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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

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    Does "For all $x\in X,p(x)\in P$ if $q$ mean "For all $x\in X, [p(x)\in P\text{ if }q]$" or does it mean "$[\text{For all }x\in X,p(x)\in P]$ if $q$"? – 2017-01-16
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    I mean the first one. For any $x\in X$, if $q$ is satisfied, then $p(x) \in P$ – 2017-01-16

3 Answers 3

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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.

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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 $$\forall x\in X\ [q\rightarrow p(x)\in P]\ \land\ \neg q\rightarrow\exists x\in X\ [p(x)\notin P]$$

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    Answer has been edited. – 2017-01-16
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    I am sorry but I am confused. How is $q$ linked to $p(x)$ in the above example? Take your last line for instance. If $x=1$, then $p(x)=1\notin P$ regardless of $q$ – 2017-01-16
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    @tsm I'm not sure I understand your comment. Are you questioning why Statement 1 is false in my example, or why Statement 2 is true? – 2017-01-16
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    Both. Take statement 2. I think you are saying that if any false statement exists (i.e. if $q$) then $p(x)\in \{0\}$. But this is not true because $p(1)\notin P$. Maybe the clarifying example I added will help? – 2017-01-16
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    Your example is pretty much the same as mine. Suppose the teacher is in a bad mood, and some students pass and some fail. Statement 1 is false, because Sam passes, although teacher is not in a good mood. Statement 2 is true. "If teacher is in a good mood then all students pass" is true, because the antecedent "teacher is in a good mood" is false. "If teacher is in a bad mood, at least one student fails" is true, because at least one student fails. – 2017-01-17
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    Statement 2 is equivalent to $[\forall x p(x)\in P]\leftrightarrow q.$ However, in your reply to my comment on your question, you indicated that you intended $\forall x[p(x)\in P\leftrightarrow q]$ and your example with the teacher seems to agree with this. **Please add brackets to the formulas in your question to remove the ambiguity.** – 2017-01-17
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On the edited version, the two statements are equivalent:

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

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    I read "For any student $x$ in the classroom, $x$ passes the class if and only if the teacher is in a good mood" as $\forall x\ (P(x)\leftrightarrow q),$ not $(\forall x\ P(x))\leftrightarrow q.$ – 2017-01-17
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    @bof. Yeah, it's ambiguous, I agree! – 2017-01-17
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    Is there a difference between: $p(x)\in P$ for all $x\in X$ vs For any $x\in X$, $p(x)\in P$? Is this where the ambiguity comes from? – 2017-01-17
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    @tsm Yes, it matters where the parentheses go! $\forall x P(x) \leftrightarrow q$ is not equivalent to $\forall x( P(x) \leftrightarrow q)$. And as my answer shows, the former is equivalent to your statement 2, but as bof's answer shows, the latter is not equivalent to your statement 2. – 2017-01-17
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    Ok. Think I see it now! Thank you both Bram28 and @bof – 2017-01-17