[proofplan] We first verify that derivations form a vector subspace of $\operatorname{End}_F(\mathfrak h)$ because the Leibniz identity is linear in the operator. We then prove closure under the commutator by expanding $(D_1 \circ D_2 - D_2 \circ D_1)([u,v]_{\mathfrak h})$ and cancelling the mixed terms. The remaining Lie algebra axioms for the commutator follow from bilinearity and associativity of composition. Finally, the Jacobi identity in $\mathfrak h$ gives exactly the Leibniz rule for each adjoint map $\operatorname{ad}_x$. [/proofplan]

[step:Show that derivations form a vector subspace of $\operatorname{End}_F(\mathfrak h)$] Let $0_{\operatorname{End}}: \mathfrak h \to \mathfrak h$ denote the zero $F$-[linear map](/page/Linear%20Map). For all $u,v \in \mathfrak h$, \begin{align*} 0_{\operatorname{End}}([u,v]_{\mathfrak h}) = 0_{\mathfrak h} = [0_{\operatorname{End}}(u),v]_{\mathfrak h} + [u,0_{\operatorname{End}}(v)]_{\mathfrak h}, \end{align*} so $0_{\operatorname{End}} \in \operatorname{Der}(\mathfrak h)$. Let $D,E: \mathfrak h \to \mathfrak h$ be elements of $\operatorname{Der}(\mathfrak h)$, and let $a,b \in F$. Define the $F$-linear map $aD+bE: \mathfrak h \to \mathfrak h$ by \begin{align*} (aD+bE)(w) := aD(w)+bE(w) \end{align*} for $w \in \mathfrak h$. For all $u,v \in \mathfrak h$, using the derivation identities for $D$ and $E$ and the bilinearity of $[\cdot,\cdot]_{\mathfrak h}$, \begin{align*} (aD+bE)([u,v]_{\mathfrak h}) &= aD([u,v]_{\mathfrak h})+bE([u,v]_{\mathfrak h})\\ &= a\bigl([D(u),v]_{\mathfrak h}+[u,D(v)]_{\mathfrak h}\bigr) + b\bigl([E(u),v]_{\mathfrak h}+[u,E(v)]_{\mathfrak h}\bigr)\\ &= [aD(u)+bE(u),v]_{\mathfrak h} + [u,aD(v)+bE(v)]_{\mathfrak h}\\ &= [(aD+bE)(u),v]_{\mathfrak h} + [u,(aD+bE)(v)]_{\mathfrak h}. \end{align*} Thus $aD+bE \in \operatorname{Der}(\mathfrak h)$, and $\operatorname{Der}(\mathfrak h)$ is an $F$-vector subspace of $\operatorname{End}_F(\mathfrak h)$. [/step]

[step:Prove that the commutator of two derivations is a derivation]Let $D,E: \mathfrak h \to \mathfrak h$ be elements of $\operatorname{Der}(\mathfrak h)$. Define the commutator map $[D,E]_{\operatorname{End}}: \mathfrak h \to \mathfrak h$ by \begin{align*} [D,E]_{\operatorname{End}} := D \circ E - E \circ D. \end{align*} This map is $F$-linear because $D$, $E$, and composition of $F$-linear maps are $F$-linear. Fix $u,v \in \mathfrak h$. Applying the derivation rule for $E$ and then for $D$ gives \begin{align*} D(E([u,v]_{\mathfrak h})) &= D\bigl([E(u),v]_{\mathfrak h}+[u,E(v)]_{\mathfrak h}\bigr)\\ &= [D(E(u)),v]_{\mathfrak h} + [E(u),D(v)]_{\mathfrak h} + [D(u),E(v)]_{\mathfrak h} + [u,D(E(v))]_{\mathfrak h}. \end{align*} Similarly, \begin{align*} E(D([u,v]_{\mathfrak h})) &= E\bigl([D(u),v]_{\mathfrak h}+[u,D(v)]_{\mathfrak h}\bigr)\\ &= [E(D(u)),v]_{\mathfrak h} + [D(u),E(v)]_{\mathfrak h} + [E(u),D(v)]_{\mathfrak h} + [u,E(D(v))]_{\mathfrak h}. \end{align*} Subtracting the second identity from the first, the two mixed terms cancel: \begin{align*} [D,E]_{\operatorname{End}}([u,v]_{\mathfrak h}) &= [D(E(u))-E(D(u)),v]_{\mathfrak h} + [u,D(E(v))-E(D(v))]_{\mathfrak h}\\ &= [[D,E]_{\operatorname{End}}(u),v]_{\mathfrak h} + [u,[D,E]_{\operatorname{End}}(v)]_{\mathfrak h}. \end{align*} Therefore $[D,E]_{\operatorname{End}} \in \operatorname{Der}(\mathfrak h)$.[/step]

