= Solution
This part proves $(c)\Rightarrow(a)$. Write the <Weyl algebra> generators as $x_1,\ldots,x_n,D_1,\ldots,D_n$, with $[D_i,x_j]=\delta_{ij}$ and the other generator <commutators> zero. The ordered <monomials> $x^\alpha D^\beta$ form a <basis> over any <field>: one may construct the <algebra> by successively adjoining the commuting derivations $D_i$ to $k[x_1,\ldots,x_n]$, with multiplication $D_if=fD_i+\partial f/\partial x_i$, obtaining unique ordered forms. This also makes $A_n(k)$ infinite-dimensional for $n\ge1$.
Assume $\operatorname{char}k=0$ and let $I$ be a nonzero two-sided <ideal>. Choose a nonzero element $a=\sum c_{\alpha\beta}x^\alpha D^\beta\in I$, and a nonzero term of maximal total degree, say $(\alpha_0,\beta_0)$. The relations give
$$
[D_i,x^\alpha D^\beta]=\alpha_i x^{\alpha-e_i}D^\beta,\qquad
[x_i,x^\alpha D^\beta]=-\beta_i x^\alpha D^{\beta-e_i}.
$$
Apply $(\operatorname{ad}D)^{\alpha_0}(\operatorname{ad}x)^{\beta_0}$ to $a$. Every lower-degree term vanishes. A term of the same degree survives only if all its exponents dominate $(\alpha_0,\beta_0)$, which forces equality. Thus the resulting element of $I$ is
$$
(-1)^{|\beta_0|}\alpha_0!\beta_0!c_{\alpha_0\beta_0}\ne0.
$$
The factorials are nonzero in characteristic zero. This is a nonzero <scalar>, so $1\in I$ and $I=A_n(k)$. Therefore \b[$A_n(k)$ is simple in characteristic zero]. This <scalar extraction by iterated Weyl commutators> supplies the full simplicity proof.
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