# Does a universal affirmation entail an existential affirmation?

Suppose we have a universal affirmative statement (a statement of the form: ∀x(Px → Qx)), such as "all dogs go to heaven". Does an existential affirmative statement (a statement of the form: ∃x(Px ∧ Qx)) such as "there exists a dog, such that she went to heaven" follow from that?

• No. Since ∀x(Dx → Hx) is simply ∀x(¬Dx v Hx), any model with no dogs in it (¬Dx) will make the universal vacuously true. But &exist;x(Dx &and; Hx) is true only in models where there exists at least one dog who went to heaven. That suggests a counterexample: pick an arbitrary empty model and it will satisfy the universal vacuously and will fail to satisfy the existential. – Hunan Rostomyan May 23 '14 at 4:26
• Vaguely related: What do “universal” and “existential” mean in logic? – DBK May 24 '14 at 23:03

I think that we need a more precise discussion of the above topic.

First : are we assuming classical logic ? From your question, it is not so clear.

But, see comment above, if we "equate" ∀x(Px → Qx) with ∀x(¬Px v Qx), then we are assuming so, because (P → Q) is equivalent to (¬P v Q) in classical logic.

Second, it seems to me that you are mixing" two different questions :

(i) Does a universal affirmation entail an existential affirmation?

with :

(ii) does an universal affirmative statement, like : ∀x(Px → Qx) entails an existential affirmative statement, like : ∃x(Px ∧ Qx) ?

For (i), in classical logic the general answer is : YES. We have : ∀xDx |= ∃xDx, because the model-theoretic semantics for first-order logic assume that every interpretation has a not-empty domain.

Thus, if in the interpretation I with domain D we have that all things are Dogs, being the domain D not-empty, for sure there is at least one Dog.

The case of (ii) is different. Again, if we assume classical logic, we are licensed to "traslate" ∀x(Px → Qx) with ∀x(¬Px v Qx).

Now, the question is :

does ∀x(¬Px v Qx) |= ∃x(Px ∧ Qx) ?

Of course : NO. As per Hunan's comment above :

any interpretation with no dogs in it will make the universal vacuously true, because (¬Dx) is true. But if there are no dogs, then (Dx ∧ Hx) is always false.

But obviously [see (i)] we have : ∀x(Px → Qx) |= ∃x(Px → Qx).

Domain: set of dogs

H__: __ goes to heaven

'For all d, Hd' is true for every d that exists. It doesn't state that any d does exist. So no, the universal quantifier doesn't imply the existential quantifier.

However, for arbitrary domains X,Y and arbitrary predicate Fxy, 'there exists y such that for all x, Fxy' implies that 'for all x, there exists y such that Fxy' since the existing y is the same for every x, but 'for all x, there exists y such that Fxy' does not imply 'there exists y such that for all x, Fxy' since it is possible that the existing y is different for every x.

• Glad to see you participating! – Hunan Rostomyan May 23 '14 at 4:55
• You have to specify that you are working with so-called Free Logic, otherwise the "standard" semantics for first-order logic assumes that every domain of interpreattion must be not empty. Thus, the domain = set of dogs include at least one dog and if the sentence 'For all d, Hd' is true then, being that all dogs go to heaven, for sure at least one dog goes; i.e. 'For all d, Hd' entails 'Exists d, Hd'. – Mauro ALLEGRANZA May 23 '14 at 13:41

Mathematically, your first statement means: for all x, when x is P, then it is also Q (e.g. for all things x, when x is a unicorn, its skin has white color). From this it does not follow that there exists a unicorn.