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Let $K_0$ be a consistent $n$-th order theory, for some fixed $n \in \{1, 2, \dots\}$. Suppose $K_1$ is a consistent extension of $K_0$, and suppose $K_2$ is a conservative extension of $K_0$.

Is $K_1 \cup K_2$ necessarily consistent? If so, is $K_1 \cup K_2$ necessarily a conservative extension of $K_1$?

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    I can't answer this question because I do not know enough about the rest of the content. But if $K_1$ is consistent; then $K_1\lor K_2$ will be as well. This is because the question of whether the union of the two is consistent will be true; regardless of the truth value of whether $K_2$ is true, when $K_1$ is consistent. – 2017-02-04
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    @user400188: The first-order-with-equality theory with the axiom $\forall x\forall y, x=y$ is consistent, since any singleton set is a model for it. The first-order-with-equality theory with the axiom $\exists x\exists y, x\neq y$ is consistent, since any set of cardinality $\geq 2$ is a model for it. However the union of these theories is inconsistent. – 2017-02-04

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No, $K_1\cup K_2$ is not necessarily consistent, hence neither is it necessarily a conservative extension of $K_1$.

As a counter-example consider the first order formal language $L_0$ with signature consisting of the constant $0$ (zero), the binary function $+$ (addition) and the binary predicate $=$ (equality).

Denote by $K_0$ the $L_0$-theory consisting of the equality axioms, of the (additive) Abelian group axioms, and of the axiom $\exists x, x \neq 0$. Any Abelian group with cardinality $\geq 2$ is a model for $K_0$, hence $K_0$ is consistent.

Denote by $L_1$ the language obtained from $L_0$ by adding to its signature a unary function $-$ (minus).

Denote by $K_1$ the $L_1$-theory obtained by augmenting $K_0$ with the following axiom: $\forall x, -x=x$. Any model for $K_0$ can be transformed into a model for $K_1$ by interpreting $-$ as the identity function. Hence $K_1$ is consistent.

Denote by $K_2$ the $L_1$-theory obtained by augmenting $K_0$ with the following axiom: $\forall x, x+(-x)=0$. $K_2$ is a conservative extension of $K_0$ by the conservativity theorem (for a proof, see [1] Proposition 2.28, pp. 102-103 (link)).

However, the theory $K_1\cup K_2$ is inconsistent.

[1] Elliott Mendelson (2015). Introduction to Mathematical Logic (6th ed.) CRC Press.

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    That's the right idea, but you should exclude $-$ from the language of $K_0$: see the Wikipedia page you cite: $-$ has to be a **new** constant: otherwise statements that could be made but not proved in $K_0$ will become provable in $K_2$ (so $K_2$ won't be a conservative extension). – 2017-02-04
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    @RobArthan: Thanks. I've corrected my answer. – 2017-02-04