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The answer is no. It is enough to find a model of MA which is an integral domain of characteristic $0$ (whence O1 is true and E1 false) such that $2$ is not invertible (whence E2 is true and O2 false).

One example of such a model is the ring of dyadic $2$-adic integers $\mathbb Z_2$. This is clearly a domain, and $2$ is not a unit, hence it suffices to show

Theorem: For any prime $p$, the ring $\mathbb Z_p$ is a model of MA.

Proof: The only problem is to show verify that induction holds. Assume $\mathbb Z_p\models\phi(0)\land\forall x\,(\phi(x)\to\phi(x+1))$, where $\phi$ is an arithmetic formula with parameters from $\mathbb Z_p$, and put $\phi(\mathbb Z_p):=\{a\in\mathbb Z_p:\mathbb Z_p\models\phi(a)\}$.

Since $\phi(\mathbb Z_p)$ is definable in $\mathbb Z_p$, it is also definable in the field $\mathbb Q_p$ endowed with a unary predicate for $\mathbb Z_p$. Macintyre [1] proved that such structures admit a form of quantifier elimination, and as a corollary (Thm. 2 on p. 609), every infinite definable set has a nonempty interior. Thus, there is $a_0\in\phi(\mathbb Z_p)$ and $k\ge0$ such that $a_0+p^k\mathbb Z_p\subseteq\phi(\mathbb Z_p)$. Let $a\in\mathbb Z_p$ be arbitrary, and let $b< p^k$ be a natural number such that $b\equiv a-a_0\pmod{p^k}$. Since $\phi(\mathbb Z_p)$ is closed under successor, we have $a\in b+a_0+p^k\mathbb Z_p\subseteq\phi(\mathbb Z_p)$. Thus, $\phi(\mathbb Z_p)=\mathbb Z_p$, i.e., $\mathbb Z_p\models\forall x\,\phi(x)$.   QED

I suspect the following may work as additional countermodels (they are domains where $2$ is not a unit, the issue is whether they satisfy induction):

• The ring of algebraic integers $\tilde{\mathbb Z}$. A form of quantifier elimination for $\tilde{\mathbb Z}$ was proved by van den Dries [2] and Prestel and Schmid [3], but the basic formulas are somewhat messy, so it is not immediately clear to me whether this implies induction.

• The localization of $\tilde{\mathbb Z}$ at a maximal ideal containing $2$. Elimination of quantifiers for this (and similar) rings is reported as Fact 3 in [2], where it is attributed to [4]. It seems it could imply induction by a similar argument as for $\mathbb Z_p$.

[1] Angus Macintyre, On definable subsets of $p$-adic fields, Journal of Symbolic Logic 41 (1976), no. 3, pp. 605–610.

[2] Lou van den Dries, Elimination theory for the ring of algebraic integers, Journal für die reine und angewandte Mathematik 388 (1988), pp. 189–205.

[3] A. Prestel and J. Schmid, Existentially closed domains with radical relations, Journal für die reine und angewandte Mathematik 407 (1990), pp. 178–201.

[4] Angus Macintyre, Kenneth McKenna, Lou van den Dries, Elimination of quantifiers in algebraic structures, Advances in Mathematics 47 (1983), no. 1, pp. 74–87.

1

The answer is no. It is enough to find a model of MA which is an integral domain of characteristic $0$ (whence O1 is true and E1 false) such that $2$ is not invertible (whence E2 is true and O2 false).

One example of such a model is the ring of dyadic integers $\mathbb Z_2$. This is clearly a domain, and $2$ is not a unit, hence it suffices to show

Theorem: For any prime $p$, the ring $\mathbb Z_p$ is a model of MA.

Proof: The only problem is to show that induction holds. Assume $\mathbb Z_p\models\phi(0)\land\forall x\,(\phi(x)\to\phi(x+1))$, where $\phi$ is an arithmetic formula with parameters from $\mathbb Z_p$, and put $\phi(\mathbb Z_p):=\{a\in\mathbb Z_p:\mathbb Z_p\models\phi(a)\}$.

Since $\phi(\mathbb Z_p)$ is definable in $\mathbb Z_p$, it is also definable in the field $\mathbb Q_p$ endowed with a unary predicate for $\mathbb Z_p$. Macintyre [1] proved that such structures admit a form of quantifier elimination, and as a corollary (Thm. 2 on p. 609), every infinite definable set has a nonempty interior. Thus, there is $a_0\in\phi(\mathbb Z_p)$ and $k\ge0$ such that $a_0+p^k\mathbb Z_p\subseteq\phi(\mathbb Z_p)$. Let $a\in\mathbb Z_p$ be arbitrary, and let $b< p^k$ be a natural number such that $b\equiv a-a_0\pmod{p^k}$. Since $\phi(\mathbb Z_p)$ is closed under successor, we have $a\in b+a_0+p^k\mathbb Z_p\subseteq\phi(\mathbb Z_p)$. Thus, $\phi(\mathbb Z_p)=\mathbb Z_p$, i.e., $\mathbb Z_p\models\forall x\,\phi(x)$.   QED

I suspect the following may work as additional countermodels (they are domains where $2$ is not a unit, the issue is whether they satisfy induction):

• The ring of algebraic integers $\tilde{\mathbb Z}$. A form of quantifier elimination for $\tilde{\mathbb Z}$ was proved by van den Dries [2] and Prestel and Schmid [3], but the basic formulas are somewhat messy, so it is not immediately clear to me whether this implies induction.

• The localization of $\tilde{\mathbb Z}$ at a maximal ideal containing $2$. Elimination of quantifiers for this (and similar) rings is reported as Fact 3 in [2], where it is attributed to [4]. It seems it could imply induction by a similar argument as for $\mathbb Z_p$.

[1] Angus Macintyre, On definable subsets of $p$-adic fields, Journal of Symbolic Logic 41 (1976), no. 3, pp. 605–610.

[2] Lou van den Dries, Elimination theory for the ring of algebraic integers, Journal für die reine und angewandte Mathematik 388 (1988), pp. 189–205.

[3] A. Prestel and J. Schmid, Existentially closed domains with radical relations, Journal für die reine und angewandte Mathematik 407 (1990), pp. 178–201.

[4] Angus Macintyre, Kenneth McKenna, Lou van den Dries, Elimination of quantifiers in algebraic structures, Advances in Mathematics 47 (1983), no. 1, pp. 74–87.