# $T\vDash\psi$ equivalences

$T\vDash\psi$ means $T$ satifies $\psi$ from Tarski's definition of truth, it simply means that the sentence $\psi$ is valid in $M$. I call a sentence $\psi$ universally if it is valid in every structure, i.e $\emptyset\vDash \psi$

In my logic book there are two equivalences, which are intuitively clear to me, but I do not know how to prove them formally.

1. $T\vDash\psi$ if and only if $T\cup\{\neg\psi\}$ has no model.
2. If $T=\{\phi_1,\dots,\phi_n\}$ is a set of finite sentences, then $T\vDash \psi$ is equivalent that $(\phi_1\wedge \dots \wedge\phi_n)\rightarrow\psi$ is universally.

For (1) I think it has to something with the allocation $\beta:\{\phi\mid \phi \mbox{ is a$\sigma$-formula}\}\rightarrow \{0,1\}$ where $\sigma$ is the signature for $M$. A $\sigma$-sentence $\phi$ in a $\sigma$-structure $M$ is valid if for one (therefore for every) allocation $\beta$: $\beta(\phi)=1$

For (2) I have no idea.

Just chase the definitions. For (1)

$T\vDash\psi$

says that

every model $M$ which makes all of $T$ true makes $\varphi$ true

which is trivially equivalent to

no model $M$ which makes all of $T$ true also makes $\neg\varphi$ true

which is in turn trivially equivalent to

no model $M$ makes all of $T \cup \{\neg\varphi\}$ true

For (2)

$\{\phi_1,\dots,\phi_n\} \vDash \psi$

says that

every model $M$ which makes all of $\phi_1,\dots,\phi_n$ true makes $\psi$ true.

which (given the truth-table for $\land$) is trivially equivalent to

every model $M$ which makes $\phi_1 \land \phi_2 \land \ldots \land \phi_n$ true makes $\psi$ true.

which (given the truth-table for $\to$) is trivially equivalent to

every model $M$ makes $\phi_1 \land \phi_2 \land \ldots \land \phi_n \to \psi$ true.

• Just one question (a lingustic question). Do you mean with "Every model M which makes all of T true", that "In every model M in which the sentences from T are true, in this model phi also true" ? – Babla Apr 24 '13 at 21:29
• Yes that's what I mean. – Peter Smith Apr 24 '13 at 21:34