Quantifiers with if and only if

Question

Are these two statements the same?

Statement 1: For any $x\in X$, $p(x)\in P $ if, and only if, $q$

Statement 2: For all $x\in X$, $p(x)\in P$ if $q$. Furthermore, if $\neg q$, then $\exists x\in X$ such that $p(x)\notin P$.

EDIT:

Here is an example to clarify:

Let $X$ be the set of students in a classroom, $P=\{pass\}$ and $p(x)$, which can take values $\{pass, fail\}$, is the outcome for each student $x\in X$. Let $q$ represent "teacher is in a good mood".

Then statement 1 says: For any student $x$ in the classroom, $x$ passes the class if and only if the teacher is in a good mood.

Statement 2 says: If teacher is in a good mood, then all students pass. Furthermore, if teacher is in a bad mood, there will be at least one student who fails.

Edited question:

Are these two statements the same?

Statement 1: $p(x)\in P$ $ \forall x\in X $ if, and only if, $q$

Statement 2: For all $x\in X$, $p(x)\in P$ if $q$. Furthermore, if $\neg q$, then $\exists x\in X$ such that $p(x)\notin P$.

I believe the conclusion is the edited statements are equivalent

Answer 1

On the edited version, the two statements are equivalent:

$\forall x P(x) \leftrightarrow q \Leftrightarrow (\forall x P(x) \rightarrow q) \land (q \rightarrow \forall x P(x)) \Leftrightarrow (\neg q \rightarrow \neg \forall x P(x)) \land (q \rightarrow \forall x P(x)) \Leftrightarrow (\neg q \rightarrow \exists x \neg P(x)) \land (q \rightarrow \forall x P(x))$

Answer 2

The statements are not equivalent. To see this, let $X=\{0,1\},$ let $P=\{0\},$ let $p(x)=x,$ and let $q$ be any false statement, say $0=1.$ Under this interpretation, Statement 1 is false and Statement 2 is true.

Statement 1 is equivalent to each of the following statements: $$\forall x\in X\ [p(x)\in P\leftrightarrow q]$$ $$\forall x\in X\ [(q\rightarrow p(x)\in P)\ \land\ (p(x)\in P\rightarrow q)]$$ $$\forall x\in X\ [q\rightarrow p(x)\in P]\ \land\ \forall x\in X\ [p(x)\in P\rightarrow q]$$ $$\forall x\in X\ [q\rightarrow p(x)\in P]\ \land\ \forall x\in X\ [\neg q\rightarrow p(x)\notin P]$$ $$\forall x\in X\ [q\rightarrow p(x)\in P]\ \land\ \neg q\rightarrow\forall x\in X\ [p(x)\notin P]$$ Compare the last statement with Statement 2, which is equivalent to $$\forall x\in X\ [q\rightarrow p(x)\in P]\ \land\ \neg q\rightarrow\exists x\in X\ [p(x)\notin P]$$

Answer 3

Then statement 1 says: For any student $x$ in the classroom, $x$ passes the class if and only if the teacher is in a good mood.

$$\newcommand{\getsto}{\gets\!\!\!\!\to} \forall x{\in}X:(p(x){\in}P\getsto q)$$

This means: if the teacher is in a bad mood, no student passes the exam.

Statement 2 says: If teacher is in a good mood, then all students pass. Furthermore, if teacher is in a bad mood, there will be at least one student who fails.

$$(p\to \forall x{\in}X:p(x){\in}P)~\wedge~(\neg p\to\exists x{\in}X:p(x){\notin}P)$$

Or more concisely: $(\forall x{\in}X:p(x){\in}P)\getsto p$

This means: If the teacher is in a bad mood, there may be a student who passes.   "At least one fails" does not require all to do so.

Therefore the statements are not equivalent.