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I am currently trying to write (and prove) a statement and I am having some trouble figuring out if the statement needs to be of the form \begin{gather} p\Rightarrow(q\Leftrightarrow r) \end{gather} or of the form \begin{gather} q\Leftrightarrow (p\wedge r) \end{gather} Using truth tables, one can see that see that the truth values of these statements only differ in the following two scenarios (sorry for the quality of the handwritting):

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Further, suppose that I can show (or know) that \begin{gather} \neg p\Rightarrow\neg q\quad\text{ (i.e., }q\Rightarrow p\text{)} \end{gather} Can I then rely on this fact to con conclude that my statement is of the following form? \begin{gather} q\Leftrightarrow (p\wedge r) \end{gather} My intuition is that since $\neg p\Rightarrow \neg q$, the fact that $p\Rightarrow(q\Leftrightarrow r)$ is true when $p$ is false and $q$ is true, the statement $p\Rightarrow(q\Leftrightarrow r)$ cannot be the form of my statement.

More generally, I guess that my question is: how to tell statements $(p\Rightarrow(q\Leftrightarrow r))$ and $(q\Leftrightarrow (r\wedge p))$ apart?

Thank you all very much for your time.

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    $\begingroup$ Why don't you write the full truth table (all $8$ rows)? If those two expressions produce the same function, they are equivalent and it doesn't matter which one you take. If for some choice of $p,q,r$ one of those expressions is true and the other is false, then, well, that is how you can tell them apart! $\endgroup$
    – user700480
    Commented Jan 5, 2022 at 19:59
  • $\begingroup$ Thank you for your comment. Unfortunately, I don’t see how it gets me closer to an answer. Could you please expand it further? $\endgroup$
    – EoDmnFOr3q
    Commented Jan 5, 2022 at 20:05
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    $\begingroup$ @Héctor "the truth values of these statements only differ in the following two scenarios" $\;-\;$ Once they differ, the statements are not equivalent. It doesn't matter if they differ for one combination of inputs, or two, or many. You know the statements are different, so what exactly do you mean by "tell them apart"? $\endgroup$
    – dxiv
    Commented Jan 5, 2022 at 20:26
  • $\begingroup$ Thank you for your clarifying comment. I know (and knew since the beginning) that both statements are not equivalent. Instead, my situation is this one: I am trying to write a theorem and I know that this theorem is of one of these two forms —my problem is that I do not know which one! Then, I was wondering if the fact that $\neg p\Rightarrow \neg q$ could be use to figure it out. $\endgroup$
    – EoDmnFOr3q
    Commented Jan 5, 2022 at 20:42
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    $\begingroup$ But ... if $\lnot p\implies\lnot q$ then those two formulae are equivalent. $\lnot p\implies\lnot q$ (or $q\implies p$) means that those two rows where they are different - simply cannot happen! Or, look at it this way: let's say you prefer $q\iff(p\land r)$. Assume $p$ is true. Then $p\land r$ is just $r$ and you have proven $q\iff r$. In other words, you have proven $p\implies(q\iff r)$ as well! $\endgroup$
    – user700480
    Commented Jan 5, 2022 at 21:00

1 Answer 1

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I am having some trouble figuring out if the statement needs to be of the form \begin{gather} p\Rightarrow(q\Leftrightarrow r) \tag1\end{gather} or of the form \begin{gather} q\Leftrightarrow (p\wedge r)\tag2\end{gather}

Further, suppose that I can show (or know) that \begin{gather} \color{red}{\neg p\Rightarrow\neg q}\quad\text{ (i.e., }q\Rightarrow p\text{)}\tag{*}\end{gather} Can I then rely on this fact to conclude that my statement is of the following form? \begin{gather} q\Leftrightarrow (p\wedge r)\tag2\end{gather}

Here are the relevant facts ($\models$ means logically implies): \begin{align} q↔ (p\wedge r) \quad&\normalsize\models\quad \color{red}{\neg p→\neg q}\\ p→(q↔ r) \quad&\not\normalsize\models\quad \color{red}{\neg p→w\neg q}\\ \color{red}{\neg p→\neg q} \quad&\not\normalsize\models\quad \Big(p→(q↔ r) \quad\text{or}\quad q↔ (p\wedge r) \Big)\\ \color{red}{\neg p→\neg q} \quad&\normalsize\models\quad \Big(p→(q↔ r) \quad\equiv\quad q↔ (p\wedge r) \Big). \end{align}

(The brute-force way to verify these four claims is to replace each entailment or non-entailment symbol with $\to,$ then use a truth table to check whether the conditional is a tautology.)

So:

  • $(*)$ is a necessary condition for statement $(2).$
  • However, $(*)$ is a sufficient condition for neither statement $(1)$ nor statement $(2).$
  • Fortunately, given that $(*)$ is true, statements $(1)$ and $(2)$ are equivalent to each other (i.e., they are both true or both false).

More generally, I guess that my question is: how to tell statements $(p\Rightarrow(q\Leftrightarrow r))$ and $(q\Leftrightarrow (r\wedge p))$ apart?

Your truth table reveals that for the valuation $(p,q,r)=(F,T,T),$ statement $(1)$ is true whereas statement $(2)$ is false.

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