# Section 2.4: Problem 6 Solution

Working problems is a crucial part of learning mathematics. No one can learn... merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
James R. Munkres
(a) Show that if $\vdash\alpha\rightarrow\beta$ , then $\vdash\forall x\alpha\rightarrow\forall x\beta$ .
(b) Show that it is not in general true that $\alpha\rightarrow\beta\vDash\forall x\alpha\rightarrow\forall x\beta$ .
(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, $\alpha\rightarrow\beta\not\vDash\forall x\alpha\rightarrow\forall x\beta$ .