[proofplan] We work entirely in the $t$-adic formal [power series](/page/Power%20Series) ring $R[[t]]$. The complete series is defined as the exponential of a power-sum exponent, so the rules for alphabet addition and subtraction turn the exponent for $A+B$ into a sum and the exponent for $A-B$ into a difference. The formal exponential converts sums of commuting series into products, and $H[B;t]$ is invertible because its constant term is $1$. [/proofplan] [step:Introduce the formal exponents for the complete series] For each alphabet expression $X \in \{A,B,A+B,A-B\}$, define \begin{align*} E_X: \{t\text{-adic formal power series variable}\} &\to tR[[t]] \\ t &\mapsto \sum_{r=1}^{\infty}\frac{p_r[X]t^r}{r}. \end{align*} This is an element of $tR[[t]]$, since every term has positive $t$-degree. By definition of the complete symmetric generating series, \begin{align*} H[X;t] = \exp(E_X(t)). \end{align*} The formal exponential is well-defined on $tR[[t]]$ because, for each fixed degree in $t$, only finitely many powers of $E_X(t)$ can contribute. [/step] [step:Convert alphabet addition into addition of exponents] Using the defining rule $p_r[A+B]=p_r[A]+p_r[B]$ for every $r \geq 1$, we compute in $tR[[t]]$: \begin{align*} E_{A+B}(t) &= \sum_{r=1}^{\infty}\frac{p_r[A+B]t^r}{r} \\ &= \sum_{r=1}^{\infty}\frac{(p_r[A]+p_r[B])t^r}{r} \\ &= \sum_{r=1}^{\infty}\frac{p_r[A]t^r}{r} + \sum_{r=1}^{\infty}\frac{p_r[B]t^r}{r} \\ &= E_A(t)+E_B(t). \end{align*} Since $R[[t]]$ is commutative, the formal identity $\exp(F+G)=\exp(F)\exp(G)$ applies to $F=E_A(t)$ and $G=E_B(t)$. Therefore \begin{align*} H[A+B;t] &= \exp(E_{A+B}(t)) \\ &= \exp(E_A(t)+E_B(t)) \\ &= \exp(E_A(t))\exp(E_B(t)) \\ &= H[A;t]H[B;t]. \end{align*} [guided] The point of introducing $E_X(t)$ is that the complete series is multiplicative only after we pass through the exponential form. For the sum alphabet $A+B$, the defining plethystic rule says that each power sum splits additively: \begin{align*} p_r[A+B] = p_r[A]+p_r[B]. \end{align*} Substituting this into the exponent gives \begin{align*} E_{A+B}(t) &= \sum_{r=1}^{\infty}\frac{p_r[A+B]t^r}{r} \\ &= \sum_{r=1}^{\infty}\frac{(p_r[A]+p_r[B])t^r}{r}. \end{align*} Because addition of formal power series is coefficientwise, we may separate the two sums: \begin{align*} E_{A+B}(t) &= \sum_{r=1}^{\infty}\frac{p_r[A]t^r}{r} + \sum_{r=1}^{\infty}\frac{p_r[B]t^r}{r} \\ &= E_A(t)+E_B(t). \end{align*} Now both $E_A(t)$ and $E_B(t)$ lie in $tR[[t]]$, and $R[[t]]$ is commutative. Hence the formal exponential satisfies \begin{align*} \exp(E_A(t)+E_B(t))=\exp(E_A(t))\exp(E_B(t)). \end{align*} This identity is valid formally because each coefficient of $t^n$ depends on only finitely many products of coefficients from the two exponents. Therefore \begin{align*} H[A+B;t] &= \exp(E_{A+B}(t)) \\ &= \exp(E_A(t)+E_B(t)) \\ &= \exp(E_A(t))\exp(E_B(t)) \\ &= H[A;t]H[B;t]. \end{align*} [/guided] [/step] [step:Convert alphabet subtraction into subtraction of exponents] Using the defining rule $p_r[A-B]=p_r[A]-p_r[B]$ for every $r \geq 1$, we compute: \begin{align*} E_{A-B}(t) &= \sum_{r=1}^{\infty}\frac{p_r[A-B]t^r}{r} \\ &= \sum_{r=1}^{\infty}\frac{(p_r[A]-p_r[B])t^r}{r} \\ &= E_A(t)-E_B(t). \end{align*} Thus \begin{align*} H[A-B;t] &= \exp(E_{A-B}(t)) \\ &= \exp(E_A(t)-E_B(t)) \\ &= \exp(E_A(t))\exp(-E_B(t)). \end{align*} Since $E_B(t) \in tR[[t]]$, the series $\exp(E_B(t))$ has constant term $1$ and is therefore invertible in $R[[t]]$, with inverse $\exp(-E_B(t))$. Hence \begin{align*} H[A-B;t] &= \exp(E_A(t))\exp(E_B(t))^{-1} \\ &= \frac{H[A;t]}{H[B;t]}. \end{align*} [/step] [step:Conclude both identities in the completed formal series ring] The preceding two steps prove \begin{align*} H[A+B;t] &= H[A;t]H[B;t], \\ H[A-B;t] &= \frac{H[A;t]}{H[B;t]}. \end{align*} All manipulations took place in $R[[t]]$, where each coefficient of a formal exponential or product is determined by finitely many terms. Therefore both displayed formulas are identities of formal power series in $R[[t]]$. [/step]