discrete math quantifier equality I came across this problem in a textbook -- I have tried to solve it completely but upon checking the answers, I realized that there might me more than one way to write this statement.
Are these two statements equivalent?
Express the statement that no one has more than three grandmothers.
G(x, y) : x is the grandmother of y
$$
\exists y ( (\exists _a \exists _b \exists _c \exists _d, (G(a, y) \land G(b, y) \land G(c, y)\land G(d, y))) \to (a=b \space \lor a=b \space \lor a=c \space \lor a=d \space \lor b=c \space \lor b=d \space \lor c=d))
$$
This is my solution. What I am trying to say is that if there exists a person y (anyone) who has four grandmothers then at least two of those grandmothers must be the same.
Is this correct?
The books solution is this: 
$$
\forall y ( \lnot \exists _a \exists _b \exists _c \exists _d, (a \neq b \space \lor a\neq b \space \lor a\neq c \space \lor a\neq d \space \lor b\neq c \space \lor b\neq d \space \lor c\neq d \land (G(a, y) \land G(b, y) \land G(c, y)\land G(d, y))
$$
I am thinking this means:
For all persons y, there does not exist four different people each of whom is the grandmother of y.
It seems as if mine is simple the negation of his statement, where $$ \neg p \to q = \neg q \land p $$...
I am not sure, would love some guidance .. thanks!
 A: I believe you are using Discrete Mathematics and Its Applications by Kenneth H. Rosen as textbook, but there's some typos in the question. The book's solution for "no one has more than three grandmothers" is actually $$\forall_y\lnot\exists_a\exists_b\exists_c\exists_d\\\big(a\ne b\land a\ne c\land a\ne d\land b\ne c\land b\ne d\land c\ne d\land\\G(a,y)\land G(b,y)\land G(c,y)\land G(d,y)\big).\ (*)$$
It means that:


For every person $y$, there do not exist four different people that each of them is a grandmother of $y$.

or

For everyone, it's impossible for him/her to have four different grandmothers.

or

No one has four different grandmothers.


And it is equivalent to $$\forall_y\forall_a\forall_b\forall_c\forall_d\\\big(G(a,y)\land G(b,y)\land G(c,y)\land G(d,y)\rightarrow\\a=b\lor a=c\lor a=d\lor b=c\lor b=d\lor c=d\big),$$ which means that:


For every person $y$, no matter who $a, b, c, d$ are, if each of them is a grandmother of $y$, then at least two of them are the same.
    (Since $y$ has at most 3 grandmothers, there must be at least one "redundant" grandmother(s) if $a, b, c, d$ are all $y$'s grandmothers.)

or

For everyone who has four grandmothers, at least one of them is "redundant".



Now let's look at your answer. If we write $$\exists_y\exists_a\exists_b\exists_c\exists_d\\\big(G(a,y)\land G(b,y)\land G(c,y)\land G(d,y)\rightarrow\\a=b\lor a=c\lor a=d\lor b=c\lor b=d\lor c=d\big),$$ which is equivalent to $$\lnot\forall_y\forall_a\forall_b\forall_c\forall_d\\\big(G(a,y)\land G(b,y)\land G(c,y)\land G(d,y)\land\\a\ne b\land a\ne c\land a\ne d\land b\ne c\land b\ne d\land c\ne d\big).\\(**)$$ Well, in my opinion, $\forall_a\forall_b\forall_c\forall_d$ should be regarded as a sampling process with put-back, so the case $C_{Dif}:=(a\ne b\land a\ne c\land a\ne d\land b\ne c\land b\ne d\land c\ne d)$ is possible. Therefore, it's impossible to ask every "sample" to be like that. We can only say $\exists_a\exists_b\exists_c\exists_dC_{Dif}$. On the other hand, $(**)$ is the negation of a contradiction, which is not what we want to express.

(Proved wrong. See Apr 14 part)
I think it's most efficient to write it as $$\forall_y\exists_a\exists_b\exists_c\forall_d\left({G(d,y)\leftrightarrow{d=a\lor d=b\lor d=c}}\right),\ (***)$$ which means that:

$d$ is $y$'s grandmother, if and only if $d$ is one of $a, b, c$.

But I haven't figured out how to prove their equivalence yet. Discussion is welcome. :)
The second part is hard to explain and I tried my best, but it still seems messy. The point is that $\rightarrow$ shouldn't be used with $\exists$.

Added on Apr. 14th, 2019:
I just figured out that $(***)\nLeftrightarrow(*)$. It in addition implies that $y$ has at least one grandmother, which is not expected. But if we write $$\forall_y\exists_a\exists_b\exists_c\forall_d\left({G(d,y)\rightarrow{d=a\lor d=b\lor d=c}}\right),$$ then it's completely equivalent to $(*)$, which can be proved by expanding quantifiers into conjunctions and disjunctions. :)
A: Your quantifiers aren’t quite right for what you’re trying to say. You can see that there’s a problem if you notice that the $a,b,c$, and $d$ to the right of the implication are not within the scopes of the corresponding quantifiers. Also, starting with $\exists y$ means that even if the rest were correct, you’d only be saying that there is at least one person who has no more than three grandmothers. What you were trying for, I think, is this version:
$$\forall y\,\forall a\,\forall b\,\forall c\,\forall d\Big(\big(G(a,y)\land G(b,y)\land G(c,y)\land G(d,y)\big)\to\varphi\Big)\;,$$
where $\varphi$ is
$$a=b\lor a=c\lor a=d\lor b=c\lor b=d\lor c=d\;.$$
This is logically equivalent to the book’s answer.
