# Section 2.4: Problem 7 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 $\vdash\exists x(Px\rightarrow\forall xPx)$ .
(b) Show that $\{Qx,\forall y(Qy\rightarrow\forall zPz)\}\vdash\forall xPx$ .
(a) Working backward (each line is sufficient to show the previous one). So, we need to show $\vdash\exists x(Px\rightarrow\forall xPx)$ or, equivalently, $\vdash\neg\forall x\neg(Px\rightarrow\forall xPx)$ .
1. $\Gamma=\{\forall x\neg(Px\rightarrow\forall xPx)\}$ is inconsistent, by RAA.
2. Two things to show, by the definition of inconsistency:
1. $\forall x\neg(Px\rightarrow\forall xPx)\vdash\forall xPx$ ,
2. $\forall x\neg(Px\rightarrow\forall xPx)\vdash\neg\forall xPx$ . We will show this one directly.
3. One thing to show:
1. $\vdash\forall x\neg(Px\rightarrow\forall xPx)\rightarrow\forall xPx$ , by MP.
4. One thing to show:
1. $\vdash\forall x(\neg(Px\rightarrow\forall xPx)\rightarrow Px)$ , by A3 and MP.
5. One thing to show:
1. $\vdash\neg(Px\rightarrow\forall xPx)\rightarrow Px$ , by the Generalization Theorem. But this is just a tautology.
To show 2(b), we apply MP-rule twice to and the tautology $\vdash\neg(Px\rightarrow\forall xPx)\rightarrow\neg\forall xPx$ .
(b) We need to show $\{Qx,\forall y(Qy\rightarrow\forall zPz)\}\vdash\forall xPx$ . Working backward (each line is sufficient to show the previous one).
1. $\{Qx,\forall y(Qy\rightarrow\forall zPz)\}\vdash\forall wPw$ , by EAV.
2. $\{Qx,\forall y(Qy\rightarrow\forall zPz)\}\vdash Pw$ , by the Generalization Theorem. We show that this holds directly.
By axiom group 2, $\vdash\forall y(Qy\rightarrow\forall zPz)\rightarrow Qx\rightarrow\forall zPz$ and $\vdash\forall zPz\rightarrow Pw$ . By MP$^{3}$ , $\{Qx,\forall y(Qy\rightarrow\forall zPz)\}\vdash Pw$ .