명제의 증명과 절대부등식

Conjecture: For all integers $a$, $b$, and $c$ with $a \ne 0$, if $a$ divides $b$ and $a$ divides $c$, then $a$ divides $b + c$.

A Know-show table for a proof of the conjecture in Part (3).

| Step | Know | Reason | | --- | --- | --- | | $P$ | $a \mid b$ and $a \mid c$ | Hypothesis | | $P1$ | $\left( {\exists s \in \mathbb{Z}} \right) \left( {b = a \cdot s} \right)$ $\left( {\exists t \in \mathbb{Z}} \right)\left( {c = a \cdot t} \right)$ | Definition of “divides” | | $P2$ | $b + c = as + at$ | Substituting for $b$ and $c$ | | $P3$ | $b + c = a\left( {s + t} \right)$ | Distributive property | | $Q1$ | $s + t$ is an integer | $\mathbb{Z}$ is closed under addition | | $Q$ | $\left. {a } \right\| \left( {b + c} \right)$ | Definition of “divides” | | Step | Show | Reason |