The distributive property of division over addition reads
$\forall x\in\mathbb{R}, \forall y\in\mathbb{R}, \forall z\in\mathbb{R}\setminus\{0\} : \frac{x+y}{z}=\frac{x}{z}+\frac{y}{z}$
We know that $\forall w\in A:p(w)$ is equivalent to $\forall w: w\in A\implies p(w)$. Using this, we have
$\forall z\in\mathbb{R}\setminus\{0\} : \frac{x+y}{z}=\frac{x}{z}+\frac{y}{z} \iff \forall z: z\in\mathbb{R}\setminus\{0\} \implies\frac{x+y}{z}=\frac{x}{z}+\frac{y}{z}$
$\iff \forall z: z\in\mathbb{R}\land z\notin \{0\} \implies\frac{x+y}{z}=\frac{x}{z}+\frac{y}{z}$
Next, Im going to use the exportation identity: $(P\land Q) \implies R$ is equivalent to $P\implies(Q\implies R)$
$\iff \forall z:z\in\mathbb{R}\implies\left( z\notin \{0\} \implies\frac{x+y}{z}=\frac{x}{z}+\frac{y}{z}\right)$
$\iff \forall z\in\mathbb{R} :z\notin \{0\} \implies\frac{x+y}{z}=\frac{x}{z}+\frac{y}{z}$
We get that the distributive property of division over addition is
$\forall x\in\mathbb{R}, \forall y\in\mathbb{R}, \forall z\in\mathbb{R}:z\notin \{0\} \implies\frac{x+y}{z}=\frac{x}{z}+\frac{y}{z}$
$\forall x\in\mathbb{R}, \forall y\in\mathbb{R}, \forall z\in\mathbb{R}:z\neq 0\implies\frac{x+y}{z}=\frac{x}{z}+\frac{y}{z}$
Now, the implication $q(x,y,z)$ given by
$z\neq 0\implies\frac{x+y}{z}=\frac{x}{z}+\frac{y}{z}$
must be true for all real values. I want to check that.
For $x=1,y=1,z=1$
$1\neq 0$ is true, and $\frac{1+1}{1}=\frac{1}{1}+\frac{1}{1}$ is true, so the implication is true.
The problem is for cases like $x=1,y=1,z=0$
$0\neq 0$ is false, and I dont know the truth value of $\frac{1+1}{0}=\frac{1}{0}+\frac{1}{0}$. Of course, I know that false implies true is true and that false implies true is true. So $\frac{1+1}{0}=\frac{1}{0}+\frac{1}{0}$ should be false or true (if $q(1,1,0)$ is true).
Note that if $\frac{1+1}{0}=\frac{1}{0}+\frac{1}{0}$ is false, then $\frac{1+1}{0}\neq\frac{1}{0}+\frac{1}{0}$ is true.
There is something else I tried, the contrapositive. $q(x,y,z)$ is equivalent to $q'(x,y,z)$ given by
$\frac{x+y}{z}\neq\frac{x}{z}+\frac{y}{z}\implies z=0$
Here I have the similar problem for $x=1,y=1,z=0$. I want to verify that $q'(1,1,0)$ is true. I dont know truth value of $\frac{1+1}{0}\neq\frac{1}{0}+\frac{1}{0}$, but I know that $0=0$ is true.
Is there some error in my reasoning?
what's going on with $\frac{1+1}{0}=\frac{1}{0}+\frac{1}{0}$? true? false?
what's going on with $\frac{1+1}{0}\neq\frac{1}{0}+\frac{1}{0}$? true? false?
Are $q(x,y,z)$ and $q'(x,y,z)$ equivalent?
Is the proposition
$\forall x\in\mathbb{R}, \forall y\in\mathbb{R}, \forall z\in\mathbb{R}:z\neq 0\implies\frac{x+y}{z}=\frac{x}{z}+\frac{y}{z}$
true?
The way I think about it is that the notation $x/y$ is "defined conditionally" with the condition $y\ne0$. This means whenever we use this notation, it must be clear that $y\ne0$ or that the value of the expression $x/y$ ultimately doesn't affect the value of some surrounding expression in the case $y=0$. (Technically we might define $x/0$ to be some dummy value like the empty set, if we're using a formal system that requires this.)
This is the case in your examples because if $P$ is false, then $P\implies Q$ is true regardless of $Q$, and if $Q$ is true, then $P\implies Q$ is true regardless of $P$.