Logical Entailment: Understanding F $\models \omega$

[gnome]

Please help me understand this:

F $$\models \omega \:\text{(where}\: \omega\: \text{is any wff!)}$$

(That comes from Nilsson's "Artificial Intelligence, A New Synthesis", pg 225)

How does that make any sense? There is no interpretation for which F is true.

 

[selfAdjoint]

False implies anything is a standard law of logic.

 

[gnome]

Yes, clearly, if it said
$$F \implies \omega$$
that would always be true.

But apparently there is a distinction between implication and entailment, and I'm trying to understand what that distinction is.

This is how he defines entailment:

If a wff ω has value True under all of those interpretations for which each of the wffs in a set Δ has value True, then we say that Δ logically entails ω and that ω logically follows from Δ and that ω is a logical consequence of Δ.

 

[Hurkyl]

Consider this:

There are no interpretations in which F is true.

Thus, it is trivial that ω is true for all interpretations in which F is true.

 

[gnome]

Thanks Hurkyl. It's taking me a long time to respond because I'm trying to figure out what possible use there is to a statement like that $$\text{F}\:\models \omega$$

Can you explain the distinction between
$$\text{P} \wedge \text{Q}\: \models \text{P}$$
and
$$\text{P} \wedge \text{Q}\: \implies \text{P}$$

Edit: added a related question:
Is
$$\text{P} \wedge \text{Q}\: \models \text{P}$$
true only because
$$\text{P} \wedge \text{Q}\: \implies \text{P}$$
is a tautology?

 

[Hurkyl]

You would like

$$P \wedge Q \models P$$

to be true right? What if P and Q are both false statements? ...


I'm a little fuzzy in the formal logic department, but if I recall correctly, $\Rightarrow$ and $\models$ work out to be roughly equivalent.

 

[gnome]

I don't think "what if P and Q are both false statements" is relevant. As I read that definition, (P and Q) logically entails P because P is true whenever (P and Q) is true.

Unfortunately, "roughly equivalent" doesn't cut it on a final.

Thanks anyway. I'll post back if I find out anything to clarify the difference.

 

[Hurkyl]

P and Q can be any statements. It would be awkward (and somewhat redundant) to state "Whenever P and Q is satisfiable, $P \wedge Q \models P$," would it not?


I don't have my reference at the moment, so I may be wrong, but I seem to recall there being a theorem that says $A \wedge B \wedge \ldots \Rightarrow P$ if and only if $A, B, \ldots \models P$. I don't remember it precisely, which is why I said "roughly" as a qualification.

 

[gnome]

This is probably the theorem you were thinking of:

$${\phi_1, ... \phi_n} \models \phi \:\textrm{iff} \:\models (\phi_1, ... \phi_n) \Rightarrow \phi$$

(where
$$\models \omega$$
by itself means $$\omega$$ is a tautology)