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Logical entailment

  1. May 1, 2004 #1
    Please help me understand this:

    F [tex]\models \omega \:\text{(where}\: \omega\: \text{is any wff!)}[/tex]

    (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.
     
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  3. May 1, 2004 #2

    selfAdjoint

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    False implies anything is a standard law of logic.
     
  4. May 1, 2004 #3
    Yes, clearly, if it said
    [tex]F \implies \omega[/tex]
    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:
     
  5. May 1, 2004 #4

    Hurkyl

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    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.
     
  6. May 1, 2004 #5
    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 [tex]\text{F}\:\models \omega [/tex]

    Can you explain the distinction between
    [tex] \text{P} \wedge \text{Q}\: \models \text{P}[/tex]
    and
    [tex] \text{P} \wedge \text{Q}\: \implies \text{P}[/tex]

    Edit: added a related question:
    Is
    [tex] \text{P} \wedge \text{Q}\: \models \text{P}[/tex]
    true only because
    [tex] \text{P} \wedge \text{Q}\: \implies \text{P}[/tex]
    is a tautology?
     
    Last edited: May 1, 2004
  7. May 1, 2004 #6

    Hurkyl

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    You would like

    [tex]P \wedge Q \models P[/tex]

    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, [itex]\Rightarrow[/itex] and [itex]\models[/itex] work out to be roughly equivalent.
     
  8. May 1, 2004 #7
    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.
     
  9. May 2, 2004 #8

    Hurkyl

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    P and Q can be any statements. It would be awkward (and somewhat redundant) to state "Whenever P and Q is satisfiable, [itex]P \wedge Q \models P[/itex]," 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 [itex]A \wedge B \wedge \ldots \Rightarrow P[/itex] if and only if [itex]A, B, \ldots \models P[/itex]. I don't remember it precisely, which is why I said "roughly" as a qualification. :smile:
     
  10. May 2, 2004 #9
    This is probably the theorem you were thinking of:

    [tex]{\phi_1, ... \phi_n} \models \phi \:\textrm{iff} \:\models (\phi_1, ... \phi_n) \Rightarrow \phi[/tex]

    (where
    [tex] \models \omega [/tex]
    by itself means [tex]\omega [/tex] is a tautology)
     
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