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Let $\Gamma$ be $\{p\}$ and $\Delta$ be $\{q\}$ and $\varphi$ be $p\lor q$, then $\Gamma\vDash\varphi$ and $\Delta\vDash\varphi$. However, $\Gamma\cap\varphi$ is the empty set $\{\}$ and $\{\} \not\vDash p\lor q$.

The answer sheet said this, but doesn't an empty set entail everything? Shouldn't this be true?

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    $\begingroup$ Why do you think an empty set entails everything? $\endgroup$
    – Karl
    Commented May 19 at 8:12
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    $\begingroup$ The false formula $\bot$ entails everything. Contexts are interpreted as conjunctions of formulas, and the conjunction of an empty set is $\top$. $\endgroup$ Commented May 19 at 8:16
  • $\begingroup$ And it's easy to construct non-empty counterexamples as well. For example, $\{p,p\rightarrow r, q\rightarrow r\} \models r$ and $\{q,p\rightarrow r, q\rightarrow r\} \models r$ both hold, but $\{p\rightarrow r, q\rightarrow r\} \models r$ does not hold. $\endgroup$
    – Z. A. K.
    Commented May 19 at 12:36

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