About the first order logic (valid, Unsatisfiable, Syntactically wrong)

I am in trouble , I searched a lot about how to solve this kind of questions but I did not get any answers.

I understand how can I know when the sentences is valid and Unsatisfiable in propositional logic, but in FOL I can't.

Can someone help me how can I solve this kind of questions? Because it comes always in the past exams in AI.

Let us have the following:

1- ¬(Speed(Processor(MyPC))=Speed(Processor(MyPC)))
2- ¬ ∃ x ( Speed(x) = Speed(MyPC) )
3- Speed(MyPC)=MyPC
4- ∀ y ∃ x (Speed(x) ∧ Speed(y))= Speed(MyPC)


Note :

    MyPC:
a constant symbols that represents my PC.

Speed(x):
a function symbol that refers to speed of x


The answer for 1 is Syntactically wrong
2 is Unsatisfiable
3 is Neither valid nor unsatisfiable
4 is Syntactically wrong

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– Raphael
Jan 1 '17 at 20:04

1 uses a function symbol Processor which is not in the logic signature. Hence, it's outside the syntax.
4 tries to equate a proposition with Speed(MyPC). You can't equate propositions in FOL. Outside the syntax.
3 is not universally true for every function Speed and every argument MyPC. For instance interpret MyPC as the natural $0$, and Speed as the successor function to obtain $\mathsf{succ}(0)=0$ which is false. Hence the formula is not valid. However, it is satisfiable: now interpret MyPC as zero and Speed as the identity function to obtain $\mathsf{id}(0)=0$ which is true. The formula has a model and a countermodel: satisfiable and invalid.
2 ∃ x ( Speed(x) = Speed(MyPC) ) is always true, since we can take x to be MyPC. This holds no matter how we interpret Speed and MyPC. Hence is it valid. (Alternatively, we can even show it is a theorem of FOL using a proof system, and then rely on the correctness theorem to claim it is valid.) Since formula 2 is the negation of a valid formula, it is unsatisfiable.