# Logical entailment

1. May 1, 2004

### gnome

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.

2. May 1, 2004

Staff Emeritus
False implies anything is a standard law of logic.

3. May 1, 2004

### 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:

4. May 1, 2004

### Hurkyl

Staff Emeritus
Consider this:

There are no interpretations in which F is true.

Thus, it is trivial that &omega; is true for all interpretations in which F is true.

5. May 1, 2004

### 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
$$\text{P} \wedge \text{Q}\: \implies \text{P}$$

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

Last edited: May 1, 2004
6. May 1, 2004

### Hurkyl

Staff Emeritus
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.

7. May 1, 2004

### 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.

8. May 2, 2004

### Hurkyl

Staff Emeritus
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.

9. May 2, 2004

### 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)