(∀a, b, c ∈ N) a < b ⇒ a < bc
This is what I did :
Suppose that a, b, c ∈ N and a < b.
Since c ∈ N, we have c ≥ 1.
Thus, b = b · 1 ≤ b · c.
Therefore, a < b ≤ bc, which implies that a < bc.
I know what I did works but I want to prove it only assuming the axioms and nothing else. I don't want to use the fact that if c is a natural number then it is at least one since this is not an axiom but a proposition proved by the axioms. You guys have an idea how to do this? I've been stuck for more than 40 minutes.
These are the axioms that I am allowed to use :
N is closed under multiplication and addition.
0 is not an element of the naturals
For each a ∈ Z, a = 0 or a ∈ N or -a ∈ N
a < b implies b -a ∈ N , a > b implies a -b ∈ N, a ≥ b implies a > b or a = b
I am allowed to use all the arithmetic in Z. thus Ican use the axioms and propositions in Z
So given $b-a \in \mathbb{N}$ if $c =1$ we are done because $b=b \cdot 1$. Now because addition is closed in the natural number we also have $\sum_{k=1}^cb-a = cb -a \in \mathbb{N}$ and so $a < bc$.