[proofplan] We prove the two implications directly from the definition of a submodule. If $N$ is already a submodule, then its additive subgroup structure gives closure under subtraction and its module structure gives closure under scalar multiplication. Conversely, the assumed subtraction closure and nonemptiness first produce the zero element, then additive inverses and sums, so $N$ is an additive subgroup of $M$; the scalar closure hypothesis then makes it a submodule. [/proofplan] [step:Derive subtraction and scalar closure from the submodule axioms] Assume $N \le M$. By the definition of a submodule, $N$ is an additive subgroup of the underlying abelian group of $M$ and is closed under the left action of $R$. Let $x,y \in N$ and let $r \in R$. Since $N$ is an additive subgroup, the additive inverse $-y$ belongs to $N$, and hence $x-y=x+(-y)$ belongs to $N$. Since $N$ is closed under scalar multiplication by elements of $R$, the element $rx$ belongs to $N$. This proves the stated closure conditions. [/step] [step:Use nonemptiness and subtraction closure to recover the additive subgroup structure] Assume that for all $x,y \in N$ and all $r \in R$, one has $x-y \in N$ and $rx \in N$. Let $0_M$ denote the zero element of the underlying abelian group of $M$. Because $N$ is nonempty, choose an element $a \in N$. Applying the subtraction closure assumption to $a,a \in N$ gives $a-a=0_M \in N$. Now let $x \in N$. Since $0_M \in N$ and $x \in N$, subtraction closure gives $0_M-x=-x \in N$. Thus $N$ is closed under additive inverses. Finally, let $x,y \in N$. From the previous paragraph, $-y \in N$. Applying subtraction closure to $x,-y \in N$ gives \begin{align*} x-(-y)=x+y \in N. \end{align*} Therefore $N$ contains $0_M$, is closed under additive inverses, and is closed under addition. Hence $N$ is an additive subgroup of the underlying abelian group of $M$. [guided] We need to prove that $N$ is an additive subgroup of $M$, but the hypothesis gives closure under subtraction rather than closure under addition and additive inverses separately. The nonemptiness hypothesis is what lets us start the subgroup verification. Let $0_M$ denote the zero element of the underlying abelian group of $M$. Since $N$ is nonempty, there exists an element $a \in N$. The hypothesis applies to every pair of elements of $N$, so applying it to the pair $a,a$ gives \begin{align*} a-a=0_M \in N. \end{align*} Next we recover additive inverses. Let $x \in N$. We have just proved that $0_M \in N$, so the subtraction closure hypothesis applies to the pair $0_M,x$. Hence \begin{align*} 0_M-x=-x \in N. \end{align*} Thus every element of $N$ has its additive inverse in $N$. Now we recover addition from subtraction. Let $x,y \in N$. From the previous paragraph, $-y \in N$. Applying subtraction closure to $x$ and $-y$ gives \begin{align*} x-(-y)=x+y \in N. \end{align*} So $N$ contains $0_M$, is closed under additive inverses, and is closed under addition. These are exactly the additive subgroup conditions for a subset of the underlying abelian group of $M$. Therefore $N$ is an additive subgroup of $M$. [/guided] [/step] [step:Combine scalar closure with the additive subgroup structure to obtain a submodule] It remains to check compatibility with the $R$-module structure. Let $r \in R$ and $x \in N$. The assumed closure condition gives $rx \in N$ directly. Thus $N$ is an additive subgroup of $M$ closed under scalar multiplication by every element of $R$. By the definition of a submodule, $N \le M$. Combining this reverse implication with the forward implication proves the equivalence. [/step]