so I wanted to ask if there is anything that allows for this case:
I have $3$ statements $A, B$, and $C$ and I wish to change them from the form:
$(A \operatorname{xor} B)$,
$(A \operatorname{xor} C)$
Into
$A \operatorname{xor} B \operatorname{xor} C$
Is this possible and how does one do this? Preferably with references as I wish to increase my knowledge.
-- Edit for clarity --
A little context, my friend is making a board game he wants to check that the rules won't be missing interpreted as being able to do any action $A,B$ or $C$ simultaneously.
So far a player may have access to two different cards written something like, "You may roll $3$ dice instead of $1$ this turn" and "You may double all dice rolled until the end of the turn instead of rolling $1$ dice." both of these rules change how someone takes their turn. A player will roll a die that determines far they can move, in this case actions $B$ and $C$ replace the action $A$.
$A$ $=$ "Roll $1$ die"
$B$ $=$ "Roll $3$ dice this turn" (the $3$ here is determined by some other roll)
$C$ $=$ "Double all dice rolled until the end of the turn"
The way each card has been written we get two equations.
$(A \operatorname{xor} B)$, $(A \operatorname{xor} C)$
Now the situation I want as a whole is:
$A \operatorname{xor} B \operatorname{xor} C$
Basically, with the way the rules are written currently, would someone be able to play both $B$ and $C$?
You may wish to do so, but please note that the conjunction of the first two statement is not equivalent to $A \operatorname{xor} B \operatorname{xor} C$. In fact, $(A \operatorname{xor} B)$ and $(A \operatorname{xor} C)$ don't even imply $A \operatorname{xor} B \operatorname{xor} C$
Consider: $A$ is False ($F$), and $B$ and $C$ are True ($T$).
Then $(A \operatorname{xor} B)$ and $(A \operatorname{xor} C)$ are clearly both true.
However, $A \operatorname{xor} B \operatorname{xor} C$ is False, since:
$$F \operatorname{xor} T \operatorname{xor} T = F \operatorname{xor} (T \operatorname{xor} T) = F \operatorname{xor} F = F$$
You actually have to be very careful in using expressions where you apply what is normally a binary operator to more than two operands. This may work well for operators like $\land$ and $\lor$, but it is not immediately clear it works for other operators. Indeed, it does not work for the material conditional: since $A \to (B \to C)$ is not equivalent to $(A \to B) \to C$, you can't just drop parentheses and start using $A \to B \to C$.
Of course, you say, but that is because the $\to$ is not 'symmetrical': $P \to Q$ is not equivalent to $Q \to P$. But the $\land$ and $\lor$ are, and therefore so is the $\text{xor}$. But no, that argument doesn't fly! The $\text{NAND}$ is 'symmetrical' (the technical word is commutative), but it is not associative: $A \ \text{NAND} \ (B \ \text{NAND} \ C)$ is not equivalent to $(A \ \text{NAND} \ B) \ \text{NAND} \ C$, and so we can't drop parentheses there either. So, it isn't even clear we can even drop parentheses for the $\text{XOR}$ either.
Well, it turns out the $\text{xor}$ is associative, and so we can drop parentheses. OK, but what do you think does an expression like $A \operatorname{xor} B \operatorname{xor} C$ means? Many believe that it says that exactly one of $A$, $B$,and $C$ is true ...
OK, well, that's in fact not what it means, but let's suppose that is what it means. Note that we could still not infer $A \operatorname{xor} B \operatorname{xor} C$ from $(A \operatorname{xor} B)$ and $(A \operatorname{xor} C)$, since I can use the same counterexample as before: with $A$ False and $B$ and $C$ True, we do have that exactly one of $A$ and $B$ is true, and also exactly one of $A$ and $C$ is true, but it is not the case that exactly one of $A$, $B$, and $C$ is true). Also, the other way around does not work either: If $B$ is True, and $A$ and $C$ are false, then $A \operatorname{xor} B$ is True, but $A \operatorname{xor} C$ is False.
But that's not even what the generalized $xor$ means! It turns out that a general $xor$ expression is true if and only if an odd number of its operands are true. So yes, it would be true if exactly one of the operands is true, but when you have more than two operands, there will be other situations where it is true. Consider: if $A$, $B$, and $C$ are all true, then $A \operatorname{xor} B \operatorname{xor} C$ turns out to be true as well, which is easily verified:
$$T \operatorname{xor} T \operatorname{xor} T = T \operatorname{xor} (T \operatorname{xor} T) = T \operatorname{xor} F = T$$
But note: If $A$, $B$, and $C$ are all true, then $A \operatorname{xor} B$ and $A \operatorname{xor} C$ are both false.
OK, so the moral is: you have to be real careful when it comes to generalizing binary operators to more than two operands!!