Question

So, this is sort of a 2-in-1 question.

There is no sentence W of FOL such that: ∀yB(y)⟛¬W True or False?

Now, my interpretation is: There is no sentence W in FOL where for all y in the extension of B, they are all not equivalent to A. I think this is False because of course, since we are talking about all the vast possibilities both W and A can be, there has to be at least one sentence that disproves this. Right?

Then, for the formula D, ∃z(G(x)->G(z)) Both ∃xD and ∀xD are theorems.

∃xD is a theorem, but ∀xD is not.

∀xD is a theorem, but ∃xD is not.

Neither are theorems

Now, my interpretation is that there exists a z such that if x is in the extension of G, then z is in the extension of G.

From my understanding, a theorem is something that can be logically derived from the axioms of FOL. By that definition, I think none of them are theorems since in logic, the consequent can not prove anything about the antecedent. But tbh, I am not quite sure how theorems work in FOL(all textbooks I can find focus on TFL but I read somewhere that the laws of TFL also apply to FOL since FOL is "an extension of TFL" which is why I think this answer is correct.)

So, I just want to know if I am thinking about this correctly?

Answer 1

The statement 'There is no sentence W of FOL such that: ∀yB(y)⟧¬W' is false, as it is possible to construct such a sentence in FOL. The question about whether both ∃xD and ∀xD are theorems is more complex, as their status as theorems depends on the axioms of FOL and the specific logical system.

Step-by-step explanation:

When addressing the first part of the question, we are examining the statement: 'There is no sentence W of FOL such that: ∀yB(y)⟧¬W'. The claim is concerning the existence of a universal statement in FOL (First-Order Logic) and its relationship to a particular non-equivalent sentence. This is a question rooted in logical equivalence and the properties of universal statements in logic. The statement is actually false because it is possible to construct a sentence W in FOL that is not logically equivalent to all the instances of B(y). As for the second part of the question involving the formula D, ∃z(G(x)->G(z)), and whether ∃xD and ∀xD are theorems, the inference rules of disjunctive syllogism, modus ponens, and modus tollens are involved. The claim that both are theorems could be incorrect because they depend on the structure of the particular logical system and its axioms. For ∃xD to be a theorem, it must be possible to logically derive that there exists at least one instance based on the axioms of FOL. Conversely, for ∀xD to be a theorem, it has to be universally necessary, which is not guaranteed by the given logical construct ∃z(G(x)->G(z)).