# What does the notation $\Gamma \vDash \phi$ mean (in Mathematical Logic)?

What does the notation $\Gamma \vDash \phi$ mean (in Mathematical Logic)?

I understand that $\Gamma$ represents a system of formulas, and that $\phi$ represents an individual formula.

I also know that $\Gamma \vDash \phi$ means " $\phi$ is a semantic consequence of $\Gamma$" - but I do not understand what this actually means. Can someone explain this idea to me in layman's terms?

Furthermore, suppose I had the following question:

Let $\Gamma \cup \{ \phi \}$. Does $\Gamma \vDash \phi$ hold for the set of formulae $$\Gamma = \{ p \rightarrow q, q \rightarrow r, r \rightarrow s \} \hspace{10 mm} \text{ where } \phi \text{ is } p \rightarrow s$$

Would I attempt this question by trying to prove that, for every formula $\psi \in \Gamma$, the valuation $v(\psi) = T$, based on the assumption that $v(\phi) = T$?

• Regarding the second part, the answer is yes; with the said $\Gamma$ and $\phi$ it is true that $\Gamma \vDash \phi$. They are propositional calculus formukae; thus, to chek it it is enough to use valuations. Simple check: assume $v$ such that $v(\psi)=T$ for every $\psi \in \Gamma$ and see what happens to $\phi$: is it possible that $v(\phi)=F$? – Mauro ALLEGRANZA Feb 7 '17 at 12:19

Are you familiar with the notation $\mathfrak A\vDash \phi$, meaning that the structure $\mathfrak A$ satisfies $\phi$, or (in other words) $\phi$ is true in $\mathfrak A$?
$\Gamma\vDash \phi$ is then shorthand for:
$$\forall \mathfrak A \bigl[ (\forall \psi\in\Gamma: \mathfrak A\vDash \psi) \;\to\; \mathfrak A\vDash\phi \bigr ]$$
Or, in words, $\phi$ is satisfied by every structure that satisfies all of $\Gamma$.
In yet other words, $\phi$ is true in every model of $\Gamma$.
If we're not speaking about ordinary first-order logic, something else may take the place of "structure" above -- for example, for propositional calculus, instead of $\forall\mathfrak A$ we would quantify over all truth assignments for the propositional variables in $\Gamma$ and $\phi$.