(a) If $\vdash\alpha\rightarrow\beta$ then, according to the Generalization Theorem, $\vdash\forall x(\alpha\rightarrow\beta)$ , and, according to the axiom group 3, $\vdash\forall x(\alpha\rightarrow\beta)\rightarrow(\forall x\alpha\rightarrow\forall x\beta)$ , and, by MP, $\vdash(\forall x\alpha\rightarrow\forall x\beta)$ .

(b) The problem here is that what (a) says is that for $\alpha$ and $\beta$ such that $\alpha\rightarrow\beta$ is valid, $\forall x\alpha\rightarrow\forall x\beta$ is valid as well (well, it does not say anything about validity, but it will follow). However, in (b), $\alpha\rightarrow\beta\vDash\forall x\alpha\rightarrow\forall x\beta$ says that for any $\alpha$ and $\beta$ , and any structure $\mathfrak{A}$ and $s:V\rightarrow|\mathfrak{A}|$ , if $\vDash_{\mathfrak{A}}\alpha\rightarrow\beta[s]$ then $\vDash_{\mathfrak{A}}\forall x\alpha\rightarrow\forall x\beta[s]$ . Therefore, in (a) the statement is restricted to only those $\alpha$ and $\beta$ for which $\alpha\rightarrow\beta$ is valid, while in (b) there is no such restriction.

As an example, consider $\alpha$ a valid formula, and $\beta$ the formula $x=c$ . Now, if $|\mathfrak{A}|$ has more than one element, and $s$ is such that $s(x)=c^{\mathfrak{A}}$ , then $\vDash_{\mathfrak{A}}\alpha\rightarrow\beta[s]$ . However, while $\vDash_{\mathfrak{A}}\forall x\alpha[s]$ , $\not\vDash_{\mathfrak{A}}\forall x\beta[s]$ , and, hence, $\not\vDash_{\mathfrak{A}}\forall x\alpha\rightarrow\forall x\beta[s]$ . Therefore, $\alpha\rightarrow\beta\not\vDash\forall x\alpha\rightarrow\forall x\beta$ .