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Prove that $\forall P$, $Q \in \mathbb{B}.$ $(P \Rightarrow Q) \Leftrightarrow(\neg P \vee Q)$ without using truth table

The grader says that my proof below is wrong: “To prove an iff statement, you should prove from two directions instead of cases. In this question, you need to prove $(P \Rightarrow Q) \Rightarrow (\neg P \vee Q)$ AND $(\neg P \vee Q) \Rightarrow (P \Rightarrow Q).$” I don't think that the grader is right about my approach. Is my proof correct or wrong?

To prove that $(P \Rightarrow Q)$ and $(\neg P \vee Q)$ are equivalent, we first consider the ways for the equivalence to fail; there are two such cases:

Case 1. Suppose that $(P \Rightarrow Q)$ is true and $(\neg P \vee Q)$ is false.
In this case, since $(\neg P \vee Q)$ is false, $Q$ and $\neg P$ must all be false, which means that $P$ is true. So, $P$ is true and $Q$ is false, which means that $(P \Rightarrow Q)$ is false, which contradicts our statement. So Case 1 is impossible.

Case 2. Suppose that $(P \Rightarrow Q)$ is false and $(\neg P \vee Q)$ is true.
In this case, since $(P \Rightarrow Q)$ is false, we know that $P$ is true and $Q$ is false. So, $\neg P$ is false. Since both $\neg P$ and $Q$ are false, $(\neg P \vee Q)$ is false, which contradicts our statement. Thus, Case 2 is also impossible.

Since the only two cases for the equivalence to be false are impossible, we conclude that for all $P$, $Q \in \mathbb{B}.$ $(P \Rightarrow Q) \Leftrightarrow(\neg P \vee Q).$

ryang
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mars5am
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    Your teacher is partially right. That is, your teacher's interpretation of the if and only if statement as two different implications, running in opposite directions is correct. What your teacher failed to realize is that your work in each of Case 1 and Case 2 accomplished the same thing. For example, by showing that it is impossible for (simultaneously) the left side to be true and the right side to be false, you are showing that the left side implies the right side, as required. ...see next comment – user2661923 Sep 20 '23 at 02:25
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    So, ultimately, your work is valid and your teacher made a mistake in not realizing that. I suggest that you print out this posting, including my comments, and show it to your teacher. – user2661923 Sep 20 '23 at 02:26
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    your proof is perfectly fine. your grader is wrong. – RyRy the Fly Guy Sep 20 '23 at 04:28
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    Your proof by cases is correct. Each case is in effect a proof by contradiction for the desired sides of the if and only if statement. Note: While your proof is valid, normally when doing a proof by cases - you want to exhaust every case, claiming that these are the cases that matter needs justification. Even if the other cases are trivial, you should write them out in a proof by cases. – Michael Carey Sep 20 '23 at 05:03
  • Thank you guys so much! Will show this to the grader and my professor. – mars5am Sep 20 '23 at 05:46

1 Answers1

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Your argument is perfectly valid, and your grader is illogical to claim that that bi-drectional derivation method is the only valid proof of an equivalence.

P.S. Regarding Michael's suggestion above to belabour the trivial cases, I feel that this is a choice depending on the agreed level of detail required, and that your final sentence, “Since the only two cases for the equivalence to be false are impossible” is sufficient unless your grader wants more (so yes, there is subjectivity here, but regarding the desired level of exposition rather than the validity of your proof).

P.P.S. Is there only one way to prove trigonometric identities?

ryang
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