Construct a deductive system where $1^n$ is provable iff n is not prime I'd appreciate some help for the following exercise:
Construct a (as simple as possible) deductive system where all sequences of the form 1n
(which means 111... n-times) is provable if and only if n is not prime. (Note: As simple as possible means that the deductive rules and axioms should follow a simple schema. For example the schema "1n
is an axiom iff n is prime" isn't considered simple)
There is already an answered question to the problem of finding a similar deductive system in case n is not prime. 
Construct a deductive system where $1^n$ is provable iff $n$ is prime
I am also interested in the solution to that problem, because I don't understand the notation in the answer (i.e. I don't know, what lt, ndiv, ndivsmaller etc means) and I can't simply write a comment there to clear that, because my reputation isn't high enough to do so yet. 
Thank you in advance!
 A: Note: This system is for proving that numbers are prime, but the question was later changed to ask the opposite.
Interpret the string $1^n | 1^m$ to mean that $\gcd (n, m) = 1$
Interpret the string $1^n || 1^m$ to mean that $\gcd(n, m') = 1$ for all $1 \le m' \le m$
Intepret $1^n : 1^m :: 1^k$ to mean that there exists an integer solution in $a,b$ of $a\times n + b\times m = k$.
And of course $1^n$ means $n$ is prime.
Let $\epsilon = 1^0$ be the empty string.
Axioms:
$$ 1^n : 1^m :: \epsilon$$
$$ 1^n || \epsilon$$
Deduction rules:
1. Symmetry $1^n : 1^m :: 1^k \rightarrow 1^m : 1^n :: 1^k$
2. Adding to both sides $1^n : 1^m :: 1^k \rightarrow 1^n : 1^m :: 1^{k+n}$
3. Subtracting from both sides $1^n : 1^m :: 1^{n+k} \rightarrow 1^n : 1^m :: 1^k$
4. Bezout $1^n : 1^m :: 1 \rightarrow 1^n | 1^m$
5. $1^n || 1^m, \,1^n | 1^{m+1}$ $\rightarrow$ $1^n || 1^{m+1}$
6. $1^{n+1} || 1^{n} \rightarrow 1^{n+1}$
Here is the proof of the string $11$ which means "2 is prime".
First axiom
$$11 : 1: \epsilon$$
Rule 2
$$11 : 1: 1 $$
Rule 4
$$11 | 1$$
Second axiom
$$11 || \epsilon$$
Rule 5 + Previous line
$$11 || 1$$
Rule 6
$$11$$
Here is a proof of $111$ which means 3 is prime.
Lemma 1. $111 | 1$
$$111 : 1 : \epsilon$$
$$111 : 1 : 1$$
$$111 | 1$$
Lemma 2. $111 | 11$
$$111 : 11 : \epsilon$$
$$111 : 11 : 11$$
$$111 : 11 : 1111$$
$$111 : 11 : 1$$
$$111 | 11$$
Now apply rule 5 twice, and then rule 6
$$ 111 || 1 $$
$$ 111 || 11$$
$$ 111$$
A: To prove a number $n$ is not prime it suffices to provide a divisor greater than 1, and less than $n$.
Interpret $1^n : 1^m :: 1^k$ to mean that $n \times m = k$
Interpret $1^k$ to mean $k$ is composite.
Axiom
$$1^n : 1 :: 1^n$$
Deduction Rules
1. $ 1^n : 1^m :: 1^k \rightarrow 1^m : 1^n :: 1^k$
2. $1^n : 1^m :: 1^k \rightarrow 1^n : 1^{m+1} :: 1^{k+n}$
3. $1^{2+n} : 1^{2+m} :: 1^k \rightarrow 1^k$ with  $n \ge 0, m \ge 0$
Here is a proof that 4 is composite:
Axiom
$$11:1::11$$
Rule 2
$$11:11::1111$$
Rule 3 
$$1111$$
