Two tips: 1) It sometimes helps to rephrase the sentence into an equivalent English-sentence that looks easier to analyze. 2) Often times, you can break down the sentence to make it easier to parse. If you have trouble wrapping your head around the sentence, try phrasing it in a slightly more suggestive way. For instance:
"Every grandparent is such that either they have only daughters, or they have exactly two sons, or they have no children."
In general, "Every $\varphi$ is such that $\psi$" gets translated into the predicate calculus as $\forall x (\varphi(x) \rightarrow \psi(x))$. Your $\varphi(x)$ here is "$x$ is a grandparent", whereas your $\psi(x)$ is "$x$ either has... (etc.)". So overall, the translation should look like this:
$\forall x(x \text{ is a grandparent} \rightarrow x \text{ either has only daughters, or exactly two sons, or is childless})$
So if you can figure out how to say "$x$ is a grandparent" and "$x$ either has only daughters, or has exactly two sons, or is childless", then you'll know how to translate the sentence.
How do you say "$x$ is a grandparent"? Basically, it amounts to saying that $x$ has some child, who also has some (other) child. So this just amounts to $\exists y (C(y,x) \wedge \exists z(C(z,y)))$. This formula (which has $x$ free btw) is your $\varphi(x)$, which goes in the antecedent of the conditional of your universally quantified sentence.
How do you say "$x$ either has only daughters, or exactly two sons, or is childless"? Well, it seems to be a disjunction about $x$, so split it up into cases: if you know the whole thing is a disjunction, you can tackle each disjunct separately and then put it all together with $\vee$s at the end. So you just need to analyze "$x$ has only daughters", "$x$ has exactly two sons", and "$x$ is childless". Hopefully, things are clear enough that you can do these on your own.
Yes, you can prove it with "usual" proof systems : Resolution, Tableaux or with Natural Deduction :
1) $∀z∀x∃y(R(z,y)∨¬P(x))$ --- premise
2) $∃xP(x)$ --- assumed [a]
3) $P(w)$ --- assumed [b] for $\exists$-elimination
4) $∃y(R(z,y)∨¬P(w))$ --- from 1) by $\forall$-elimination twice
5) $R(z,y)∨¬P(w)$ --- assumed [c] for $\exists$-elimination
6) $P(w) \to R(z,y)$ --- from 5) by tautological equivalence : $(p \to q) \leftrightarrow (\lnot p \lor q)$
7) $R(z,y)$ --- from 3) and 6) by $\to$-elimination
8) $\exists yR(z,y)$ --- from 7) by $\exists$-introduction
9) $\forall x \exists yR(x,y)$ --- from 8) by $\forall$-introduction : $x$ is not free in any "open" assumptions
$y$ is not free in 9); thus, we can apply $\exists$-elimination with 4), 5) and 9) and conclude with :
10) $\forall x \exists yR(x,y)$, discharging assumption [c].
In the same way, from 2), 3) and 10), discharging assumption [b] and concluding with :
11) $\forall x \exists yR(x,y)$.
Finally :
$∃xP(x) \to \forall x \exists yR(x,y)$ --- from 2) and 11) by $\to$-introduction, discharging assumption [a].
With a final application of $\to$-introduction we conclude with :
$∀z∀x∃y(R(z,y)∨¬P(x)) \to (∃xP(x) \to ∀x∃yR(x,y))$.
Best Answer
First, please allow me to add some parentheses to properly indicate the scope of the quantifiers. Also, I cannot stand to see the symbol $\Rightarrow$ being used for the material conditional, since that is more often used for logical implication ... for material implication please use the $\to$. OK, so we have:
$$ \neg\exists x \forall y (S(x) \to (L(y) \land B(x,y))) $$
Well, this is not correct. If you bring the negation inside, you see that this is equivalent to:
$$ \forall x \exists y\ \neg (S(x) \to (L(y) \land B(x,y))) $$
and thus to:
$$ \forall x \exists y\ (S(x) \land \neg (L(y) \land B(x,y))) $$
and, bringing in the $\exists y$:
$$ \forall x \ (S(x) \land \exists y \neg (L(y) \land B(x,y))) $$
... meaning (among other things) that everything in the universe is a student!
OK, so how about your:
Again, I'll rewrite this as:
$$ \neg\exists x \forall y ((S(x) \land L(y)) \to B(x,y)) $$
Now, once again, let's bring in the negation:
$$ \forall x \exists y \ \neg ((S(x) \land L(y)) \to B(x,y)) $$
and thus:
$$ \forall x \exists y \ ((S(x) \land L(y)) \land \neg B(x,y)) $$
or simply:
$$ \forall x \exists y \ (S(x) \land L(y) \land \neg B(x,y)) $$
and thus:
$$ \forall x \ (S(x) \land \exists y (L(y) \land \neg B(x,y))) $$
So, once again this can only be true if (among other things) everything in the universe is a student! OK, so that's not right either.
However, the given answer:
is indeed correct. If we bring in the $\forall y$, we get:
$$\neg \exists x (S(x) \land \forall y(L(y) \to B(x,y))$$
which is now a bit more easily seen as saying that there is no student such that for every lecture, the student attended that lecture.
In fact, if we start with the original and bring in the negation, we get:
$$\forall x\exists y \neg (S(x) \land (L(y) \to B(x, y))$$
which is equivalent to:
$$\forall x\exists y (S(x) \to \neg (L(y) \to B(x, y))$$
thus to:
$$\forall x\exists y (S(x) \to (L(y) \land \neg B(x, y))$$
and thus to:
$$\forall x (S(x) \to \exists y (L(y) \land \neg B(x, y))$$
which translates back as: 'for every student there is a lecture that they did not attend' ... which is exactly what we want!
(By the way: do you see the advantage of not always having all quantifiers on the outside when doing translation? You often get a much more natural reading of the sentence. Thus, while having all quantifiers on the outside has some practical and theoretical importance for formal logic, when it comes to symbolization, I much prefer to place the quantifiers in the 'proper' place)
Notice how you can clearly see the difference between your Solution 2 and the Given solution at the very end of their transformation. Again, your Solution 2 is equivalent to:
$$ \forall x \ (S(x) \land \exists y (L(y) \land \neg B(x,y))) $$
which says: 'Everything is a student and ...'
while the correct solution is equivalent to:
$$\forall x (S(x) \to \exists y (L(y) \land \neg B(x, y))$$
which says: 'if something is a student, then ...'
This is how the Correct solution is a claim about all students, while your solution 2 (and 1) is a claim about everything (and forces everything to be a student!)