In the paper by John Corcoran & Hassan Masoud (2014): Existential Import Today: New Metatheorems; Historical, Philosophical, and Pedagogical Misconceptions, History and Philosophy of Logic, already in the introduction it says, as self-evident, that

"The universalized conditional $∀x(x = 0 → x = (x + x))$ implies the corresponding existentialized conjunction $∃x(x = 0 \text { & } x = (x + x))$. And $∃x(x = 0)$ is tautological (in the broad sense, i.e. logically true)."

There are two assertions here, and I have difficulty with each one. In the first assertion I do not see why a model in which there is no $0$ (e.g., $\mathsf {ZFC^*}$ obtained by negating all the axioms of $\mathsf {ZFC}$), so that the conclusion of the implication is false and the premise would be (vacuously) true, would not be a counter-example. In the second assertion, I don't see why the statement is tautological (since it is not satisfied by all models, such as $\mathsf {ZFC^*}$).


1 Answer 1


As soon as you use a constant symbol, any interpretation will need to map that symbol to some object of its domain. So even if you say something like $\neg \exists x x=0$, you still need some object that $0$ denotes ... And since that object is of course identical to itself, this statement is a logical contradiction ... Meaning that $\exists x x=0$ is a tautology.

Something similar hold for the first one: whatever the $0$ denotes, it is of course identica to itself, and so by the universal it must then also be true that $0 = 0+0$, and hence the existential is true.

  • $\begingroup$ First, all this assumes that we are by convention excluding a free logic that would allow a model having the empty universe, right? :-( That said, the excellent answers and comments (I am not too clear on the difference between the categories) to my question (which was kindly edited) pulled me out of my idée fixe of 0 being interpreted by the empty set. $\endgroup$
    – nomadreid
    Jan 8, 2017 at 17:09
  • $\begingroup$ @nomadreid Yes, this is all assuming existential import. $\endgroup$
    – Bram28
    Jan 8, 2017 at 17:47
  • $\begingroup$ I am taking "existential import" to mean that the use of a symbol requires an individual in the universe to which the symbol is assigned. So 0 is a symbol which has an assignment. However, assuming that ∃xx=0 is not necessarily an axiom, then why should "0" be any different to "unicorn"? Yes, "unicorn" has existential import, but interpretations are possible where ~∃xx=unicorn, so that "unicorn=unicorn" being true is not enough to ensure that ∃xx=unicorn is a tautology; so why should ∃xx=0 be one? (Yes, the rest of the assertions are clear once ∃xx=0 is a tautology.) $\endgroup$
    – nomadreid
    Jan 9, 2017 at 5:56
  • $\begingroup$ @nomadreid Existential import simply means that the domain of any interpretation is not empty, i.e. that there is always something that exists. Or, in your words, that the logic is not a free logic (where you can have empty domains). And it is that assumption that allows us to require that any constant denotes some object in the domain, and that is exactly what standard logics do. Hence, as soon as you use $unicorn$ in a sentence, it will denote something, meaning that $\exists x x =unicorn$ is a tautology, and hence $\neg \exists x x=unicorn$ is always a contradiction in standard logic. $\endgroup$
    – Bram28
    Jan 9, 2017 at 12:02
  • 1
    $\begingroup$ The fog is beginning to clear. So, am I correct in saying that when says that unicorns don't exist, one doesn't mean ¬∃xx=unicorn, but rather that ¬∃x(P(x)), where P means the property of physical unicornness; that is, ¬∃x(x is a member of the set Real-Unicorns), that is, the set Real-Unicorn is empty. As opposed to the individual constant unicorn, which is a meaningless symbol that would have some assignment unicorn* in the model, so that unicorn* was not a member of the set Real-Unicorn? (Clumsily expressed, but I hope my meaning is clear.) $\endgroup$
    – nomadreid
    Jan 10, 2017 at 4:55

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