I have a question about representing family relationships in first order logic.
So to translate the sentence x is a sibling of y, I would need to set up the following relation symbols:
P(x,y) iff x is a parent of y =(x,y) iff x and y are the same
The sentence would therefore be:
∃w (P(w,x)&P(w,y)&~=(x,y))
However, to translate the sentence x is the brother of y, I would need to add a new 1-ary relation symbol:
M(x) iff x is a male
The sentence would therefore be:
∃w∃x (P(w,x)&P(w,y)&~=(x,y)&M(x))
So my question is: Would x in this sentence be governed by the existential quantifier? Or would it be governed by the universal quantifier instead?
Thanks in advance!
In your first formula
$\exists w (P(w,x) \land P(w,y) \land \neg (x=y)))$
only the variable $w$ is bound by the existential quantifier $\exists$, whereas the variables $x$ and $y$ are free.
This means that the formula is not a sentence, but only a formula, i.e. its truth value depends on the variable assignment: Depending on which individuals you assign to $x$ and $y$, you will get different statements and therefore different truth values. This is not the case for sentences (in which all variables are bound), as the current variable assignment gets "overridden" by the quantification.
In any case, it would be good to add $B(x,y) \leftrightarrow$ to the beginning of the formula to make the formula itself a definition (an equivalence), rather than having the predicate you want to define excluded from the predicate logic formula. If you want to apply univeral closure (as I explain in 2)), this part needs to go inside the formula in order to get the variables in the $B$ predicate into the scope of the univeral quantification.
You have two options now:
1) Leave the variables $x$ and $y$ unbound. In this case, the resulting formula expressing "x is the brother of y" will be
$B(x,y) \leftrightarrow \exists w (P(w,x) \land P(w,y) \land \neg (x=y) \land M(x))$
In this case, both $x$ and $y$ remain unquantified.
2) Build the universal closure of the formla, i.e. prepend a univeral quantifier $\forall$ for any variable that is still free. By this, you get
$\forall x \forall y (B(x,y) \leftrightarrow \exists w (P(w,x) \land P(w,y) \land \neg (x=y) \land M(x)))$
Since the variables $x$ and $y$ are no longer free, this formula is a sentence, i.e. independent of the chosen variable assignment.
This qualifies your formula for an axiom and is usually the desired variant, because in the context of statements about truth, you'd want to deal with sentences rather than just formulas.
The point of confusion in what you wrote is that in creating the final formula, you (existentially) quantified over $x$ but not over $y$. For uniformness, you either have to leave $x$ unquantified or quantify over both variables, but then you should do so univerally (with $\forall$) because the definition applies to all individuals, not only to some of them:
You want to say "For all individuals $x$ and $y$ it holds that $x$ is the brother of $y$ if and only if there is a parent $w$ such that ...", rather than "There is a (parent) individual $w$ such that there is a (brother) individual $x$..." (which would be what you wrote).
A small remark: You defined $=(x,y)$ as a predicate.
This is not so nice because the equality symbol $=$ is normally assumed to be part of the logical symbols of any language of predicate logic already, so you shouldn't define a non-logical relation symbol that looks the same as one of the already existing logical symbols.
Instead, you should invent an own predicate name, like "$is(x,y)$" - but this is unnecessary because the default symbol $=$ already serves to express identiy; therefore, I'd just leave this additonal predicate definition away and go with $x=y$ instead.