[guided]We must check the defining Leibniz rule for the new map $D \circ E - E \circ D$. Let $D,E: \mathfrak h \to \mathfrak h$ be derivations, and define \begin{align*} [D,E]_{\operatorname{End}}: \mathfrak h &\to \mathfrak h,\\ w &\mapsto D(E(w))-E(D(w)). \end{align*} The map is $F$-linear because it is a difference of compositions of $F$-linear maps. Now fix $u,v \in \mathfrak h$. The only point is to expand carefully enough that the cancellation is visible. Since $E$ is a derivation, \begin{align*} E([u,v]_{\mathfrak h}) = [E(u),v]_{\mathfrak h} + [u,E(v)]_{\mathfrak h}. \end{align*} Applying $D$ to this identity and using that $D$ is a derivation on each bracket term, \begin{align*} D(E([u,v]_{\mathfrak h})) &= D\bigl([E(u),v]_{\mathfrak h}+[u,E(v)]_{\mathfrak h}\bigr)\\ &= [D(E(u)),v]_{\mathfrak h} + [E(u),D(v)]_{\mathfrak h} + [D(u),E(v)]_{\mathfrak h} + [u,D(E(v))]_{\mathfrak h}. \end{align*} Interchanging the roles of $D$ and $E$ gives \begin{align*} E(D([u,v]_{\mathfrak h})) &= E\bigl([D(u),v]_{\mathfrak h}+[u,D(v)]_{\mathfrak h}\bigr)\\ &= [E(D(u)),v]_{\mathfrak h} + [D(u),E(v)]_{\mathfrak h} + [E(u),D(v)]_{\mathfrak h} + [u,E(D(v))]_{\mathfrak h}. \end{align*} Subtracting is exactly where the commutator matters. The mixed terms $[E(u),D(v)]_{\mathfrak h}$ and $[D(u),E(v)]_{\mathfrak h}$ occur once in each expansion with opposite signs, so they cancel. Hence \begin{align*} [D,E]_{\operatorname{End}}([u,v]_{\mathfrak h}) &= D(E([u,v]_{\mathfrak h}))-E(D([u,v]_{\mathfrak h}))\\ &= [D(E(u))-E(D(u)),v]_{\mathfrak h} + [u,D(E(v))-E(D(v))]_{\mathfrak h}\\ &= [[D,E]_{\operatorname{End}}(u),v]_{\mathfrak h} + [u,[D,E]_{\operatorname{End}}(v)]_{\mathfrak h}. \end{align*} This is precisely the derivation identity for $[D,E]_{\operatorname{End}}$, so the commutator of two derivations is again a derivation.[/guided]

[step:Verify the Lie algebra identities for the commutator bracket] Define \begin{align*} [\cdot,\cdot]_{\operatorname{Der}}: \operatorname{Der}(\mathfrak h) \times \operatorname{Der}(\mathfrak h) &\to \operatorname{Der}(\mathfrak h),\\ (D,E) &\mapsto D \circ E - E \circ D. \end{align*} The previous step proves that this map has codomain $\operatorname{Der}(\mathfrak h)$. Bilinearity follows from the bilinearity of addition, scalar multiplication, and composition in $\operatorname{End}_F(\mathfrak h)$. For every $D \in \operatorname{Der}(\mathfrak h)$, \begin{align*} [D,D]_{\operatorname{Der}} = D \circ D - D \circ D = 0_{\operatorname{End}}, \end{align*} so the bracket is alternating. It remains to verify the Jacobi identity. Let $D,E,G \in \operatorname{Der}(\mathfrak h)$. Using associativity of composition in $\operatorname{End}_F(\mathfrak h)$, we expand: \begin{align*} [D,[E,G]_{\operatorname{Der}}]_{\operatorname{Der}} &= D \circ E \circ G - D \circ G \circ E - E \circ G \circ D + G \circ E \circ D,\\ [E,[G,D]_{\operatorname{Der}}]_{\operatorname{Der}} &= E \circ G \circ D - E \circ D \circ G - G \circ D \circ E + D \circ G \circ E,\\ [G,[D,E]_{\operatorname{Der}}]_{\operatorname{Der}} &= G \circ D \circ E - G \circ E \circ D - D \circ E \circ G + E \circ D \circ G. \end{align*} Adding the three displayed identities cancels every composition term with the same term of opposite sign. Therefore \begin{align*} [D,[E,G]_{\operatorname{Der}}]_{\operatorname{Der}} + [E,[G,D]_{\operatorname{Der}}]_{\operatorname{Der}} + [G,[D,E]_{\operatorname{Der}}]_{\operatorname{Der}} = 0_{\operatorname{End}}. \end{align*} Thus $\operatorname{Der}(\mathfrak h)$ is a Lie algebra over $F$ under the commutator bracket. [/step]

[step:Use the Jacobi identity to prove that every adjoint map is a derivation] Fix $x \in \mathfrak h$. Define the adjoint map \begin{align*} \operatorname{ad}_x: \mathfrak h &\to \mathfrak h,\\ y &\mapsto [x,y]_{\mathfrak h}. \end{align*} The map $\operatorname{ad}_x$ is $F$-linear because the bracket $[\cdot,\cdot]_{\mathfrak h}$ is bilinear. Let $y,z \in \mathfrak h$. The Jacobi identity in $\mathfrak h$ gives \begin{align*} [x,[y,z]_{\mathfrak h}]_{\mathfrak h} + [y,[z,x]_{\mathfrak h}]_{\mathfrak h} + [z,[x,y]_{\mathfrak h}]_{\mathfrak h} = 0_{\mathfrak h}. \end{align*} Since the bracket is alternating and bilinear, it is skew-symmetric, so \begin{align*} [y,[z,x]_{\mathfrak h}]_{\mathfrak h} &= -[y,[x,z]_{\mathfrak h}]_{\mathfrak h},\\ [z,[x,y]_{\mathfrak h}]_{\mathfrak h} &= -[[x,y]_{\mathfrak h},z]_{\mathfrak h}. \end{align*} Substituting these identities into Jacobi and rearranging gives \begin{align*} \operatorname{ad}_x([y,z]_{\mathfrak h}) &= [x,[y,z]_{\mathfrak h}]_{\mathfrak h}\\ &= [[x,y]_{\mathfrak h},z]_{\mathfrak h} + [y,[x,z]_{\mathfrak h}]_{\mathfrak h}\\ &= [\operatorname{ad}_x(y),z]_{\mathfrak h} + [y,\operatorname{ad}_x(z)]_{\mathfrak h}. \end{align*} This is the derivation identity for $\operatorname{ad}_x$, hence $\operatorname{ad}_x \in \operatorname{Der}(\mathfrak h)$. The two assertions of the theorem follow. [/step]