# Prove that $\forall x(\phi(x)\longleftrightarrow \psi(x))\iff \{x\mid \phi(x)\}=\{x\mid \psi(x)\}$ (by P. Suppes definition)

Context: I'm reading "Axiomatic set theory" by P. Suppes. Compared to other authors, say Hrbaceck or Enderton for example, the definition of $\{x\mid \phi(x)\}$ is rather different. While these authors worry only about the case when such a set actually exists, Suppes define it as $\emptyset$ when that is not the case so that it is always defined (For example $\{x\mid x=x\}=\emptyset$ for Suppes, whereas such notation doesn't make sense for the other authors). This is his definition:

\begin{equation*} y=\{x\mid\phi(x)\}\iff \{\forall x[(x\in y\longleftrightarrow \phi(x))\wedge y \text{ is a set}]\vee [y=\emptyset \wedge \neg \exists B\forall x(x\in B\longleftrightarrow \phi(x))]\} \end{equation*}

In this case: $\text{y is a set}\iff \exists x(x\in y \vee y=\emptyset)$

Then there's this theorem whose proof is left as an exercise:

\begin{equation*} \forall x(\phi(x)\longleftrightarrow \psi(x))\iff \{x\mid \phi(x)\}=\{x\mid \psi(x)\}.\end{equation*}

I have no problem in showing the $\implies$ part, but to the converse I have this "counterexample": Let $A$ be a non empty set. Define $\phi(x):=x=x$ and $\psi(x):=x\notin A$. Then $\{x\mid \phi(x)\}=\{x\mid \psi(x)\}=\emptyset$ but it's not true that $\forall x(\phi(x)\longleftrightarrow\psi(x))$, for example if $x\in A$ we have $x=x \wedge x\in A$, meaning $\phi(x)\wedge\neg\psi(x)$.

Where am I wrong?

• @user18921 Thanks for the comment. Here I'm assuming that $\neg \exists B\forall x(x\in B)$, meaning there's no set that includes everything. – Daniela Diaz Sep 17 '13 at 2:52
• I do not see any mistakes. Are there restrictions on the behavior of the free variables of the formulas involved; are they allowed to include parameters? – Andrés E. Caicedo Sep 17 '13 at 2:56
• @AndresCaicedo No. I read it again but there are no restrictions over the free variables. – Daniela Diaz Sep 17 '13 at 3:06

In the edition of the book I have, Theorem Schema 54 (exercise 7) does not state an equivalence, only the implication you have already proved. What you have done is to verify that the equivalence is not true, your counterexample works. So, all is well with the world.

(Exercises 8 and 9 are also related, and) it is curious that Suppes does not directly ask for a counterexample to the converse implication.

(Both exercise 7 and theorem 54 appear in page 36. The copy I have is a reproduction of the 1960 edition.)

• Your clarification about exercise 9 is everything. And yes, we are with the same version. I misread the theorem but now everything is clear. Thank you so much!!! – Daniela Diaz Sep 17 '13 at 3:37
• Now that I think about it, the negation in exercise 9 is not a typo because in that case it would be asking the same as Theorem Schema 54 (exercise 7) and it wouldn't make sense to ask the same question again. Rather, as this make reference to similarity with theorems 49 and 50, which are $\{x\mid x=x\}=\emptyset=\{x\mid x\neq x\}$, then the question make sense and that's not a typo, though it's obviously false. – Daniela Diaz Sep 17 '13 at 4:32
• Ah, you are right, I hadn't noticed Theorem 54 was exercise 7. Silly exercise, then, but not a typo. – Andrés E. Caicedo Sep 17 '13 at 5:57