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Proofs involving negations and conditionals
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[QUOTE="jfierro, post: 4970167, member: 146828"] [B]0. Background[/B] First and foremost, this is a proof-reading request. I'm going through Velleman's "How To Prove It" because I found that writing and understanding proofs is a prerequisite to serious study of mathematics that I did not meet. Unfortunately, the book is very light on answers to its exercises and there is no solution manual available for purchase. Also, I understand that the best (and only?) way of learning to write proofs is by getting feedback from actual human beings. For what it's worth, I have an engineering degree's curriculum worth of math. [B] 1. Homework Statement [/B] Exercise 3.2.2. This problem could be solved by using truth tables, but don't do it that way. Instead, use the methods for [URL="https://www.physicsforums.com/insights/how-most-proofs-are-structured-and-how-to-write-them/"]writing proofs[/URL] discussed discussed so far in this chapter. (See Example 3.2.4.) [LIST=1] [*]Suppose ## P \rightarrow Q ## and ## R \rightarrow \neg Q ## are both true. Prove that ## P \rightarrow \neg R ## is true. [*]Suppose that ## P ## is true. Prove that ## Q \rightarrow \neg (Q \rightarrow \neg P) ##. [/LIST] [h2]Homework Equations[/h2] At this point Velleman has introduced sentential and quantificational logic. He has talked about a few techniques that can be used to tackle proofs involving conditionals and negations, including proof by contradiction and contrapositive, how one can use the [I]givens [/I]of a problem to infer other givens via inference rules such as [I]modus ponens [/I]and [I]modus tollens[/I], etc. For example, he says that to prove a statement of the form ##P \rightarrow Q##, one can assume that P is true and then prove Q. [h2]The Attempt at a Solution[/h2] The first one is rather simple: [LIST] [*][SIZE=4]Suppose ##P## is true. Then, since ##P \rightarrow Q##, it follows that ##Q## is true. By the contrapositive of ## R \rightarrow \neg Q ##, we know that ## Q \rightarrow \neg R ##. Therefore, ## P \rightarrow \neg R ##.[/SIZE] [/LIST] [SIZE=4]The second one I found at least 3 ways of proving, so I'm most interested in input on this one. Which one would be better? Are there other, better ways of proving it?[/SIZE] [LIST] [*][SIZE=4][I](By assuming the antecedent is true and proving the consequent. Contrapositive.)[/I] Suppose ##Q## is true. Then ## \neg (Q \rightarrow \neg P) ## is also true since it is equivalent to ## Q \wedge P ## and we know ##P## is true. Therefore, if ## Q ##, then ## \neg (Q \rightarrow \neg P)##.[/SIZE] [/LIST] [LIST] [*][SIZE=4][I](By contrapositive.)[/I] We will prove ##Q \rightarrow \neg (Q \rightarrow \neg P)## by its contrapositive ##(Q \rightarrow \neg P) \rightarrow \neg Q##. Suppose ##Q \rightarrow \neg P## is true. Since we know ##P## is true, it cannot be that ##Q## is true (again by contrapositive). Therefore, ##(Q \rightarrow \neg P) \rightarrow \neg Q##.[/SIZE] [/LIST] [LIST] [*][SIZE=4][I](By contradiction.) [/I]Assume ##\neg (Q \rightarrow \neg (Q \rightarrow \neg P))## is true. This is equivalent to saying that ## Q \wedge (Q \rightarrow \neg P) ## is true. Thus, both ## Q ## and ## Q \rightarrow \neg P ## must be true. But since ## Q \rightarrow \neg P ##, we have ## \neg P ##, which is a contradiction. Therefore, it cannot be the case that ##Q \rightarrow \neg (Q \rightarrow \neg P)## is false.[/SIZE] [/LIST] Thanks. [/QUOTE]
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Proofs involving negations and conditionals
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