[proofplan] We use the additive characterization of the usual strict order on $\mathbb{N}$: $x<y$ means that $y=x+k$ for some positive natural increment $k \in \mathbb{N}$. After choosing this increment for $a<b$, the addition statement follows by rewriting $b+c$ as $(a+c)+k$. The multiplication statement follows by distributing $c$ across $b=a+k$, giving $bc=ac+kc$, and then observing that $kc$ is again a positive natural increment. [/proofplan] [step:Extract the positive natural increment from $a<b$] Since $a<b$ in the usual strict order on $\mathbb{N}$, there exists $k \in \mathbb{N}$ such that \begin{align*} b=a+k. \end{align*} Fix such a [natural number](/page/Natural%20Number) $k$. Because $\mathbb{N}=\{1,2,3,\dots\}$, this increment is positive. [/step] [step:Rewrite $b+c$ as $(a+c)+k$] Let $c \in \mathbb{N}$. Using $b=a+k$ and the associativity and commutativity of addition from [citetheorem:9514], we compute \begin{align*} b+c=(a+k)+c. \end{align*} By associativity, \begin{align*} (a+k)+c=a+(k+c). \end{align*} By commutativity of addition, $k+c=c+k$, and hence \begin{align*} a+(k+c)=a+(c+k). \end{align*} By associativity again, \begin{align*} a+(c+k)=(a+c)+k. \end{align*} Therefore \begin{align*} b+c=(a+c)+k. \end{align*} Since $k \in \mathbb{N}$ is a positive natural increment, the additive characterization of the strict natural order gives \begin{align*} a+c<b+c. \end{align*} [guided] Fix $c \in \mathbb{N}$. The goal is to prove $a+c<b+c$. For the usual strict order on $\mathbb{N}$, this means we must exhibit a positive natural number which, when added to $a+c$, gives $b+c$. From the previous step we have chosen $k \in \mathbb{N}$ with $b=a+k$. Substitute this expression for $b$ into $b+c$: \begin{align*} b+c=(a+k)+c. \end{align*} Now we want the expression to have the form $(a+c)+(\text{positive increment})$. The arithmetic laws from [citetheorem:9514] allow us to reassociate and commute the summands. First, associativity gives \begin{align*} (a+k)+c=a+(k+c). \end{align*} Next, commutativity of addition gives $k+c=c+k$, so \begin{align*} a+(k+c)=a+(c+k). \end{align*} Finally, associativity gives \begin{align*} a+(c+k)=(a+c)+k. \end{align*} Combining these equalities yields \begin{align*} b+c=(a+c)+k. \end{align*} Since $k \in \mathbb{N}$, it is a positive natural number. Thus $b+c$ is obtained from $a+c$ by adding the positive natural increment $k$, and the additive characterization of the usual strict order gives \begin{align*} a+c<b+c. \end{align*} [/guided] [/step] [step:Rewrite $bc$ as $ac$ plus a positive natural increment] Let $c \in \mathbb{N}$ satisfy $c>0$. Since $\mathbb{N}=\{1,2,3,\dots\}$, this positivity condition is automatic, but we keep it in force. Using $b=a+k$ and distributivity from [citetheorem:9514], we obtain \begin{align*} bc=(a+k)c=ac+kc. \end{align*} Because $k \in \mathbb{N}$ and $c \in \mathbb{N}$, closure of multiplication in $\mathbb{N}$ gives $kc \in \mathbb{N}$. Hence $kc$ is a positive natural increment. Therefore $bc=ac+kc$ implies, by the additive characterization of the strict natural order, \begin{align*} ac<bc. \end{align*} This proves both compatibility assertions. [/step]