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Definition: "A set of sentences $\Delta$ logically entails a sentence $\varphi$ (written $\Delta \vDash \varphi$) if and only if every truth assignment that satisfies $\Delta$ also satisfies $\varphi$."

Question 1: If there is no truth assignment that satisfies a set of sentences $\Delta$, then what it means for $\varphi$?

Question 2: If a set of sentences $\Delta$ is empty, then what it means for $\varphi$?

Stefan Hansen
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Victor Victorov
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1) Then for any interpretation $I$, the following conditional is true (by virtue of having a false antecedent): if $I$ satisfies $\Delta$ it satisfies $\varphi$. So $\Delta \vDash \varphi$. But that doesn't settle the value of $\varphi$ on any given interpretation. [Example: $p, \neg p \vDash q$, but that doesn't settle the value of $q$!]

2) If $\Delta$ is empty, then $\Delta \vDash \varphi$ is equivalent to $\vDash \varphi$. [Why? Any interpretation $I$ makes true the wffs in $\Delta$ -- that's the null task! -- so $\Delta \vDash \varphi$ comes to this: every interpretation makes $\varphi$ true, i.e. $\vDash \varphi$.]

Peter Smith
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    "... any interpretation makes φ true, i.e. ⊨φ." How an interpretation can makes φ true? My understanding from Stephen Kleene "Mathematical Logic" that ⊨φ is true iff is true for every interpretation.     – Victor Victorov Apr 15 '13 at 12:45
  • The original version illustrated one of those occasions in English where "any" means every. But I've rephrased to avoid misunderstanding. – Peter Smith Apr 15 '13 at 12:56
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    Dear Peter, Thank you for your help. Finally, I understood Logical Entailment and have no problem with p,¬p⊨q or Δ⊨φ for every φ if Δ is empty set. BUT, one thing steel bothers me. You wrote: "If Δ is empty, then Δ⊨φ is equivalent to ⊨φ." I question this equivalence. According to Kleene: if ⊨φ than φ MUST BE tautology and in your example it is obviously that φ might be not a tautology. – Victor Victorov Apr 16 '13 at 03:39
  • I agree with Victor. It's true that every interpretation satisfies $\Delta$ , because $\Delta$ is equivalent to $\top$ . On the other hand, only a subset of all possible interpretations satisfy $\phi$ . Therefore not all interpretations that satisfy $\Delta$ satisfy $\phi$ . So the empty set of sentences does not entail $\phi$ . – Electro-blob Jun 29 '24 at 05:59
  • I should clarify that in the latter case $\Delta \vDash \phi$ if and only if $\Delta \equiv \phi$. – Electro-blob Jun 29 '24 at 06:13
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Question 1: $\phi$ can be anything. If no truth assignment satisfies $\Delta$, then there are no truth assignments to check whether they satisfy $\phi$. Another way to say this is that an unsatisfiable $\Delta$ logically implies any $\phi$.

Question 2: $\phi$ is a logical truth, for example, "$p$ or not $p$". Every truth assignment satisfies the empty set of propositions, so that if $p$ is entailed by the empty set, than $p$ holds under any truth assignment. So $p$ will hold no matter truth values we assign to propositional variables. In fact for a connection with philosophy, the logician Alfred Tarski in the early 20th century defined the philosophical concept of a "logical truth" using this model-theoretical definition of a statement which is implied the empty set.

The answers to questions 1 and 2 are two extremes. If we make $\Delta$ very large, the set of truth assignments satisfying it will be as small as possible, and then the set of $\phi$ following from $\Delta$ will be as large as possible. Conversely if we make $\Delta$ very small, the set of truth assignments satisfying it will be very large, and then the set of $\phi$ following from $\Delta$ will be as small as possible, the set of logical truths. Hope that helps!

logic_runs_this_town
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If no assignment satisfies $\Delta$ , then $\Delta \equiv \bot$. By the principle of explosion, $\bot \vDash \phi$.

If the set of sentences in $\Delta$ is empty, then any assignment satisfies $\Delta$. So $\Delta \equiv \top$. A tautology only entails another tautology, so $\Delta \vDash \phi$ if and only if $\Delta \equiv \phi$

Electro-blob
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