[proofplan] We prove the formula $\varphi(x)$ defined by $\varphi(x) :\equiv 0\cdot x=0$ by the induction axiom schema of PA. The base case is the defining multiplication axiom $a\cdot 0=0$ instantiated at $a=0$. For the successor step, the recursive multiplication axiom rewrites $0\cdot Sx$ as $(0\cdot x)+0$, and the induction hypothesis together with the addition axiom $y+0=y$ reduces this term to $0$. [/proofplan] [step:Define the induction formula] Let $\varphi(x)$ be the formula of the language of PA given by \begin{align*} \varphi(x) \;:\equiv\; 0\cdot x=0. \end{align*} To prove $\forall x\,\varphi(x)$ in PA, it is enough, by the induction axiom schema, to prove the two PA-derivable formulas $\varphi(0)$ and $\forall x\,(\varphi(x)\implies \varphi(Sx))$. [/step] [step:Prove the base case from the multiplication zero axiom] The PA axiom for multiplication by zero is \begin{align*} \forall a\,(a\cdot 0=0). \end{align*} Instantiating this axiom with $a=0$ gives \begin{align*} 0\cdot 0=0. \end{align*} This is exactly $\varphi(0)$. [/step] [step:Derive the successor case using the recursive multiplication axiom] We prove in PA that $\forall x\,(\varphi(x)\implies \varphi(Sx))$. Let $x$ be arbitrary and assume the induction hypothesis \begin{align*} 0\cdot x=0. \end{align*} The recursive PA axiom for multiplication is \begin{align*} \forall a\,\forall b\,(a\cdot S b=a\cdot b+a). \end{align*} Instantiating it with $a=0$ and $b=x$ gives \begin{align*} 0\cdot Sx=(0\cdot x)+0. \end{align*} By equality substitution applied to the induction hypothesis $0\cdot x=0$, we obtain \begin{align*} (0\cdot x)+0=0+0. \end{align*} The PA axiom for right zero of addition is \begin{align*} \forall y\,(y+0=y). \end{align*} Instantiating it with $y=0$ gives \begin{align*} 0+0=0. \end{align*} By transitivity of equality, \begin{align*} 0\cdot Sx=0. \end{align*} This is $\varphi(Sx)$. Since $x$ was arbitrary, PA proves $\forall x\,(\varphi(x)\implies \varphi(Sx))$. [/step] [step:Apply induction to obtain the universal statement] From the base case $\varphi(0)$ and the successor implication $\forall x\,(\varphi(x)\implies \varphi(Sx))$, the induction axiom schema of PA yields \begin{align*} \forall x\,\varphi(x). \end{align*} Substituting the definition of $\varphi$ gives \begin{align*} \forall x\,(0\cdot x=0). \end{align*} Therefore PA proves $\forall x\,(0\cdot x=0)$. [/step]