Let $a, b, c \in \mathbb Z$. I know that if $c|a$ and $c|b$, then $c|a \pm b$. I was working on some elementary number theory and I began to wonder if the following were true:$$\text{if }c|a \text{ and } c|a+b \text{, then } c|b$$
I managed a very simple proof:
Suppose $c|a$ and $c|a+b$. It follows that $a=c \cdot d$ and $a+b=c \cdot d'$, where $d, d' \in \mathbb Z$. Note that if $a=c \cdot d$, then $c \cdot d +b=c \cdot d'$. Hence $b=c \cdot d' -c \cdot d$, or $b=c(d-d')$. Letting $d''=d-d' \in \mathbb Z$, we have $b=c \cdot d''$, or $c|b$, as desired.
However, while trying to prove this the good ol' fashioned way (using propositional logic), I'm hitting a barrier that has me second-guessing if my above proof is correct or not. For the life of me, I cannot fathom how $$(P \land Q) \Rightarrow R \equiv (P \land R) \Rightarrow Q$$ for $P:=c|a$, $Q:=c|b$ and $R:=c|a+b$.
Sorry for the silly question. My propositional logic is a little rusty and I'm probably just overlooking something. Any insight would be greatly appreciated!
Your number theory proof is correct, but your proposed equivalence in propositioal logic is (in general) not correct, so you can't use it to proof what you want just with your initial knowledge and "propositional magic", you need actual number theory for that.
To see that the propositional formula is generally incorrect, set for example
$$P:=2|x, Q:=4|x, R=2|x$$
Of course, I may have misunderstood what you wanted to achieve with your propositional formula. If that's true, please explain.