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If a formula $P$ is a tautology then we can write $\emptyset \models P$, and it makes sense, since by definition a set of formulas semantically entail another if there does not exist a valuation where all members of the set are true and the other formula is false. Since the formula is a tautology and it's always true then it makes sense.

My question is then, why does it not make sense for the cases where it is not a tautology? For example, suppose we have a formula $P$ that is not a tautology, and our set is the empty set. Since the empty set doesn't have any premises then there does not exist any valuation where the premises are true and the conclusion false, thus $\emptyset \models P$, but apparently this is wrong and the empty set only semantically entail tautologies. Can someone help me understand why is my reasoning incorrect?

Please don't provide an alternative proof of why $\emptyset \models P$ iff $P$ is a tautology, instead explain exactly which line of my reasoning is wrong, I've already seen an alternative proof of why is that the case and the proof makes sense but it doesn't explain why I'm wrong.

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  • $\begingroup$ Apply def: there is no truth assign that sat all formulas in the empty set of premises and falsifies the conclision P. The sat of the empty set is vaccuous and thus we can apply this fact also to the truth assign that falsifies P, when P is not taut. $\endgroup$ Commented Feb 3, 2023 at 11:30
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    $\begingroup$ The "nitpicking issue" is that every truth assignment $v$ satisfies $\emptyset$ because we have $\forall \varphi (\varphi \in \emptyset \to v(\varphi)= \text T)$. This pseudo-formula is TRUE. $\endgroup$ Commented Feb 3, 2023 at 11:53
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    $\begingroup$ Thus, spelling in detail the definition of logical consequence, we have (for an atom $p$ whatever) that: "for every truth assignment $v$, and for every formula $\varphi \in \emptyset$, we have that if (if $\varphi \in \emptyset$, then $v(\varphi)= \text T$), then $v(p)= \text T$." Now consider the truth assignment $v_0$ such that $v_0(p)= \text F$. $\endgroup$ Commented Feb 3, 2023 at 11:58
  • $\begingroup$ @MauroALLEGRANZA I understand why it is vacuously true that all valuations satisfy all formulas of the empty set (none), but by that same logic, don't all valuations also not satisfy the empty set? since for any valuation $v$ we have $\forall \phi (\phi \in \emptyset \to v(\phi)=F)$ is true. $\endgroup$
    – zlaaemi
    Commented Feb 3, 2023 at 13:28
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    $\begingroup$ Correct, but the valuations NOT satisfying the premises do NOT matter for the relation of logical consequence. $\endgroup$ Commented Feb 3, 2023 at 13:37

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The error is in “Since the empty set doesn't have any premises then there does not exist any valuation where the premises are true and the conclusion false”. In natural language, one might argue that “the premises” cannot be true if there are no premises, but here the required proposition is that all premises are true, and if there are no premises, then by definition all premises are true (and all premises are false). All elements of the empty set are white ravens.

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    $\begingroup$ I understand now why all premises are true and also why all premises are false, but now my question it: Are these two statements not contradictory? All premises of the empty set are true and all premises of the empty set are false? How can they both be true at the same time? $\endgroup$
    – zlaaemi
    Commented Feb 3, 2023 at 13:20
  • $\begingroup$ @zlaaemi: There's no contradiction. One and the same statement cannot be true and false – but there's no statement here. All unicorns are rainbow-coloured and all unicorns are white. As long as there are no unicorns, there's no contradiction, because no statement of the form "Unicorn $X$ is rainbow-coloured and unicorn $X$ is white" is implied. Perhaps this logically equivalent form makes more sense to your intuition: There's no premise that isn't true, and there's no premise that isn't false. Both are true (because there's no premise). $\endgroup$
    – joriki
    Commented Feb 3, 2023 at 13:30
  • $\begingroup$ I understand now, The unicorn analogy was a very good one. But there's something bugging me, can the empty set be a semantically consistent set? By definition a set is semantically consistent if there exist a valuation where all formulas are true, so I guess yes, is that correct? $\endgroup$
    – zlaaemi
    Commented Feb 3, 2023 at 15:31
  • $\begingroup$ @zlaaemi: Yes, that's correct! $\endgroup$
    – joriki
    Commented Feb 3, 2023 at 17:02

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