0
- $0 < b$ if $b \ne 0$
Now your definition $a < b$ if $a + 1 < b + 1$ ... isn't good because it is taking you away from the base case. We need a definition that either a) takes you from the base case to $a < b$ or b) takes you from $a < b$ to the base case.
Either
2a. $a < b $ if $a - 1 < b -1$ (allows the user to start at $a<b$ and work down to $0 < b'$)
Or
2b. if $a < b$ then $a + 1 < b + 1$ (allows the user to start at $0<b'$ annd work up to $a < b$)
will do. Which one you like is a matter of taste.
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Then there is a direct definition. This is less obvious to see but more "powerful" and ,ahem, direct to use. When is $a < b$ true? It's true when $0 < b- a$ which, as these are natural numbers rather than integers, is true whenever $b - a \ne 0$ and $b - a$ is a legitimate natural number.
So
- $a < b$ if $b - a \ne 0$ and $b - a \in \mathbb N$.