In a lecture I attended, the instructor used the phrases
- $S$ has no model
- $S\models \bot$
interchangeably. While I get that (2) implies (1) because if there is a valuation $v$ such that $v(s)=1 \,\forall s\in S$ then $v(\bot)=1$. Contradiction. I don't get why (1) should also imply (2).
Or could someone explain how I should interpret the word "having no model".
'$S$ has no model' means that there is no valuation $v$ such that $v(s)=1$ for all $s \in S$
$S \vDash \bot$ means that for all valuation $s$: if $v(s)=1$ for all $s \in S$ then $v(\bot) = 1$
So, if $S$ has no model, then there is no valuation $v$ such that $v(s)=1$ for all $s \in S$. But that means that for any valuation $v$, it is not true that $v(s)=1$ for all $s \in S$. And that means that for any valuation $v$, the 'if' part of 'if $v(s)=1$ for all $s \in S$ then $v(\bot) = 1$' is false, meaning that the whole 'if $v(s)=1$ for all $s \in S$ then $v(\bot) = 1$' is true (because in this particular mathematical context, we use the 'if ... then .. ' just like a material conditional in logic: if the 'if part is false, the whole 'if .. then ..' is true). So we therefore get that for all valuation $s$: if $v(s)=1$ for all $s \in S$ then $v(\bot) = 1$, i.e. $S \vDash \bot$