Modular Arithmetic (MA) has the same axioms as first order Peano Axioms (PA) except $\forall x (Sx \ne 0)$ is replaced with $\exists x(Sx = 0)$. (http://en.wikipedia.org/wiki/Peano_axioms#First-order_theory_of_arithmetic)
MA has arbitrarily large finite models based on modular arithmetic. All finite models of MA have either an even or odd number of elements. I call a model of MA "even" if it satisfies either of these two sentences:
E1) $\exists x(x \ne 0 \land x+x = 0)$
E2) $\forall x(x+x \ne S0)$
A model of MA is odd if it satisfies either of:
O1) $\forall x(x = 0 \lor x+x \ne 0)$
O2) $\exists x(x+x = S0)$
We can use compactness to prove MA has infinite "even" size models by adding the even definitions above as axioms. We can similarly prove there are infinite "odd" size models of MA. Some infinite sets, like the integers, are both even and odd. The integers are not the basis for a model of MA. For example, the four square theorem (every number is the sum of at most four squares) is a theorem of both MA and PA. The four square theorem is false in the integers. It has been conjectured the complex numbers are a basis for a model of MA. If so, the complex numbers would be an "odd" model of MA.
My question is whether every model of MA must be exclusively even or exclusively odd? Are the following statements theorems of MA?
$$\exists x(x \ne 0 \land x+x = 0) \ \overline{\vee}\ \exists x(x+x = S0)$$
$$\forall x(x+x \ne S0) \ \overline{\vee}\ x(x = 0 \lor x+x \ne 0)$$
I included the ring theory tag because all of the axioms of ring theory can be derived from the axioms of MA. Every model of MA is a commutative ring with unity. I have found that the 1-element model of MA (the trivial ring) can cause a lot of problems in proofs. I would be happy to prove these statements are true for all models with two or more elements.
