Show that a nonempty set of integers that is closed under subtraction must also be closed under addition So this is what I have so far: 
Let X be a nonempty set of integers 
Let $a,b\in X$ and we need to show that $a+b\in X$
Because $b\in X$ and X is closed under subtraction, than $b-b\in X$ 
Once again, by closure under subtraction $b-(b-b)\in X$
Since $a\in X$ as well, by closure under subtraction, $a-[(b-b)-b]\in X$
$a-[(b-b)-b]=a-(b-2b)=a-(-b)=a+b$
$\therefore a+b\in X$
But how would I show that the empty set is closed under addition as well? 
 A: It would help readability of your argument if you said something like this:
$1.$ Let $b$ be in $X$. Then $b-b$ is in $X$. So $0$ is in $X$.
$2.$ Because $0$ is in $X$, for any $b$ in $X$ we have $0-b$ is in $X$. So $-b$ is in $X$.
$3.$ For any $a$ and $b$ in $X$, $a+b=a-(-b)$, so by $(2)$, $a+b$ is in $X$.  
A: In the fifth line I think that you meant to have $(b-b)-b\in X$; otherwise the argument’s fine. The empty set is vacuously closed under addition: there is no counterexample to the statement that if $x,y\in\varnothing$, then $x+y\in\varnothing$, simply because the antecedent is never true.
A: Any statement of the form $\forall x(x\in A\rightarrow\varphi(x))$ is vacuously true if $A$ is empty.
In this case we have the statement $\forall x(x\in A\rightarrow(\forall y(y\in A\rightarrow x+y\in A)))$, for every $x$ and $y$ in $A$ their sum is in $A$. As the above says, this is automatically true when $A=\varnothing$ because $x\in\varnothing$ is false and the implication is therefore true.
A: Theorem: Let $\varnothing \subsetneq A \subseteq \mathbb{Z}$ be closed under subtraction. Then it is also closed under addition.
Proof: We will show that: $a \in A \Longrightarrow (-a)\in A$.
By $A$ being nonempty then exists some $n \in A$.
By $A$ being closed under subtraction $(n - n) \in A$. That is, $0 \in A$.
Considering $n \in A$ we have that by $A$ being closed under subtraction $(0 - n) \in A$. That is, $-n \in A$.
Hence: $a \in A \Longrightarrow (-a)\in A$
Now considering $n, m \in A$ we have that $-m \in A$ and by $A$ being closed under subtraction $(n - (-m)) \in A$. That is, $(n + m) \in A$.
$\blacksquare$
For $A = \varnothing$ we consider the conditional:
$n, m \in A \Longrightarrow (n+m) \in A$.
Because the antecedent is false (that's the part: $n, m \in A$) the entire statement is always true.
For more information on how to find the truth value of a conditional read [this].1
