[proofplan] The sentence $\forall x\, (x \cdot 0 = 0)$ is one of the defining recursive axioms for multiplication in Peano Arithmetic. A formal proof from $\mathrm{PA}$ may therefore consist of citing this axiom directly. Since axioms are available as derivable lines in any Hilbert-style, natural-deduction, or sequent presentation of $\mathrm{PA}$, this immediately yields $\mathrm{PA} \vdash \forall x\, (x \cdot 0 = 0)$. [/proofplan] [step:Identify the multiplication axiom with zero on the right] By the axiomatization of $\mathrm{PA}$ in the language $\mathcal{L}_{\mathrm{PA}} = \{0, S, +, \cdot\}$, multiplication is governed by the recursive axiom \begin{align*} \forall x\, (x \cdot 0 = 0). \end{align*} This is exactly the sentence whose provability is asserted in the theorem. [/step] [step:Conclude provability from availability as an axiom] Every axiom of $\mathrm{PA}$ is, by definition, available as a theorem of $\mathrm{PA}$. Since \begin{align*} \forall x\, (x \cdot 0 = 0) \end{align*} is an axiom of $\mathrm{PA}$, there is a formal derivation of this sentence from $\mathrm{PA}$ consisting of that axiom as a line. Hence \begin{align*} \mathrm{PA} \vdash \forall x\, (x \cdot 0 = 0). \end{align*} This proves the stated right zero law for multiplication in Peano Arithmetic. [/step]