In Enderton's "Mathematical Introduction to Logic" (2ed p.88), $\Gamma\models\phi$ is defined to mean that for every model $M$ and every assignment $s$ such that $M\models\Gamma[s]$, $M\models \phi[s]$.

By contrast, in Bilaniuk's "Problem Course in Mathematical Logic" definition 6.6 on p.38, $\Gamma\models\phi$ is defined to mean that for every model $M$ such that $M\models\Gamma$, $M\models\phi$. Here, $M\models \Gamma$ means $M\models \gamma[s]$ for every assignment $s$ and every $\gamma\in\Gamma$, similarly for $M\models \phi$.

$\Gamma\models\phi$ in symbols:

- Enderton: $\forall M \forall s (M\models \Gamma[s]\rightarrow M\models\phi[s])$
- Bilaniuk: $\forall M ((\forall s M\models \Gamma[s])\rightarrow (\forall s M\models\phi[s]))$

According to the former, $\lbrace x=y\rbrace\not\models\forall x\forall y (x=y)$. According to the latter, $\lbrace x=y\rbrace\models\forall x\forall y(x=y)$.

What do the logicians here at Math Overflow think about this conundrum?

NOTinclude this rule. Plot hole?) $\endgroup$ – Sam Alexander Jul 29 '11 at 7:28sentences, in which case the distinction disappears. In any case, in your particular example,`$\forall x\forall y\,x=y$`

is provable by itself without any assumptions. $\endgroup$ – Emil Jeřábek Jul 29 '11 at 10:15`$\forall x\forall y\,x=y$`

is not provable,`$\forall x\,x=x$`

is.) $\endgroup$ – Emil Jeřábek Jul 29 '11 at 17:35