If x and y are natural numbers and x > y, then x-y is also a natural number.
how to prove this by induction?
is it becuase x-1 is a natural number,
and then for a number k, x - (k+1) = x-k-1=(x-k)-1 is also a natural number?
I don't understand the definition $\displaystyle S=\{x\in\mathbb{N}\mid y\in\mathbb{N}\lor x-y\in\mathbb{N}\}$. The variable $\displaystyle y$ is not bound (by a quantifier), so it is not clear whether $\displaystyle y\in\mathbb{N}\lor x_0-y\in\mathbb{N}$ is true or false for each particular $\displaystyle x_0$.
I guess, a simple answer is to do induction on x with the base case on x=y+1. Then if for x there is a number z (= x-y) such that y+z = x, then a number with the same property exists for x+1.If x and y are natural numbers and x > y, then x-y is also a natural number.
S is the set of all natural numbers, y, such that "if x is a natural number and x> y then x-y is a natural number".
Your induction must start with y= 1. Show that if x is a natural number and x> 1 then x- 1 is a natural number.
But you still haven't done what emakarov asked you to do in his first reponse: what is your definition of ">" and what is your definition of "x- y" for natural numbers?