Section 2.4: Problem 13 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

Show that $\vdash Py\leftrightarrow\forall x(x=y\rightarrow Px)$ .

Remarks: More generally, if $t$ is substitutable for $x$ in $\phi$ and $x$ does not occur in $t$ , then $[\phi_{t}^{x}\leftrightarrow\forall x(x=t\rightarrow\phi)].$ Thus the formula $\forall x(x=t\rightarrow\phi)$ offers an alternative of sorts to the substitution $\phi_{t}^{x}$ .

We need to show that $\vdash Py\rightarrow\forall x(x=y\rightarrow Px)$ and $\vdash\forall x(x=y\rightarrow Px)\rightarrow Py$ .

For the first formula, by the deduction and generalization theorems, it is sufficient to show that $Py\vdash x=y\rightarrow Px$ . By MP, it is sufficient to show that $\vdash Py\rightarrow x=y\rightarrow Px$ . Now, by axiom group 6, we have $\vdash x=y\rightarrow Py\rightarrow Px$ . By applying MP twice, we obtain $x=y;Py\vdash Px$ , and by using the deduction theorem twice, we get $\vdash Py\rightarrow x=y\rightarrow Px$ .

The second formula, by Rule T, follows from $\vdash\forall x(x=y\rightarrow Px)\rightarrow\forall xPy$ and $\vdash\forall xPy\rightarrow Py$ , where the second one belongs to axiom group 2. Now, for the first one, by axiom group 3, MP and generalization theorem, it is sufficient to show that $\vdash(x=y\rightarrow Px)\rightarrow Py$ . Now, $\vdash x=y\rightarrow Px\rightarrow Py$ belongs to axiom group 6, $\vdash(x=y\rightarrow Px\rightarrow Py)\rightarrow(x=y\rightarrow Px)\rightarrow Py$ is a tautology (axiom group 1), and, by Rule T, we obtain $\vdash(x=y\rightarrow Px)\rightarrow Py$ .

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Section 2.4: Problem 13 Solution

Section 2.4: Problem 13 Solution