# Is this a sentence of FOL?

Let F(x) = x is a footballer, M(x) = x is a man.

Consider this sentence of FOL: ∃y∀x(F(x)↔M(x))

Even though there is no occurrence of y variable in the scope of the existential quantifier, is this still a sentence of FOL?

If this is a sentence of FOL, would the translation into English be something like: "There exists an object y such that all and only men are footballers"?

A quantifier that "acts" on no variable - like in your example, where there is no occurrences of y in the scope of the leading existential quantifier, i.e. in the subformula ∀x(F(x) ↔ M(x)) - is not "formally" wrong but it does not add anything to the meaning of the formula.

Thus, the formula must be read as:

"all and only men are footballers."

• Just pondering, might the longer formula Toru mentioned also require that the domain of discourse contains at least one object? If in fact there are no objects that can be bound to y, in theory the statement would be false, while the subformula you mention would be true. An absurd technicality, perhaps, but sometimes it could matter. – Cort Ammon Feb 15 '16 at 20:16
• @CortAmmon - I agree, but the "basic" semantics for f-o logic (that used in many textbooks) needs non-empty domains. – Mauro ALLEGRANZA Feb 15 '16 at 20:20

Noam Chomsky once wrote a sentence that went like this:

Colourless green ideas sleep furiously

This is not formally incorrect, in so far we think of the grammar of the English language; but it is incorrect when we take into account semantics; then we see that the sentence has no content.

Similarly, your sentence is formally correct; one would say, formally, that it is a well-formed sentence, given the grammar/syntax of First Order Logic (FOL); but when we take into account semantics, we see that the existential prefix for y is unnecessary; in this higher level of understanding, then the sentence should be construed as being not quite right - but easily fixable.

Consider this sentence of FOL: ∃y∀x(F(x)↔M(x))

∃yP <-> ∀yP <-> P, for all P is a theorem.

i.e. ∃y∀x(F(x)↔M(x)) <-> ∀y∀x(F(x)↔M(x)) <-> ∀x(F(x)↔M(x)).

• Only if y does not occur as a free variable in P. – user2953 Jun 9 '16 at 13:51
• This doesn't answer the question if it is a valid sentence, it just presupposes it. – user2953 Jun 9 '16 at 13:52
• Any chance we can persuade you to write an answer involving words? (∀y(~W(y) -> ~A(y))) (W being a function to test whether it uses words and A being a function that tests whether it qualifies as a good answer) – virmaior Jun 10 '16 at 0:40