I was asked to express the following using existential and universal quantifiers: "There is at least one and at most two students in this class who have sent email to every student in the class."
We were given the prepositional function $S(x,y)$ which is true if $x$ sent an email to $y$, and false otherwise. Where $x$ and $y$ are part of the domain of all students in the class.
The answer I produced was: $∃x∃y∀c( S(x,c) ∧ S(y,c) ) ∧ ∃x∃y∃z∀m( (S(x,m) ∧ S(y,m) ∧ S(z,m) → ((z=x) ∨ (z=y)) )$.
Through the above statement I tried to communicate the idea that: That there are students x and y s.t. they sent an email to every student in the class, and there is a student c s.t. c sent an email to every student in the class and is equal to either x or y.
I was told this was wrong and that the right answer is: $∃x[∀yS(x,y) ∧ ∀u∀v(u=x ∨ v=x ∨ u=v ∨ ∃y(¬S(u,y) ∨ ¬S(v,y)))] $
This is the equivalent of saying: "I can find one student who has sent an email to every student, but there cannot be a pair of students, both of which are different from the original student I picked and from each other, and both of which have sent an email to every student."
Can anyone tell me where I went wrong?
There are always many ways to express the desired concept, the given solution is just one of them, see below. So in principle, it could be the case that both your attempt and the given solution are correct. However, this i snot the case
Your attempt is of the form $\Phi\land\Psi$, where the first part $$\Phi:\equiv∃x∃y∀c( S(x,c) ∧ S(y,c) ) $$ can be simplified to $$∃x∀c( S(x,c) ) $$ (and so stands for "There is at least one student who ...") because nothing prevents $x=y$. The main problem is with your second part $$ \Psi:\equiv∃x∃y∃z∀m( (S(x,m) ∧ S(y,m) ∧ S(z,m) → ((c=x) ∨ (c=y)) ) $$ because it has a free $c$ in it. I suppose you wanted $m$ there (and an extra closing parenthesis), i.e., $$ ∃x∃y∃z∀m( (S(x,m) ∧ S(y,m) ∧ S(z,m)) → ((m=x) ∨ (m=y)) ) $$ However, this expresses nothing really helpful. Indeed, this $\Psi$ is satisfied for example if there exists a student who sent no mail at all. Thus your attempt would also be satisfied by four students $A,B,C,D$ where $A,B,C$ write to all students and $D$ writes to no one. This is not supposed to be the case.
Fixing your idea: I like your first part however because it allows $x,y$ to be either different and be the two writing-to-all students or $x=y$ is the one writing-to-all student. What is missing, however, is to express (within the scope of the same $\exists x\exists y$!) that not other apart from these two (or this one) is a writing-to-all student; that is: if $z$ is a writing-to-all student then $z=x$ or $z=y$: $$\forall z((\forall m S(z,m))\to (z=x\lor z=y)) $$ so in full (recycling the $c$ for $z$): $$∃x∃y∀c( S(x,c) ∧ S(y,c) \land ((\forall m S(c,m)) \to (c=x\lor c=y)))$$
I find my solution easier to read because it is shorter and less convoluted.
In addition, if you prefer all quantifiers at the beginning, you can transform mine to
$$∃x∃y∀c\exists m( S(x,c) ∧ S(y,c) \land (S(c,m) \to (c=x\lor c=y)))$$