Regarding definite quantification in predicate logic, these are the final two sentences that I am struggling to symbolise:
"Ada hates at least two clever persons"
and
"The only clever one is Ada"
(Where a: Ada, Hxy: x hates y, Cx: x is clever)
I have translated them to:
Ada hates at least two clever persons: ∃x∃y((Cx^Cy)^x≠y)^∀x∀y(Hax^Hay)
Broken Down: There are at least two clever persons: ∃x∃y((Cx^Cy)^x≠y) ...and Ada hates them: ^∀x∀y(Hax^Hay)
The only clever one is Ada: (∃x)(Cx^∀y(Cy→x=y)∀x(x=a))
I believe that my main problems are in ordering the sentences and knowing where it is appropriate to add quantifiers. For example, I am unsure whether it was necessary for me to use the universal quantifiers: ∀x∀y(Hax^Hay) (in the first sentence) and ∀x(x=a)) (in the second sentence).
I would appreciate any help clarify this, thank you.
(I am beginner in logic so please don't judge my attempt too harshly!)
Both are not quite correct and indeed your mistakes are at the quantifiers.
Statement one:
First, $∃x∃y((Cx\wedge Cy)\wedge x≠y)$ is completely fine (assuming you're allowed the $≠$) and it exactly means "there are at least two clever persons.
However, $∀x∀y(Hax\wedge Hay)$ is not correct. This statement means "Ada hates everyone". Or technically even "for every pair of two people, Ada hates both of them", as you have two all quantifiers. Note that those $\forall$ are not in the scope of the $\exists$ anymore. The $x$ and the $y$ in the $\forall$ and $\exists$ statements have nothing to do with each other. Therefore your first part only says something about clever persons and your second part only says something about who Ada hates. Connecting them with an AND does not imply that the groups you described in the two parts (clever people / people that Ada hates) coincide or even have any members in common.
So now for the correct statement: There are at least two clever people that Ada hates. $∃x∃y((Cx\wedge Cy)\wedge x≠y)$ as said states that there are two persons that are clever. Now we must make sure to model that it's exactly those two people that Ada hates and to ensure that we must make another statement about $x$ and $y$, that must be in the scope of $\exists$ to make sure we really mean the two clever persons with $x$ and $y$. We do that by simply saying:
$∃x∃y((Cx\wedge Cy)\wedge x≠y \wedge Hax\wedge Hay)$.
"There exist to people, such that both are clever, they are not the same person and Ada hates both of them."
Note that you do not need (and in fact can't have) and $\forall$, because you don't want to say something about all persons with certain traits.
Statement two:
(I think you missed a $\wedge$?) $∃x(Cx\wedge ∀y(Cy→x=y)\wedge ∀x(x=a)$
Never nest quantifiers over the same variable. I think depending on who you ask, it's not even well defined. Something like $\exists x \forall x$ would mean "You can pick an $x$, such that for all $x$..." which does not make any sense. What would "For all $x$" mean here? You just picked one specific $x$. There might be some definitions which allow this, but then both quantifiers would mean different $x$ which gets confusing, so just don't do it.
The problem is your statement $∀x(x=a))$. On its own it just says that everyone is Ada and as said, inside the $\exists x$ it can't be used at all. However, without the quantifier it works:
$∃x(Cx\wedge ∀y(Cy→x=y)\wedge (x=a))$
"There is a person $x$ that is clever and if a person is clever, then that person is $x$ (=$x$ is the only clever person) and that person $x$ is Ada."
However, the $\exists x$ is a bit unnecessary, as it means something like "there is some person", but in this case, you know the person your statement is about, namely Ada. So it could be shortened to: $Ca \wedge ∀y(Cy→a=y)$ which would mean "Ada is clever and all clever persons are Ada (i.e. Ada is the only clever person)".