logical entailment

[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. :smile:

[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)