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?

1 Answer 1


For your first sentence: ∀yB(y) - its negation is simply ¬∀yB(y) so substituting that for W satisfies your logical equivalence relation.

∀yB(y) ⟛ ¬¬∀yB(y)

Classical logic is bivalent (at least in its standard form) so for any given pair of sentences A, ¬A the truth of one entails the falsehood of the other and vice versa. This seems rather too simple: are you sure you have the question right?

With your second formula, you are asking whether these are theorems:

A) ∃x∃z(G(x) → G(z))

B) ∀x∃z(G(x) → G(z))

A in effect says: if something is G then something is G. It is a theorem. In fact it is always satisfiable quite straightforwardly because the value of x may be identical to the value of z.

B in effect says: if anything is G then something is G. It is also a theorem.

When considering sentences like this, it is always a good idea to pay attention to the case where the predicates are empty. What if there are no G's? Then B is true, since there is no way for G(x) to be true and hence the conditional is trivially true. With A, it appears on the face of it that we are asserting the existence of some G(x) or G(z), so A should come out false. But in fact we are not. As with B, the conditional is trivially true because its antecedent G(x) is unsatisfied. This illustrates the point that existentially quantified sentences whose main connective is a material conditional are unintuitive and easy to misunderstand.

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